Programming Interview Questions

aonecoding

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of 2 vote

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Solution:

public TreeNode closestLeaf(TreeNode root, TreeNode target) {
        if(root == null || target == null) {
            return null;
        }
        Map<TreeNode, List<TreeNode>> mapping = new HashMap();
        buildAdjList(mapping, root, null);

        Queue<TreeNode> queue = new LinkedList();
        Set<TreeNode> visited = new HashSet();

        for (TreeNode node: mapping.keySet()) {
            if (node != null && node == target) {
                queue.add(node);
                visited.add(node);
                break;
            }
        }

        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node != null) {
                if (mapping.get(node).size() <= 1) {
                    return node;
                }
                for (TreeNode n: mapping.get(node)) {
                    if (!visited.contains(n)) {
                        visited.add(n);
                        queue.add(n);
                    }
                }
            }
        }
        return null;
    }

    public void buildAdjList(Map<TreeNode, List<TreeNode>> mapping, TreeNode node, TreeNode p) {
        if (!mapping.containsKey(node)) {
            mapping.put(node, new ArrayList<>());
        }
        if (!mapping.containsKey(p)) {
            mapping.put(p, new ArrayList<>());
        }
        if (node != null) {
            mapping.get(p).add(node);
            mapping.get(node).add(p);
            if (node.left != null) {
                buildAdjList(mapping, node.left, node);
            }
            if (node.right != null) {
                buildAdjList(mapping, node.right, node);
            }
        }
    }

- aonecoding August 24, 2018 | Flag Reply

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of 10 vote

public int shortestNonSeq(int[] a, int k) {
        int result = 0;
        Set<Integer> flag = new HashSet();
        for (int i = 0; i < a.length; i++) {
            if (!flag.contains(a[i])) {
                flag.add(a[i]);
                if (flag.size() == k) {
                    result++;
                    flag = new HashSet<>();
                }
            }
        }
        return result + 1;
    }

- aonecoding July 12, 2018 | Flag Reply

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of 1 vote

public int connectedComponents(int n, int[][] edges) {
        int[] uf = new int[n];
        for(int i = 0; i < n; i++) {
            uf[i] = i;
        }

        for(int[] e : edges) {
            int x = e[0];
            int y = e[1];

            while (uf[x] != x){
                x = uf[x];
            }

            while (uf[y] != y){
                y = uf[y];
            }

            if(x != y) {
                uf[x] = y;
                n--;
            }
        }
        return n;
    }

- aonecoding June 25, 2018 | Flag Reply

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of 6 vote

int assignBalls(int m, int n) {
        if (m == 0 || n == 1) {
            return 1;
        }
        if (n > m) {
            return assignBalls(m, m);
        } else {
            return assignBalls(m, n - 1) + assignBalls(m - n, n);
        }
    }

- aonecoding June 24, 2018 | Flag Reply

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of 5 vote

int smallestFactors(int n) {     
if (n < 10) {    
     return n;     
}     
List<Integer> list = new ArrayList<>();    
     for (int i = 9; i > 1; i--) {         
	while ((n % i) == 0) {             
		n /= i; 
          list.add(i);  
       }
     }
     if (n > 10) {
         return 0; 
    }     
    int result = 0;     
    for (int i : list) {
         result = result * 10 + i;     
    } 
    return result; 
}

followup: what if we need to worry about the integer overflow?

- aonecoding June 06, 2018 | Flag Reply

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of 1 vote

boolean find(String s1, String s2) {
        if(s1.isEmpty()) return false;
        int l1 = s1.length(), l2 = s2.length();
        if(l1 >= l2) {
            return kmp(s1 + s1, s2); //find if s2 is a substring of s1 + s1
        } else {
            for(int i = 0; i < l1; i++) {
                if(match(s1, i, s2)) return true; //find if s2 is made of multiple rotated s1
            }
        }
        return false;
    }

    boolean match(String s1, int j, String s2) {
        int idx = 0;
        while(idx < s2.length()) {
            if(s2.charAt(idx) != s1.charAt(j)) {
                return false;
            }
            idx++;
            j = j % s1.length();
        }
        return true;
    }

    boolean kmp(String haystack, String needle) { 
        if (haystack.length() < needle.length()) { 
            String t = haystack; 
            haystack = needle; 
            needle = t; 
        } 
        int[] table = new int[needle.length() + 1]; 
        for (int i = 1; i < needle.length(); i++) { 
            int j = table[i]; 
            while (j > 0 && needle.charAt(i) != needle.charAt(j)) { 
                j = table[j]; 
            } 
            table[i + 1] = (needle.charAt(i) == needle.charAt(j)) ? j + 1 : 0; 
        }  
        haystack += haystack; 
        int j = 0; 
        for (int i = 0; i < haystack.length(); i++) { 
            while (j > 0 && needle.charAt(j) != haystack.charAt(i)) { 
                j = table[j]; 
            } 
            if (needle.charAt(j) == haystack.charAt(i)) { 
                j++; 
            } 
            if (j == needle.length()) { 
                return true; 
            } 
        } 
        return false;
     }

- aonecoding May 28, 2018 | Flag Reply

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of 4 vote

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class RingBuffer:
    def __init__(self, capacity):
        self.size = capacity
        self.capacity = 0
        self.Buffer = [None] * capacity
        self.Head = 0
        self.Tail = 0

    def full(self):
        return self.size == self.capacity

    def empty(self):
        return self.capacity == 0

    def add(self, obj):
        self.Buffer[self.Tail] = obj
        self.Tail = (self.Tail + 1) % self.size
        Self.capacity += 1

    def pop(self):
        self.Head = (self.Head + 1) % self.size
        Self.capacity -= 1
        return self.Buffer[self.Head - 1]

- aonecoding May 24, 2018 | Flag Reply

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of 2 vote

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public void backButton(String[] cmds) {
          Stack<String> forwardStack=new Stack();
          Stack<String> backwardStack=new Stack();
          String currentPage="about:blank";

          for(int i = 0; i < cmds.length; i++)
          {
               String cmd = cmds[i];
               if(cmd.equals("open"))
               {
                    backwardStack.push(currentPage);
                    currentPage=cmds[++i];
                    forwardStack = new Stack();
               }
               else if(cmd.equals("back"))
               {    
                        if(!backwardStack.isEmpty()){
                             forwardStack.push(currentPage);
                             currentPage=backwardStack.pop();
                        }
               }
               else if(cmd.equals("forward"))
               {
                        if(!forwardStack.isEmpty()){
                            backwardStack.push(currentPage);
                            currentPage=forwardStack.pop();
                        }
               }
          }
         return currentPage;
     }

- aonecoding May 23, 2018 | Flag Reply

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of 2 vote

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SOLUTION:
BFS approach does it

public List<Integer> leftView(TreeNode root) {
        List<Integer> leftview = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        if(root != null) q.add(root);
        while(!q.isEmpty()) {
            leftview.add(q.peek().val);
            Queue<TreeNode> nextLevel = new LinkedList<>();
            while(!q.isEmpty()) {
                TreeNode node = q.poll();
                if(node.left != null) nextLevel.add(node.left);
                if(node.right != null) nextLevel.add(node.right);
            }
            q = nextLevel;
        }
        return leftview;
    }

