monica_shankar
BAN USERThe images are SHA'ed and their long ids are obtained, each user has a unique user id which is stored in dictionary order. Each user has a hash table of 10 recent image requests <Integer, Long ids>. If excess, the least recent one is deleted. This is just image retrieval hence I don't think synchronization is necessary. Please correct me if I am wrong.
 monica_shankar June 29, 2015You don't even need a stack in the first place.
 monica_shankar June 08, 2015/*
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* To change this template file, choose Tools  Templates
* and open the template in the editor.
*/
package Careercup;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.lang.IllegalArgumentException;
/**
*
* @author
*/
public class CountBracketBalance {
static BufferedReader br;
static StringTokenizer st;
static int head=0,N=0,count=0;
static ArrayList<Character> queue=new ArrayList<Character>();
public static void init() throws IOException
{
if(st==null)
st=new StringTokenizer(br.readLine().replace("", " "));
}
public static void push(Character ch)
{
if(N>=queue.size()) //cases where N==arraylist.size
{
queue.ensureCapacity(queue.size()*2);
for(int i=N+1;i<queue.size();i++)
queue.add(null);
}
//stack.set(N++, ch);
queue.add(N++, ch);
}
public static Character pop()
{
if(queue.isEmpty())
{ System.err.println("Underflow operation requested, exiting");
System.exit(0); }
Character temp=queue.get(head);
queue.set(head, null);
head++;
N;
return temp;
}
public static void main(String args[]) throws IOException
{
int maxcount=0;
br=new BufferedReader(new InputStreamReader(System.in));
init();
while(st.hasMoreTokens())
{
//push(st.nextToken().charAt(0));
String temp=st.nextToken();
push(temp.charAt(0));
}
while(N>0)
{
Character ch=pop();
if(ch=='(')
{ count++;
maxcount=Math.max(maxcount, count);}
if(ch==')')
count;
if(count<0)
throw new IllegalArgumentException();
}
if(count!=0)
throw new IllegalArgumentException();
System.out.println(maxcount);
}
}

monica_shankar
June 08, 2015 guilhebi, correct, it is difficult to imagine separate chaining for searching and nextfavsearch.
 monica_shankar June 01, 2015I think a bigger concern should be the order of insertion as an input with ascending order as in reality could make all operations ( search , insert , delete) O(N),, hence better to go for red black trees
 monica_shankar June 01, 2015I have come up with BST approach
it would be a binary search tree indexed by time t
1. Insert operation: same as the standard put operation. Avg case O(lg N) worst case O(N) ,, memory O(1)
2. Search operation: this is like a ceiling operation in BST where we find the node that is immediately greater than the current key value
Avg case: O(lg N) , worst case : O(N) memory: O(1)
public Node search(int t)
{
return search(root,t); // Assume that we have a pointer to the root.
}
private Node search(Node x, int t)
{
if (x==null) return null;
else if(compareTo(x.t,t)<0) // x.t less than t , focus on right
return search(x.right,t);
else if(compareTo(x.t,t)>=0)
{
Node t=min(x.left); // check to ensure that the we don't have an element which is smaller than the current element x
if(t!=null)
return t;
else
return x;
}
}
3. NextFavourite(t): similar operation as above but replace min(x.left) with minfav(x.left)
avg case : O(lg N) ,, worst case O(N) , memory O(1)
minfav(Node x)
{
Node y;
if(x==null) return null;
y=minfav(x.left);
if(y==null)
{
if(x.fav==1);
return x
else
y=minfav(x.right);
}
return y;
}