- aonecoding May 13, 2018 | Flag Reply

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of 2 vote

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SOLUTION:
ExpTree build() is not required. Solution from equals() below

from collections import defaultdict
class TreeNode():
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        if val != '+' and val != '-':
            self.isLeaf = True
        else:
            self.isLeaf = False
            
class ExpressionTree():
    def __init__(self, exp):
        self.root = self.build(exp)
        
    def build(self, exp):
        if not exp:
            return 
        if len(exp) == 1:
            return TreeNode(exp)
        mid = (len(exp) - 1) / 2
        if exp[mid] != '+' and exp[mid] != '-':
            mid += 1
        if exp[mid] == '-':
            exp = self.flip(exp, mid + 1)
        node = TreeNode(exp[mid])
        node.left = self.build(exp[:mid])
        node.right = self.build(exp[mid + 1:])
        return node
    
    def flip(self, exp, idx):
        ls = list(exp)
        for i in range(idx, len(exp)):
            if ls[i] == '+':
                ls[i] = '-' 
            elif ls[i] == '-':
                ls[i] = '+'
        return "".join(ls)
    
    def equals(self, other):
        #print self.result()
        #print other.result()
        return self.result() == other.result()
    
    def result(self):
        d = self.calculate(self.root)
        for key, value in d.items():
            if value is 0:
                del d[key]
        return d
    
    def calculate(self, node):
        if not node:
            return defaultdict(lambda:0)
        if node.isLeaf:
            d = defaultdict(lambda:0)
            d[node.val] += 1
            return d
       
        left = self.calculate(node.left)
        right = self.calculate(node.right)

        for entry in right:
            left[entry] = left[entry] + (right[entry] if node.val == '+' else -right[entry])
        return left

A tester

A = 'a+b-c-a-b-c-a-b'
B = 'b+a+c+a+b-c+b'
print ExpressionTree(A).equals(ExpressionTree(B))

- aonecoding May 06, 2018 | Flag Reply

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of 3 vote

from heapq import heappush, heappop
import numpy as np
def minCostPath(maze):
    if len(maze) is 0 or len(maze[0]) == 0 or maze[0, 0] == -1:
        return
    
    m, n = len(maze), len(maze[0])
    distances = np.zeros(shape = (m, n))
    distances.fill(-1)
    
    heap = []
    heappush(heap, (maze[0, 0], 0, 0))
    value = maze[0, 0]
    distances[0, 0] = value
    neighbors = [(1, 0), (-1, 0), (0, 1), (0, -1)]
    while heap and distances[m - 1, n - 1] == -1:
        dist, x, y = heappop(heap)
        for neighbor in neighbors:
            i, j = x + neighbor[0], y + neighbor[1]
            if i >= 0 and j >= 0 and i < m and j < n and maze[i, j] != -1 and distances[i, j] == -1:
                heappush(heap, (maze[i, j] + dist, i, j))
                distances[i, j] = maze[i, j] + dist
    print distances
    
    if distances[m - 1][n - 1] == -1:
        return
 
    x, y = m - 1, n - 1
    path = []
    while x != 0 or y != 0:
        path.append((x, y, distances[x, y]))
        x, y = getMinParent(distances, x, y, neighbors)
    path.append((0, 0, maze[0, 0]))
    
    return path

def getMinParent(distances, x, y, neighbors):
    minDist = distances[x, y]
    fromx, fromy = None, None
    for neighbor in neighbors:
        i, j = x + neighbor[0], y + neighbor[1]
        if i >= 0 and j >= 0 and i < len(distances) and j < len(distances[0]) and distances[i, j] >= 0 and distances[i, j] <= minDist:
            fromx, fromy = i, j
            minDist = distances[i, j]
    return fromx, fromy

A tester

minCostPath(np.array([[4,2,-1,1,1],
                      [1,-1,6,7,1],
                      [4,-1,3,0,1],
                      [4,7,3,10,1],
                      [4,8,6,-1,1],
                                ]))

- aonecoding May 06, 2018 | Flag Reply

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of 8 vote

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SOLUTION:
Q1 - DFS to count the number of connected components in graph.

Followup -
What if board is too big?
Consider dividing the board and process each part separately.

Divide into blocks or row by row?
Row by Row. It's hard to deal with the border of blocks.
Union find by each row

What's the memory consumption?
Each time load 1 row of input into memory. Create a union find array for two rows (current row and previous row). Memory = 3 * row.

public class NumberOfIslands {
    int rowLen;
    int rowCounter;
    int numberOfIslands;

    Integer[] islands;

    public NumberOfIslands(int rowLen) {
        this.rowLen = rowLen;
        islands = new Integer[(rowLen + 1) * 2];
    }

    public int loadRow(int[] row) {
        rowCounter++;
        for(int i = 1; i <= rowLen; i++) {
            if(row[i - 1] != 0) {
                int j = i + rowLen + 1;
                Integer top_root = find(i);
                Integer left_root = find(j - 1);

                //System.out.println("idx " + i );
                //System.out.println("top " + top_root + " left " + left_root);
                if(top_root == null && left_root == null) { //new island found
                    numberOfIslands++;
                    islands[j] = j;
                } else if(top_root == null) { //no top neighbor. join left neighbor
                    islands[j] = left_root;
                } else if(left_root == null){ //no left neighbor. join top neighbor
                    if(top_root <= rowLen) {  //top neighbor had no connection with new row
                        islands[top_root] = j;
                        islands[j] = j;
                    } else {                  //top neighbor's ancestor is in new row
                        islands[j] = top_root;
                    }
                } else {
                    if(top_root == left_root) { //no union since left neighbor and top neighbor were in same island
                        islands[j] = left_root;
                    } else {                    //union left and top neighbors
                        islands[top_root] = left_root;
                        islands[j] = left_root;
                        numberOfIslands--;
                    }
                }
            }
        }

        compression();

        for(int i = 0; i <= rowLen; i++) {
            int j = i + rowLen + 1;
            if(islands[j] != null) {
                islands[i] = islands[j] % (rowLen + 1);
                islands[j] = null;
            } else {
                islands[i] = null;
            }
        }
        return numberOfIslands;
    }

    private Integer find(int idx) {
        if(islands[idx] == null) return null;
        while(idx != islands[idx]) {
            idx = islands[idx];
        }
        return idx;
    }

    private void compression() {
        for(int i = rowLen + 1; i < islands.length; i++) {
            if(islands[i] != null) {
                islands[i] = find(i);
            }
        }
    }

A tester

public static void main(String[] args) {
        NumberOfIslands instance = new NumberOfIslands(10);
        System.out.println("number of islands " + instance.loadRow(new int[]{0,0,1,1,0,1,0,1,0,1}) + '\n');
        System.out.println("number of islands " + instance.loadRow(new int[]{1,1,1,1,1,1,0,1,1,1}) + '\n');
        System.out.println("number of islands " + instance.loadRow(new int[]{0,0,0,0,0,0,0,0,0,1}) + '\n');
        System.out.println("number of islands " + instance.loadRow(new int[]{1,0,0,1,0,1,0,0,0,1}) + '\n');
        System.out.println("number of islands " + instance.loadRow(new int[]{1,1,1,1,0,1,0,1,0,1}) + '\n');
    }

- aonecoding April 21, 2018 | Flag Reply

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of 4 vote

public class KClosestPoints {

    public class Point {
        public int x;
        public int y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public List<Point> findKClosest(Point[] p, int k) {
        PriorityQueue<Point> pq = new PriorityQueue<>(k, new Comparator<Point>() {
            @Override
            public int compare(Point a, Point b) {
                return (b.x * b.x + b.y * b.y) - (a.x * a.x + a.y * a.y);
            }
        });

        for (int i = 0; i < p.length; i++) {
            if (i < k)
                pq.offer(p[i]);
            else {
                Point temp = pq.peek();
                if ((p[i].x * p[i].x + p[i].y * p[i].y) - (temp.x * temp.x + temp.y * temp.y) < 0) {
                    pq.poll();
                    pq.offer(p[i]);
                }
            }
        }

        List<Point> x = new ArrayList<>();
        while (!pq.isEmpty())
            x.add(pq.poll());

        return x;
    }
}

- aonecoding April 07, 2018 | Flag Reply

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of 5 vote

public static String balanceParenthesis(String s) {
        StringBuilder str = new StringBuilder(s);
        int left = 0;
        for(int i = 0; i < str.length(); i++) {
            if(str.charAt(i) == '(') {
                left++;
            } else if(str.charAt(i) == ')') {
                if(left > 0) {
                    left--;
                } else {
                    str.deleteCharAt(i--);
                }
            }
        }
        int right = 0;
        for(int i = str.length() - 1; i >= 0; i--) {
            if(str.charAt(i) == ')') {
                right++;
            } else if(str.charAt(i) == '('){
                if(right > 0) right--;
                else {
                    str.deleteCharAt(i);
                }
            }
        }
        return str.toString();
    }

- aonecoding April 07, 2018 | Flag Reply

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of 1 vote

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SOLUTION:
Seems like there are 3 cases:
1. s1 and s2 share no mutual characters. Then there is no way to further minimize the distance.
2. s1 and s2 share a pair of characters on crossed positions, such as 'b' and 'c' in "aabaaac" and "aacaaab". Then swap 'b' and 'c'.
3. s1 and s2 has no crossed pairs but has the same character on different positions. Then swap to align the character.

//@return the pair of indices in s2 to swap with.
    public int[] Swap(String s1, String s2) {
        if(s1.length() > s2.length()) {
            return Swap(s2, s1);
        }
        int swap_idx1 = -1, swap_idx2 = -1;
        for(int i = 0; i < s1.length(); i++) {
            if(s1.charAt(i) != s2.charAt(i)) {
                for (int j = i + 1; j < s1.length(); j++) {
                    if(s1.charAt(i) == s2.charAt(j) && s1.charAt(j) == s2.charAt(i)) {
                        return new int[]{i, j};  //case 2 found
                    } else if(s1.charAt(i) == s2.charAt(j) && s1.charAt(j) != s2.charAt(j)) {
                        swap_idx1 = i;
                        swap_idx2 = j;
                    }
                }
                for(int j = s1.length(); j < s2.length(); j++) {
                    if(s1.charAt(i) == s2.charAt(j)) {
                        swap_idx1 = i;
                        swap_idx2 = j;
                    }
                }
            }
        }
        if(swap_idx1 == -1) return null; //case1
        else return new int[]{swap_idx1, swap_idx2};
    }

- aonecoding April 01, 2018 | Flag Reply

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of 2 vote

import java.util.List;
import java.util.Map;
import java.util.HashMap;
class TreeNode {
    int val;
    List<TreeNode> kids;
}
public class HouseRobTree {
    public int houseRob(TreeNode root) {
        Map<TreeNode, Integer> money_in_tree = new HashMap<>();
        return houseRobHelper(root, money_in_tree);
    }
    
    private int houseRobHelper(TreeNode root, Map<TreeNode, Integer> money_in_tree) {
        if(root == null) return 0;
        if(root.kids == null || root.kids.isEmpty()) return root.val;
        if(money_in_tree.containsKey(root)) return money_in_tree.get(root);
        
        int max = 0;
        for(TreeNode kid: root.kids) {
            max = Math.max(max, houseRobHelper(kid, money_in_tree));
            if(kid.kids != null) {
                for(TreeNode grandKid: kid.kids) {
                    max = Math.max(max, houseRobHelper(grandKid, money_in_tree) + root.val);
                }
            }
        }
        money_in_tree.put(root, max);
        return max;
    }
}

- aonecoding April 01, 2018 | Flag Reply

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of 2 vote

//O(N) time and space for processing the board and lamps
//O(1) for finding if a cell is illuminated
public class GridIllumination {

    int N; //board size
    Set<Integer> illuminated_x = new HashSet<>();
    Set<Integer> illuminated_y = new HashSet<>();
    Set<Integer> illuminated_diag0 = new HashSet();
    Set<Integer> illuminated_diag1 = new HashSet();


    //@param lamps - a list of (x,y) locations of lamps
    public GridIllumination(int N, int[][] lamps) {
        this.N = N;
        for(int[] lamp: lamps) { //this lamp illuminates 4 lines of cells
            illuminated_x.add(lamp[0]);               //the entire column
            illuminated_y.add(lamp[1]);               //the entire row
            illuminated_diag0.add(lamp[1] - lamp[0]); //diagonal line with a slope of 1
            illuminated_diag1.add(lamp[0] + lamp[1]); //diagonal lines with a slope of -1
        }
    }

    public boolean is_illuminated(int x, int y) {
        if(illuminated_x.contains(x) ||
            illuminated_y.contains(y) ||
            illuminated_diag0.contains(y - x) ||
            illuminated_diag1.contains(x + y)) {
                return true;
        }
        return false;
    }
}

- aonecoding March 21, 2018 | Flag Reply

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of 3 vote

public class MovingAvg {

    int[] q;  // a circular queue of size N
    int head; //queue head
    int tail; //queue tail
    int size; // queue size
    int sum;

    public MovingAvg(int N) {
        q = new int[N];
    }

    //@param num - the next number from data stream
    //@return - new average with num included and expired number excluded
    public double getAverage(int num) {
        double avg = 0;
        sum += num;
        if(size == q.length) {
            sum -= q[head];
            head = (head + 1) % q.length;
        } else {
            size++;
        }
        q[tail] = num;
        tail = (tail + 1) % q.length;
        return 1.0 * sum / size;
    }
}

- aonecoding March 17, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

public class MatrixPuzzle {

    int column;
    int row;

    public MatrixPuzzle(int length, int width) {
        this.column = length;
        this.row = width;
    }

    //The dp formula will be M[i,j] = M[i - 1, j - 1] + M[i - 1, j] + M[i - 1, j + 1]
    //Because one could only land on current cell from the 3 cells in the upper-left, left and lower-left.
    //To make the space consumption 1d, cache the numbers in one column of the matrix at a time.
    //Follow-up 2: Just reset path-counts for blocked cells to 0
    public int numberOfPaths() {
        int[] paths = new int[row];
        paths[row - 1] = 1; //start from bottom-left corner
        for(int col = 1; col < column; col++) {
            int upper_left_value = 0;
            for(int r = 0; r < row; r++) {
                int left_value = paths[r];
                paths[r] += upper_left_value + (r == row - 1 ? 0 : paths[r + 1]);
                upper_left_value = left_value;
            }
        }

        return paths[row - 1]; //exit from bottom-right
    }
}

- aonecoding March 10, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

//Extra space O(1) Runtime O(nlogn)
    List<Integer> getFibonacciNumber(int[] nums) {
        List<Integer> fibonacciNumbers = new ArrayList<>();
        Arrays.sort(nums);
        int fib1 = 1, fib2 = 1;
        for(int i = 0; i < nums.length;) {
            if(nums[i] < fib2) {
                i++;
            } else if(nums[i] == fib2) {
                fibonacciNumbers.add(nums[i]);
                i++;
            } else {
                int fib3 = fib1 + fib2;
                fib1 = fib2;
                fib2 = fib3;
            }
        }
        return fibonacciNumbers;
    }


    //Math Solution: Extra space O(1) Runtime O(n)
    List<Integer> getFibonacciNumbers(int[] nums) {
        List<Integer> fibonacciNumbers = new ArrayList<>();
        for(int n: nums) {
            if(isFibonacci(n)) fibonacciNumbers.add(n);
        }
        return fibonacciNumbers;
    }
    boolean isFibonacci(int n)
    {
        // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both
        // is a perfect square
        return isPerfectSquare(5*n*n + 4) ||
                isPerfectSquare(5*n*n - 4);
    }

    boolean isPerfectSquare(int x)
    {
        int s = (int) Math.sqrt(x);
        return (s*s == x);
    }