monica_shankar
June 01, 2015 /*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools  Templates
* and open the template in the editor.
*/
package Dummy;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
import java.util.StringTokenizer;
/**
*
* @author
*/
public class Dummy02 {
static Stack<int[]> accesspoints=new Stack<int[]>();
public static boolean findPath(int[][] points, int[] startpt, int[] endpt)
{
int[][] pointsvisited=new int[points.length][points[0].length];
accesspoints.add(startpt);
visitpoint(points,pointsvisited,startpt);
for(int[] coord:accesspoints)
{
if((coord[0]==endpt[0])&&(coord[1]==endpt[1]))
return true;
}
return false;
}
public static void visitpoint(int[][] points, int[][] pointsvisited , int [] currentpoint)
{
if(currentpoint[0]<0currentpoint[0]>=points.lengthcurrentpoint[1]<0currentpoint[1]>=points[0].length
pointsvisited[currentpoint[0]][currentpoint[1]]==1points[currentpoint[0]][currentpoint[1]]==1)
{
return ;
}
pointsvisited[currentpoint[0]][currentpoint[1]]=1;
accesspoints.add(currentpoint);
visitpoint(points,pointsvisited,new int[]{currentpoint[0]+1,currentpoint[1]});
visitpoint(points,pointsvisited,new int[]{currentpoint[0]1,currentpoint[1]});
visitpoint(points,pointsvisited,new int[]{currentpoint[0],currentpoint[1]+1});
visitpoint(points,pointsvisited,new int[]{currentpoint[0],currentpoint[1]1});
}
static BufferedReader br;
static StringTokenizer st;
public static void main(String args[]) throws IOException
{
br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the dimensions of the value matrix");
int length=nextInt();
int breadth=nextInt();
int[][] points=new int[length][breadth];
System.out.println("Enter the matrix");
for(int i=0;i<length;i++)
{
for(int j=0;j<breadth;j++)
{
points[i][j]=nextInt();
}
}
int[] startpt=new int[2];
int[] endpt=new int[2];
System.out.println("Enter the starting point");
for(int i=0;i<2;i++)
{
startpt[i]=nextInt();
}
System.out.println("Enter the coordinates of end point");
for(int i=0;i<2;i++)
{
endpt[i]=nextInt();
}
boolean value=findPath(points,startpt,endpt);
if(value)
System.out.println("There exist a path");
else
System.out.println("There exist no path");
}
public static int nextInt() throws IOException
{
return Integer.parseInt(nextToken());
}
public static String nextToken() throws IOException
{
while(st==null!st.hasMoreTokens())
{
String line=br.readLine();
if(line==null)
return line;
st=new StringTokenizer(line.replaceAll("", " "));
}
return st.nextToken();
}
}
in input just flip for x and y
 monica_shankar May 29, 2015In a hurry, here is just the pseudocode
pseudocode
=========
1. let (a,b) be the first zero element in 2d matrix of values Value
a. If (a,b)==null, return false
b. else percolation(a,b)
//Assume that the Value matrix is 4*5 matrix as given here
ahorizontal index, b vertical index
percolation(int a, int b)
{
if(a!=3)
{
find list L=[(xa,y1),(xa,y2)....(xa,yN)] where (xa,yi)=(x(a+1)),yi)=0 // top zero, the corresponding bottom is also zero
if L==null, return false
prev a=a, prevb=b
for each coord in L:
{
if adjacent(coord,(a,b))
{
a=coord(x)+1
b=coord(y)
percolation(a,b)
}
}
if preva==a && prevb==b return false
}
else return true
}

monica_shankar
May 28, 2015 while(A[L]<A[R]) is wrong, it should be while(L<=R)
 monica_shankar May 22, 2015Assume that we are going to declare N as the maximum value of the ball we see during the pick up times.
so we pick up k times and let the value of balls during the pick up be m1,m2,m3....mk
Without replacement:
The probability of drawing the N'ed value ball at first pick is 1/N
The probability of drawing the N'ed value ball at second pick is 1/(N1)
The probability of drawing the N'ed value ball at third pick is 1/(N2)
.... The probability of drawing the N'ed value ball at Kth pick is 1/(NK)
As probability of one try does not affect the other, we use addition priciple
so with the value 1/N+1/(N1).............1/(NK) we can say that the maximum value of ball we pick would be the value N
With replacement:
The probability of picking the N'ed value ball at first try=1/N
The probability of picking the N'ed value ball at second try=1/N
.... The probability of picking the N'ed value ball at third try=1/N
so with probability of k/N, we can state that the maximum ball we pick would be the value N.
correct me if i am wrong somewhere, thanks!!!
@nikolay, explain your logic...
 monica_shankar May 21, 2015Nice one but
int j=input.indexOf(c,i+1);
returns the first substring occurrence
so for example "Today is a good day to sing and toddle" wouldn't work because 'T' of "Today" looks at 't' of "to" and not 't' of "toddle"
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how do you store users and their 10 recent requests in an efficient manner, also isn't requests for each individual user in which case you don't need monitors
 monica_shankar June 29, 2015