- aonecoding February 15, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 4 vote

final String[] largeNumbers = new String[] {""," thousand"," million"," billion"," trillion"," quintillion"};//...etc
    final String[] digits = new String[] {"", " one", " two", " three", " four", " five", " six"," seven"," eight", " nine"};
    final String[] tens = new String[] {"", " ten"," twenty"," thirty"," forty"," fifty"," sixty"," seventy"," eighty"," ninety"};
    final String[] teens = new String[] {" ten", " eleven"," twelve", " thirteen", " fourteen", " fifteen", " sixteen", " seventeen"," eighteen"," nineteen"};

    String digitsToEnglish(String num) {
        if(num.length() == 1) {
            return digits[Integer.parseInt(num)];
        }
        int len = num.length();
        StringBuilder english = new StringBuilder();
        int largeNumIdx = 0, tripletIdx = 0;
        while(tripletIdx < len) {
            StringBuilder triplet = new StringBuilder();
            if(len > tripletIdx) {
                triplet.append(num.charAt(len - 1 - tripletIdx));
            }
            if(len > tripletIdx + 1) {
                triplet.insert(0, num.charAt(len - 1 - tripletIdx - 1));
            }
            if(len > tripletIdx + 2) {
                triplet.insert(0, num.charAt(len - 1 - tripletIdx - 2));
            }
            if(Integer.parseInt(triplet.toString()) != 0) {
                StringBuilder current = new StringBuilder();
                if (triplet.length() == 3 && triplet.charAt(0) != '0') {
                    //current.append(' ');
                    current.append(digits[triplet.charAt(0) - '0']);
                    current.append(' ');
                    current.append("hundred");
                }
                if (triplet.length() >= 2 && triplet.charAt(triplet.length() - 2) == '1') {
                    //current.append(' ');
                    current.append(teens[triplet.charAt(triplet.length() - 1) - '0']);
                } else {
                    if (triplet.length() >= 2) {
                        //current.append(' ');
                        current.append(tens[triplet.charAt(triplet.length() - 2) - '0']);
                    }
                    //current.append(' ');
                    current.append(digits[triplet.charAt(triplet.length() - 1) - '0']);
                }
                //current.append(' ');
                current.append(largeNumbers[largeNumIdx]);
                english.insert(0, current);
            }
            largeNumIdx += 1;
            tripletIdx += 3;
        }
        return english.toString();
    }

- aonecoding February 15, 2018 | Flag Reply

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of 2 vote

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SOLUTION:
Two-way BFS, similar to 'WordLadder'.
s1Derives and s2Derives are the words reachable by swapping characters in s1 and s2 in any random way. If s1and s2 are anagrams, for sure s1Derives and s2Derives would join at some point. Find out all possible derivatives 'nextLevel' of s1/s2. Then find out all derivatives of words in 'nextLevel'......until the earliest joint of s1Derives and s2Derives occurs. Return the number of levels gone through.

int minNumberOfSwaps(String s1,String s2) {
        //if(!isAnagram(s1, s2)) return Integer.MAX_VALUE; //assume s1 and s2 are anagrams
        int step = 0;
        Set<String> s1Derives = new HashSet<>();
        Set<String> s2Derives = new HashSet<>();
        s1Derives.add(s1);
        s2Derives.add(s2);
        Set<String> visited = new HashSet<>();
        int len = s1.length();
        while(!containsSameString(s1Derives, s2Derives)) {
            Set<String> set = step % 2 == 0 ? s1Derives: s2Derives;
            Set<String> nextLevel = new HashSet<>();
            for(String s: set) {
                for (int i = 0; i < len; i++) {
                    for (int j = i; j < len; j++) {
                        if(s.charAt(i) != s.charAt(j)) {
                            char[] charArray = swap(s.toCharArray(), i, j);
                            String derived = new String(charArray);
                            if(!visited.contains(new String(derived))) {
                                nextLevel.add(derived);
                                visited.add(derived);
                            }
                        }
                    }
                }
            }
            if(step % 2 == 0) s1Derives = nextLevel;
            else s2Derives = nextLevel;
            step++;
        }
        return step;
    }

- aonecoding February 03, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

public Integer random(int[] weights) {
        int totalWeight = 0;
        Integer selected = null;
        Random rand = new Random();
        for(int i = 0; i < weights.length; i++) {
            int r = rand.nextInt(totalWeight + weights[i]);
            if(r >= totalWeight) {
                selected = i;
                totalWeight += weights[i];
            }
        }
        return selected;
    }

FOLLOW-UP

//main
//RandomSelectTree tree = new RandomSelectTree();
//while(tree.n > 0) tree.randomPoll();

//Segment Tree O(nlogn)
public class RandomSelectTree {

    int[] tree;
    int[] weights;
    int n; //number of remaining balls

    RandomSelectTree(int[] weights) {
        tree = new int[2 * weights.length + 1];
        this.weights = weights;
        n = weights.length;
        buildTree(0, 0, weights.length - 1);
    }
    //build segment tree for weights[]
    private void buildTree(int root, int start, int end) { 
        if(start > end) return;
        if(start == end) {
            tree[root] = weights[start];
            return;
        }
        int mid = (start + end) / 2;
        int leftChild = 2 * root + 1, rightChild = 2 * root + 2;
        buildTree(leftChild, start, mid);
        buildTree(rightChild, mid + 1, end);
        tree[root] = tree[leftChild] + tree[rightChild];
    }

    public int randomPoll() {
        if(n == 0) {
            //Exception
        }
        int rand = new Random().nextInt(tree[0]);
        int start = 0, end = weights.length - 1, node = 0;
        while(start < end) {
        //chosen ball is between start to (start + end) / 2
            if(rand < tree[node * 2 + 1]) {
               node = node * 2 + 1;
                end = (start + end) / 2;
            } else { //chosen ball between (start + end) / 2 + 1 and end
                rand -= tree[node * 2 + 1];  //!!! narrow down to remaining range
                node = node * 2 + 2;
                start = (start + end) / 2 + 1;
            }
        }
        delete(start, node);
        return start;
    }

    //remove ball at idx.
    private void delete(int idx, int node) { 
       n--;
        while(node > 0) {
            tree[node] -= weights[idx];
            node = (node - 1) / 2;
        }
        tree[0] -= weights[idx];
    }

- aonecoding January 20, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 4 vote

//Solution to question   O(n)
   boolean isPeriod(String s) {
        StringBuilder str = new StringBuilder(s + s);
        str.deleteCharAt(0);
        str.deleteCharAt(str.length() - 1);
        return strStr(str.toString(), s); //KMP strStr(T, S) to find if T has S in it.
    }

    //Solution to follow-up
    //This method looks for the repeating pattern in string
    private static String getPeriod(String string) { // O(n * n)
        //for every possible period size i, check if it's valid
        for (int i = 1; i <= string.length() / 2; i++) {
            if (string.length() % i == 0) {
                String period = string.substring(0, i);
                int j = i;
                while(j + i < string.length()) {
                    if (period.equals(string.substring(j, j + i))) {
                        j = j + i;
                        if(j == string.length()) { //period valid through entire string
                            return period;
                        }
                    } else {
                        break;
                    }
                }
            }

        }
        return null; //string is not periodic
    }

- aonecoding January 20, 2018 | Flag Reply

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of 2 vote

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SOLUTION:
Step 1, have a table to store all the visits sorted by time stamp.
or
have a queue to store the visits per second in the past 5 minute.

Follow-up:
Have two arrays hits[range] and lastupdated[range].
Range = 5 mins = 300 seconds in this case. Any hit within 'range' time from now is valid and should counted.
Index of a hit in the two arrays will be timestamp % 300.
Store the last updated time of a hit. Later if a query for the hit count arrives with a query timestamp, sum up all the valid hits from array 'hits'. Threshold for valid hits: query time - hit last updated time < 300.

public class HitCounter {

    int[] times, hits;
    int timeRange;

    /** Initialize your data structure here. */
    public HitCounter(int range) {
        times = new int[range];
        hits = new int[range];
        timeRange = range;
    }

    /** Record a hit.
     @param timestamp - The current timestamp (in seconds granularity). */
    public void hit(int timestamp) {
        int idx = timestamp % timeRange;
        if (times[idx] != timestamp) {
            times[idx] = timestamp;
            hits[idx] = 1;
        } else {
            ++hits[idx];
        }
    }

    /** Return the number of hits in the past 5 minutes.
     @param timestamp - The current timestamp (in seconds granularity). */
    public int getHits(int timestamp) {
        int res = 0;
        for (int i = 0; i < timeRange; ++i) {
            if (timestamp - times[i] < timeRange) {
                res += hits[i];
            }
        }
        return res;
    }
}

Followup2
For writing,
Concurrency update becomes an issue. Add write lock for protection. But this may slowdown the machine badly.

Move hit counters onto a distributed system. Have several machines counting together. Assign userIDs to diff hosts.
Add LB on top to make sure requests get distributed evenly.
Upon reading, sum up counts across all machines. For a read-heavy system, add cache.

- aonecoding January 18, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 6 vote

//Binary Indexed Tree (O(logN) Toggle, O(logN) Get.
public class ToggleBulbs {

    int[] bulbs;

    public ToggleBulbs(int n) { //n bulbs given

        bulbs = new int[n + 1];
    }

    public boolean isOn(int i) {
        //a bulb is on if it's toggled an odd number of times
        return read(i) % 2 == 1;
    }

    // toggle(i, j) is equivalent to
    // toggle(i, n - 1) and then toggle(j, n - 1)
    public void toggle(int i,int j) {
        toggle(i);
        toggle(j + 1);
    }

    //toggle from ith to the last bulb (a standard update in BITree)
    private void toggle(int idx){
        int node = idx + 1;
        while (node < bulbs.length){
            bulbs[node] = (bulbs[node] + 1) % 2;
            node += (node & -node);
        }
    }

    //get the number of times bulb i toggled (read prefix sum from 0 to i in BITree)
    private int read(int idx) {
        int node = idx + 1;
        int sum = 0;
        while (node > 0) {
            sum += bulbs[node];
            node -= (node & -node);
        }
        return sum;
    }

}

- aonecoding January 16, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

boolean hasFactorKSubArray(int[] arr, int k) {
        for(int i = 1; i < arr.length; i++) {
            arr[i] += arr[i - 1]; //becomes a prefix sum array of arr (can be recovered if necessary)
        }
        for(int hi = arr.length - 1; hi >= 1; hi-- ) {
            for(int lo = 0; lo <= hi - 1; lo++) {
                if((arr[hi] - lo == 0 ? 0: arr[lo - 1]) % k == 0) {
                    return true;
                }
            }
        }
        return false;
    }

- aonecoding January 13, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 4 vote

class Node:
    def __init__(self, ID, parent):
        self.ID = ID
        self.parentID = parent
        self.left = None
        self.right = None

#function to identify if given is preorder
'''
Create a stack to store nodes in the current path when traversing.
Push node[i] into stack once node[i] is verified to be valid (valid only when parent of node[i] is in stack. 
In preorder a parent must show up earlier than its child)
Whenever stack top is not the parent of node[i], pop until parent of node[i] is at stack top. Push node[i].
If all nodes popped but parent of node[i] still not found, then node[i] is not in preorder sequence.
'''
def isPreorder(nodes):
    if not nodes:
        return True
    
    st = [nodes[0].ID]
    i = 1
    while i < len(nodes):
        if not st:
            return False
        if st[-1] is nodes[i].parentID:
            st.append(nodes[i].ID)
            i += 1
        else:
            st.pop()
    return True

- aonecoding January 06, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

# The question is about identifying any window size k that its numbers at both ends bigger than numbers in between
def bloomDay(a, k): # k must be >= 0
    window = [] # elements in window maintain descending order. 
    firstDay = 10000000
    i = 0
    while i < len(a):
        if not window:
            window.append(i)
            i += 1
        elif i - window[0] <= k:
            if a[i] >= a[window[0]]:
                window = [i]
                i += 1
            elif a[i] >= a[window[-1]]:
                window.pop()
            else:
                window.append(i)
                i += 1
        else: #window size == k
            if len(window) == 1 or a[window[1]] < a[i]:#valid window with both end numbers biggest
                firstDay = min(firstDay, a[i], a[window[0]])
                window = [i]
                i += 1
            else:  #a[i] is less than a number in the middle of the window
                window.pop(0)
                if a[window[-1]] <= a[i]:
                    window.pop()
                else:
                    window.append(i)
                    i += 1
    return firstDay

- aonecoding January 05, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

class Stack:
    def __init__(self):
        self.nums = []
        self.add = []
    
    def push(self,num):
        self.nums.append(num)
        self.add.append(0)
        print num," "
        
    def pop(self):
        try:
            number_to_add = self.add.pop()
            print self.nums.pop() + number_to_add
            if self.add:
                self.add[-1] += number_to_add
        except:
            print "can't pop from an empty stack"
            
    def inc(self, n, m):
        if not self.nums:
            print "empty"
        if n > 0:
            n = min(n, len(self.nums))
            self.add[n - 1] += m
        print self.nums[-1] + self.add[-1], " "

- aonecoding January 04, 2018 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

public static List<Integer> getAnagrams(String s, String word) {
        Map<Character, Integer> letters = new HashMap<>();
        int distinct_letters = 0;
        for(char c: word.toCharArray()) {
            if(!letters.containsKey(c)) distinct_letters++;
            letters.put(c, letters.getOrDefault(c, 0) + 1);
        }

        //search for anagrams with two pointers
        List<Integer> res = new ArrayList<>();
        int lo = 0, hi = 0;
        while(hi < s.length()) {
            if(!letters.containsKey(s.charAt(hi))) {
                while(lo < hi) {
                    char c = s.charAt(lo);
                    if(letters.get(c) == 0) distinct_letters++;
                    letters.put(c, letters.get(c) + 1);
                    lo++;
                }
                hi++;
                lo = hi;
            } else if(letters.get(s.charAt(hi)) == 0){
                while(s.charAt(lo) != s.charAt(hi)) {
                    char c = s.charAt(lo);
                    if(letters.get(c) == 0) distinct_letters++;
                    letters.put(c, letters.get(c) + 1);
                    lo++;
                }
                lo++;
            } else {
                char c = s.charAt(hi);
                letters.put(c, letters.get(c) - 1);
                if(letters.get(c) == 0) distinct_letters--;
                if(distinct_letters == 0) {
                    res.add(lo);
                }
                hi++;
            }
        }
        return res;
    }

- aonecoding November 03, 2017 | Flag Reply

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of 3 vote

int sqrt(int x) {
        //check for x >= 0 if necessary
        int r = x;
        while (r * r > x)
            r = (r + x / r) / 2;
        return r ;
    }

BST Implementation:

class BinarySearchTree {
    /* Class containing left and right child of current node and key value*/
    class Node {
        int key;
        Node left, right;
 
        public Node(int item) {
            key = item;
            left = right = null;
        }
    }
 
    // Root of BST
    Node root;
 
    // Constructor
    BinarySearchTree() { 
        root = null; 
    }
 
    // This method mainly calls insertRec()
    void insert(int key) {
       root = insertRec(root, key);
    }
     
    /* A recursive function to insert a new key in BST */
    Node insertRec(Node root, int key) {
        /* If the tree is empty, return a new node */
        if (root == null) {
            root = new Node(key);
            return root;
        }
 
        /* Otherwise, recur down the tree */
        if (key < root.key)
            root.left = insertRec(root.left, key);
        else if (key > root.key)
            root.right = insertRec(root.right, key);
 
        /* return the (unchanged) node pointer */
        return root;
    }
 
    // This method mainly calls InorderRec()
    void inorder()  {
       inorderRec(root);
    }
 
    // A utility function to do inorder traversal of BST
    void inorderRec(Node root) {
        if (root != null) {
            inorderRec(root.left);
            System.out.println(root.key);
            inorderRec(root.right);
        }
    }

   // This method mainly calls deleteRec()
    void deleteKey(int key)
    {
        root = deleteRec(root, key);
    }
 
    /* A recursive function to insert a new key in BST */
    Node deleteRec(Node root, int key)
    {
        /* Base Case: If the tree is empty */
        if (root == null)  return root;
 
        /* Otherwise, recur down the tree */
        if (key < root.key)
            root.left = deleteRec(root.left, key);
        else if (key > root.key)
            root.right = deleteRec(root.right, key);
 
        // if key is same as root's key, then This is the node
        // to be deleted
        else
        {
            // node with only one child or no child
            if (root.left == null)
                return root.right;
            else if (root.right == null)
                return root.left;
 
            // node with two children: Get the inorder successor (smallest
            // in the right subtree)
            root.key = minValue(root.right);
 
            // Delete the inorder successor
            root.right = deleteRec(root.right, root.key);
        }
        return root;
    }
}

- aonecoding November 03, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

public static int[] llac(int[] array) {
        if(array.length == 0) return new int[]{};
        int maxLen = 1, //len of longest sequence
            end = array[0], //end number of sequence
            step = 0;
        Map<Integer,Map<Integer, Integer>> index_to_llac = new HashMap<>(); //Map<currentIndex, Map<difference, length>>

        for(int current = 1; current < array.length; current++) {
            index_to_llac.put(current, new HashMap<>());
            for(int prev = 0; prev < current; prev++) {
                int diff = array[current] - array[prev];
                int length = 2;
                if(index_to_llac.containsKey(prev) && index_to_llac.get(prev).containsKey(diff)) {
                    length = index_to_llac.get(prev).get(diff) + 1;
                }
                Map<Integer, Integer> diff_to_llac = index_to_llac.get(current);
                //if this is a arithmetic progression with [diff] exists before current, add 1 to the length to include current in the progression
                //otherwise, [prev, current] forms a progression of size 2
                diff_to_llac.put(diff, length);

                if(maxLen < length) {
                    maxLen = length;
                    end = array[current];
                    step = diff;
                }
            }
        }

        int[] sequence = new int[maxLen];
        for(int i = maxLen - 1; i >= 0; i--) {
            sequence[i] = end;
            end -= step;
        }
        return sequence;
    }

- aonecoding November 03, 2017 | Flag Reply

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of 3 vote

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public static int rankWays(int n) {
int[] base = new int[2]; // start from case for 0 participants
base[1] = 1;

for(int p = 2; p <= n; p++) {
int[] ways = new int[p + 1]; //index is row number, value is count of ways
for(int r = 1; r < base.length; r++) {
ways[r + 1] += base[r] * (r + 1);
ways[r] += base[r] * r;
}
base = ways;
}
int sum = 0;
for(int x: base) sum += x;
return sum;
}

With N participants [p1, p2, p3... pN], one ranking could be:
1st place: p1
2nd place: p2, p3
...
xth place: pN

In this case, if one more player pL joins the game. The new player pL can be placed into

pL can be here as new 1st place
1st place p1 pL can be here tied with 1st place
pL can be here as new 2nd place
2nd place p2, p3 pL can be here tied 2nd place
... ...
xth place pN pL can be here tied last place
pL can be here as new last

So for every rank combination of N people, adding 1 more player creates rowNumber * 2 + 1 ways of ranking.
So if we keep track of the row numbers of all previous combinations of ranking,
a ranking with R rows is gonna derive into R + 1 more cases (with R + 1 row numbers) plus R more cases (with R row numbers).

To print all possible combinations:
public static List<List<List<Integer>>> rankCombinations(int n) {
List<List<List<Integer>>> base = new ArrayList<>();
base.add(new ArrayList<>());
base.get(0).add(new ArrayList<>());
base.get(0).get(0).add(1);
for(int p = 2; p <= n; p++) {
List<List<List<Integer>>> combinations = new ArrayList<>();
for(List<List<Integer>> comb: base) {
for(int i = 0; i <= comb.size(); i++) {
List<Integer> rank = new ArrayList<>();
rank.add(p);
List<List<Integer>> newComb = new ArrayList<>(comb);
newComb.add(i, rank);
combinations.add(newComb);
}

for(int i = 0; i < comb.size(); i++) {
List<List<Integer>> newComb = deepcopy(comb);
newComb.get(i).add(p);
combinations.add(newComb);
}
}
base = combinations;
}
//print all ranking combinations. multi players in a same sublist makes a tie.
for(List<List<Integer>> ranks: base) System.out.println(ranks);
return base;
}

private static List<List<Integer>> deepcopy(List<List<Integer>> ls) {
List<List<Integer>> copy = new ArrayList<>();
for(List<Integer> sublist: ls) {
copy.add(new ArrayList<Integer>(sublist));
}
return copy;
}

- aonecoding November 03, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

int paginate(const std::vector<std::string>& addressbook, std::vector<std::string>& results) {
    std::list<std::pair<int, int> > curve; // host_id, position in addressbook
    int num_entries = addressbook.size(); // already sorted by score.
    for (int index = 0; index < num_entries; ++index) { 
        const std::string& s = addressbook[index];
        int host_id = std::stoi(s.substr(0, s.find_first_of(',')));
        curve.emplace_back(host_id, index);
    }

    while (!curve.empty()) {
        std::unordered_set<int> uniq_host_ids;
        std::vector<decltype(curve)::iterator> hosts;
        bool combo = false;
        for (auto iter = curve.begin(); iter != curve.end(); ++iter) {
            if (uniq_host_ids.find(iter->first) == uniq_host_ids.end()) {
                uniq_host_ids.insert(iter->first);
                hosts.push_back(iter);
                if (hosts.size() == resultsPerPage) {
                    combo = true;
                    break;
                }
            }
        }
        for (auto& host : hosts) {
            results.push_back(addressbook[host->second]);
        }
        for (auto& host : hosts) { 
            curve.erase(host);
        }

        if (combo) {
            if (!curve.empty()) results.push_back("");
            continue; // enter next loop
        }

        if (!curve.empty()) {
            int num_done = hosts.size();
            for (auto iter = curve.begin(); iter != curve.end(); ++iter) {
                hosts.push_back(iter);
                if (hosts.size() == resultsPerPage) {
                    combo = true; break;
                }
            }
            int num_hosts = hosts.size();
            for (int i = num_done; i < num_hosts; ++i) {
                results.push_back(addressbook[hosts[i]->second]);
            }
            for (int i = num_done; i < num_hosts; ++i) {
                curve.erase(hosts[i]);
            }
            if (combo && !curve.empty()) {
                results.push_back("");
            }
        } // end if condition
    } // end while loop
    return 0;
}

- aonecoding October 13, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

public static void main(String[] args) {
        System.out.println(lonelyPixelCount(new int[][]{ //1: black, 0: white
                {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
                {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
                {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
                {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
                {0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
        }));
    }

    static int lonelyPixelCount(int[][] picture) {
        int m = picture.length, n = picture[0].length;
        //First traversal sum up the count of black pixels by column
        for(int i = 1; i < m; i++){
            for(int j = 0; j < n; j++){
                picture[i][j] += picture[i - 1][j];
            }
        }
        int result = 0;
        //Then traverse row by row from the bottom, count the black pixels in current row.
        //If there is only 1 black pixel in current row, verify whether it is also the only in the column.
        for(int i = m - 1; i >= 0; i--) {
            int pixel_count_in_row = 0;
            boolean only_pixel_in_column = false;
            for(int j = n - 1; j >= 0; j--) {
                if(picture[i][j] > 0 && (i == 0 || picture[i - 1][j] + 1 == picture[i][j])) {	//verify if current cell number is a black pixel, by method in blue text above
                    pixel_count_in_row++;
                    if((i < m - 1 && picture[i + 1][j] == 1) || (i == m - 1 && picture[i][j] == 1)) {
                        only_pixel_in_column = true;
                    }
                }
                if(i < m - 1) {
                    //overwrite current cell with the number below it
                    picture[i][j] = picture[i + 1][j];
                }
            }
            if(pixel_count_in_row == 1 && only_pixel_in_column) {
                result++;
            }
        }
        return result;
    }

- aonecoding September 22, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 4 vote

import java.util.List;
import java.util.ArrayList;

public class IntegerCollection {
    List<Integer> collection = new ArrayList<>();
    List<Integer> divisors = new ArrayList<>();
    int factor = 1;
    int addition = 0;
    int set0 = -1;

    public void append(int x) {
        collection.add(x - addition);
        divisors.add(factor);
    }

    public int get(int index) {
        if(index >= collection.size()) throw new RuntimeException();
        if(divisors.get(index) <= set0) return addition;
        return collection.get(index) * (factor / divisors.get(index)) + addition;
    }

    public void add_to_all(int x) {
        addition += x;
    }

    public void multiply_to_all(int x) {
        if(x == 0) {
            addition = 0;
            factor = 1;
            set0 = collection.size() - 1;
        } else {
            addition *= x;
            factor *= x;
        }
    }
}

- aonecoding September 22, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

SELECT 
    Department.name, 
    COUNT(Employee.id)
FROM 
    Department
LEFT JOIN 
    Employee ON Department.dept_id = Employee.dept_id
GROUP BY 
    Department.dept_id, 
    Department.name
ORDER BY 
    COUNT(Employee.id) DESC, 
    Department.name

- aonecoding September 05, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

int sqrt(int x) {
    if (x == 0)
        return x;
    int left = 1, right = x;
    while (true) {
        int mid = (left + right) / 2;
        if (mid > x / mid)
            right = mid - 1;
        else if (mid + 1 > x / (mid + 1))  //mid < x / mid
            return mid;
        else                                      //mid < x / mid
            left = mid + 1;
    }
}

IV. is on LC

V. Lossy Counting & Sticky Sampling

- aonecoding August 30, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Solution to Closest K Nodes

vector<int> closestKValues(TreeNode* root, double target, int k) {
        vector<int> res;
        inorder(root, target, k, res);
        return res;
    }
    void inorder(TreeNode *root, double target, int k, vector<int> &res) {
        if (!root) return;
        inorder(root->left, target, k, res);
        if (res.size() < k) res.push_back(root->val);
        else if (abs(root->val - target) < abs(res[0] - target)) {
            res.erase(res.begin());
            res.push_back(root->val);
        } else return;
        inorder(root->right, target, k, res);
    }

- aonecoding August 30, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

public void solve(int[] cards) {
        boolean[] played = new boolean[4];
        for(int i = 0; i < 4; i++) {
            StringBuilder str = new StringBuilder(String.valueOf(cards[i]));
            played[i] = true;
            solve(cards, played, cards[i],str);
            played[i] = false;
        }
    }

    private void solve(int[] cards, boolean[] played, double res, StringBuilder str) {
        if(played[0] && played[1] && played[2] && played[3]) {
            if(res == 24)
                System.out.println(str.toString());
            return;
        }
        for(int i = 0; i < 4; i++) {
            int len = str.length();
            if(!played[i]) {
                played[i] = true;

                str.insert(0, '(');
                str.append("+" + cards[i] + ")");
                solve(cards, played, res + cards[i], str);

                str.setCharAt(len + 1, '-');
                solve(cards, played, res - cards[i], str);

                str.deleteCharAt(0);
                str.deleteCharAt(str.length() - 1);
                str.setCharAt(len, '*');
                solve(cards, played, res * cards[i], str);

                str.setCharAt(len, '/');
                solve(cards, played, 1.0 * res / cards[i], str);

                str.delete(len, str.length());
                played[i] = false;
            }
        }
    }

- aonecoding August 14, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Looking for interview experience sharing and coaching?

Visit aonecode.com for private lessons by FB, Google and Uber engineers

Our ONE TO ONE class offers

SYSTEM DESIGN Courses (highly recommended for candidates for FLAG & U)
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algos & Clean Coding),
latest interview questions sorted by companies,
mock interviews.

Our students got hired from G, U, FB, Amazon, LinkedIn and other top tier companies after weeks of training.

Feel free to email us aonecoding@gmail.com with any questions. Thanks!

Solution:
I. If it is a write-heavy application, create a Fenwick Tree to store for every toggle(i, j), add 1 to node i and -1 to node j (O(logN)). Upon calling isOn(k), get the cumulative sum (O(logN)) till the kth node. If sum is odd, then bulb is on. Otherwise bulb is off.

II. If it is read-heavy, create a bit map that toggles in linear time and reads in constant time.

//Fenwick Tree Solution
public class ToggleBulbs {

    int[] sum;

    public ToggleBulbs(int n) {
        sum = new int[n];
    }

    public boolean isOn(int i)
    {
        return read(i)%2==1;
    }

    public void toggle(int start,int end)
    {
        update(start,1);
        update(end+1,-1);
    }

    private int read(int idx){
        int ans = 0;
        while (idx > 0){
            ans += sum[idx];
            idx -= (idx & -idx);
        }
        return ans;
    }

    private void update(int idx ,int val){
        while (idx <= sum.length){
            sum[idx] += val;
            idx += (idx & -idx);
        }
    }
}

- aonecoding August 09, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 4 vote

public void powerOf2Paths(int N) {
        List<Integer> path = new ArrayList<>();
        path.add(0);
        powerOf2PathsHelper(4, 0, path);
    }

    public void powerOf2PathsHelper(int N, int num, List<Integer> path) {
        if(num == N) {
            System.out.println(path);
        }

        for(int i = 0; num + (1 << i) <= N; i++) {
            int sum = num + (1 << i);
            path.add(sum);
            powerOf2PathsHelper(N, sum, path);
            path.remove(path.size() - 1);
        }
    }

- aonecoding August 09, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

//get the number of connected components in a list
    public int numberOfGroups(Set<Node> nodes) {
        Set<Node> visited = new HashSet<>();
        int cnt = 0;
        for(Node node: nodes) {
            while(!visited.contains(node)) {
                if(nodes.contains(node)){
                    visited.add(node);
                    node = node.next;
                } else {
                    cnt++;
                    break;
                }
            }
        }
        return cnt;
    }

- aonecoding August 09, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Hi Fernando,

I interpret the problem this way - making someone my colleague means having him into my team (assigning my manager to him). Since it is a design problem it's open-ended and your solution would be good as well.

Thanks for the reply!

- aonecoding August 06, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Looking for interview experience sharing and coaching?

Visit aonecode.com for private lessons by FB, Google and Uber engineers

Our ONE TO ONE class offers

SYSTEM DESIGN Courses (highly recommended for candidates for FLAG & U)
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algos & Clean Coding),
latest interview questions sorted by companies,
mock interviews.

Our students got hired from G, U, FB, Amazon, LinkedIn and other top tier companies after weeks of training.

Feel free to email us aonecoding@gmail.com with any questions. Thanks!

Solution:
Breath First Search.
Create a map that stores Map<Node, [PreviousNode, Cost]> the cost to the current node.
Since the question asks for the actual shortest path, in the map also store the previous node in the path.
During the BFS, if the current node has been visited before, but the current cost < former cost, update the cost and route for the current node (only keep the min cost route). Otherwise if current cost > former cost, stop search by not including the current node into the BFS queue.

To stand for a node in the map with an integer rather than a point (x, y), do
nodeID = x * matrix[0].length + y;

To convert the ID back to a point, do
x = nodeID / matrix[0].length;
y = nodeID % matrix[0].length;

//Follow-up
    public void minCostPath(int[][] matrix) {
        Queue<Integer> queue = new ArrayDeque<>();
        Map<Integer, int[]> map = new HashMap<>(); //key: current location; value[0]: the previous node from where it gets to the current location, value[1]:total cost up to current node
        for(int j = 0; j < matrix[0].length; j++) { //put first row into queue
            if(matrix[0][j] != 0) {
                queue.add(j);
                map.put(j, new int[] {-1, matrix[0][j]});
            }
        }
        int destination = -1, minCost = Integer.MAX_VALUE;
        while(!queue.isEmpty()) {
            int id = queue.poll();
            int fromX = id / matrix[0].length, fromY = id % matrix[0].length;
            if(fromX + 1 < matrix.length) {
                int cost = moveMinCost(queue, map, matrix, id, fromX + 1, fromY);
                if (cost >= 0 && fromX + 1 == matrix.length - 1) {
                    if (cost < minCost) {
                        destination = id + matrix[0].length;
                        minCost = cost;
                    }
                }
            }
            if(fromY + 1 < matrix[0].length)
                moveMinCost(queue, map, matrix, id, fromX, fromY + 1);
            if(fromX - 1 >= 0)
                moveMinCost(queue, map, matrix, id, fromX - 1, fromY);
            if(fromY - 1 >= 0)
                moveMinCost(queue, map, matrix, id, fromX, fromY - 1);
        }
        if(destination == -1) return; //no such path from first row to last row
        while(map.containsKey(destination)) {   //print shortest path from destination to source
            int x = destination / matrix[0].length, y = destination % matrix[0].length;
            System.out.println("(" + x + ","+ y + "),");
            destination = map.get(destination)[0];
        }
    }

    private int moveMinCost(Queue<Integer> queue, Map<Integer, int[]> map, int[][] matrix, int from, int x, int y) {
        if(matrix[x][y] == 0) return -1;
        int id = x * matrix[0].length + y;
        int cost = map.get(from)[1] + matrix[x][y];
        if(!map.containsKey(id) || map.get(id)[1] > cost) {
            map.put(id, new int[]{from, cost});
            queue.add(id);
            return cost;
        }
        return -1;
    }

- aonecoding August 03, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

class Employee {

    int id;
    private String name;
    //...other personal information
    private Employee manager;
    private List<Employee> subordinates; //direct subordinates

    public Employee(int id, String name) {
        this.id = id;
        this.name = name;
        subordinates = new ArrayList<>();
    }

    boolean isManager(Employee manager) {
        Employee upperLevel = this.manager;
        while(upperLevel != null && upperLevel != manager) {
            upperLevel = upperLevel.manager;
        }
        return upperLevel.manager == manager;
    }

    void beColleague(Employee p) {
        p.setManager(this.manager);
    }

    void setManager(Employee m) {
        if(manager != null) {   //remove from subordinate's list of current manager
            manager.deleteSubordinate(this);
        }
        manager = m;
        m.addSubordinate(this); //add to new manager's subordinate's list
    }

    private void deleteSubordinate(Employee m) {
        subordinates.remove(m);
    }

    private void addSubordinate(Employee m) {
        subordinates.add(m);
    }

}

public class Employment {

    private Employee admin;
    private Map<Integer, Employee> employees;               //id: employee

    public Employment() {
        admin = new Employee(0, "ADMINISTRATOR");  //root of the employment tree, the highest level supervisor
        employees = new HashMap<>();
        employees.put(0, admin);
    }

    public void assignManager(int p1, int p2) {
        employees.get(p2).setManager(employees.get(p1));
    }

    public void beColleague(int p1, int p2) {
        employees.get(p2).beColleague(employees.get(p1));
    }

    public boolean isManager(int p1, int p2) {
        return employees.get(p2).isManager(employees.get(p1));
    }

    //public void addEmployee(int p);
    //public void deleteEmployee(int p);

}

- aonecoding August 03, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

public class RandomSelectFromRectangle {
    class Rectangle {
        int x1, x2, y1, y2; //top left (x1, y1), bottom right (x2, y2)
    }

    class Point {
        int x, y;
        public Point(int a, int b) {
            x = a;
            y = b;
        }
    }

    final Random rand = new Random();

    //Round2
    public Point randomSelectFrom(Rectangle rectangle) {
        return new Point(rectangle.x1 + rand.nextInt(rectangle.x2 - rectangle.x1 + 1),
                        rectangle.y2 + rand.nextInt(rectangle.y1 - rectangle.y2 + 1));
    }

    //Round2 follow-up
    //first of all decide which rectangle yields the point (randomly select a rectangle based on area as the weight)
    //then apply randomSelectFrom(rectangle) on the selected rectangle
    public Point randomSelectFrom(Rectangle[] rectangles) {
        int selected = -1, total = 0;
        for(int i = 0; i < rectangles.length; i++) {
            int area = (rectangles[i].x2 - rectangles[i].x1) * (rectangles[i].y1 - rectangles[i].y2);
            if(rand.nextInt(total + area) >= total) {
                selected = i;
            }
            total += area;
        }
        return randomSelectFrom(rectangles[selected]);
    }
}

- aonecoding August 03, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

It does not matter. The start time and the end time of intervals are in equal position for this problem.
e.g.
[1, 5], [2, 10], [6, 9] is currently sorted by start. You may merge them from left to right and get the result [1,10].
If they were sorted by end time instead - [1, 5], [6, 9],[2, 10]. You could merge them from right to left and get the same result [1, 10].

It just depends on if you wanna start the merge from the left or right side of the given inteval list.
A sample code for when intervals sorted by start time

public List<Interval> merge(List<Interval> intervals) {
    if (intervals.size() <= 1)
        return intervals;
    
    // Sort by ascending starting point 
    intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start));
    
    List<Interval> result = new LinkedList<Interval>();
    int start = intervals.get(0).start;
    int end = intervals.get(0).end;
    
    for (Interval interval : intervals) {
        if (interval.start <= end) // Overlapping intervals, move the end if needed
            end = Math.max(end, interval.end);
        else {                     // Disjoint intervals, add the previous one and reset bounds
            result.add(new Interval(start, end));
            start = interval.start;
            end = interval.end;
        }
    }
    
    // Add the last interval
    result.add(new Interval(start, end));
    return result;
}

- aonecoding July 31, 2017 | Flag Reply

Comment hidden because of low score. Click to expand.

of 8 vote

//Sol1: O(n) time, O(1) extra memory solution, suitable if the function is rarely called on the data set
    public int random(int n, Set<Integer> ex) {
        int idx = new Random().nextInt(n - ex.size());

        for(int num = 0; num < n; num++) {
            if(!ex.contains(num)) {
                idx--;
            }
            if(idx == -1) {
                return num;
            }
        }
        return -1; //no number is available for selection (n is 0 or every number in range is excluded
    }

    //Sol2: O(n) extra memory and O(n) time at initialization
    //create a list of numbers in [0,n) excluding the numbers in excluded set
    //O(1) time on successive calls on the same data set

- aonecoding July 31, 2017 | Flag Reply

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