Snapdeal Interview Question for Tech Leads


Country: India
Interview Type: In-Person




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0
of 0 vote

int[] combos = new int[]{-1,1,1,-1};
            int sum = 0;
            int condition = 0;
            int numberofCombos = 0;

            //twos

            for(int a = 0; a < combos.Length; a++)
            {
                for(int b = a+1; b < combos.Length; b++)
                {
                    
                    sum = combos[a] + combos[b];
                    if (sum == condition)
                    {
                        Console.WriteLine("This equals " + condition + ":{a:" + combos[a] + ",b:" + combos[b] + "}");
                        numberofCombos = numberofCombos + 1;
                    }
                }
                
            }

            // threes

            for (int a = 0; a < combos.Length; a++)
            {
                for (int b = a + 1; b < combos.Length; b++)
                {

                    for (int c = b + 1; c < combos.Length; c++)
                    {
                        if (sum == condition)
                        {
                        sum = combos[a] + combos[b] + combos[c];
                        Console.WriteLine("This equals " + condition + ":{a:" + combos[a] + ",b:" + combos[b] + ",c:" + combos[c] + "}");
                        numberofCombos = numberofCombos + 1;
                        }
                    }
                    
                }

            }

            //four
            sum = combos[0] + combos[1] + combos[2] + combos[3];
            if (sum == condition)
            {
                Console.WriteLine("This equals " + condition + ":{a:" + combos[0] + ",b:" + combos[1] + ",c:" + combos[2] + ",c:" + combos[3] + "}");
                numberofCombos = numberofCombos + 1;
                      
            }

            Console.WriteLine("total combos = " + numberofCombos);
            Console.ReadLine();

- Snowcat February 06, 2017 | Flag Reply
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0
of 0 vote

int[] combos = new int[]{-1,1,1,-1};
            int sum = 0;
            int condition = 0;
            int numberofCombos = 0;

            //twos

            for(int a = 0; a < combos.Length; a++)
            {
                for(int b = a+1; b < combos.Length; b++)
                {
                    
                    sum = combos[a] + combos[b];
                    if (sum == condition)
                    {
                        Console.WriteLine("This equals " + condition + ":{a:" + combos[a] + ",b:" + combos[b] + "}");
                        numberofCombos = numberofCombos + 1;
                    }
                }
                
            }

            // threes

            for (int a = 0; a < combos.Length; a++)
            {
                for (int b = a + 1; b < combos.Length; b++)
                {

                    for (int c = b + 1; c < combos.Length; c++)
                    {
                        if (sum == condition)
                        {
                        sum = combos[a] + combos[b] + combos[c];
                        Console.WriteLine("This equals " + condition + ":{a:" + combos[a] + ",b:" + combos[b] + ",c:" + combos[c] + "}");
                        numberofCombos = numberofCombos + 1;
                        }
                    }
                    
                }

            }

            //four
            sum = combos[0] + combos[1] + combos[2] + combos[3];
            if (sum == condition)
            {
                Console.WriteLine("This equals " + condition + ":{a:" + combos[0] + ",b:" + combos[1] + ",c:" + combos[2] + ",c:" + combos[3] + "}");
                numberofCombos = numberofCombos + 1;
                      
            }

            Console.WriteLine("total combos = " + numberofCombos);
            Console.ReadLine();

- Snowcat Coding February 06, 2017 | Flag Reply
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0
of 2 vote

O(N^2) solution

package main

import "fmt"

func CountZero(m []int, sum int) int {
	res := 0
	for i := 0; i < len(m)-1; i++ {
		sum := 0
		for j := i + 1; j < len(m); j++ {
			sum += m[j]
			if sum == -m[i] {
				res++
			}
		}
	}
	return res
}

func main() {
	fmt.Println(CountZero([]int{-1, 1, -1, 1}, 0)) // 4
	fmt.Println(CountZero([]int{-1, -1, 1, 1}, 0)) // 2
	fmt.Println(CountZero([]int{-1, 1}, 0))        // 1
}

- dmitry.labutin February 06, 2017 | Flag Reply
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0
of 0 vote

O(n2) solution in Java:

public class CountZero {

	public static int countZero(int[] a) {
		int count = 0;

		for (int i = 0; i < a.length; i++) {
			int sum = a[i];
			for (int j = i + 1; j < a.length; j++) {
				sum += a[j];
				if (sum == 0) {
					count++;
				}
			}
		}
		return count;
	}

	public static void main(String[] args) {
		System.out.println(countZero(new int[] { -1, 1, -1, 1 })); // 4
		System.out.println(countZero(new int[] { 1, -1, -1, 1, -1, -1, 1 })); // 6
		System.out.println(countZero(new int[] { 1, 1, -1, -1 })); // 2
		System.out.println(countZero(new int[] { -1 })); // 0
		System.out.println(countZero(new int[] {})); // 0
		System.out.println(countZero(new int[] { -1, -1, -1, -1, -1, -1, -1 })); // 0
	}
}

- sivabnow February 06, 2017 | Flag Reply
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of 0 vote

#include <bits/stdc++.h>

using namespace std;


/// Expected Time O(n)
int solve(){

/// read the lenght of array
int n;
scanf("%d",&n);

/// read the elements
vector<int> elements(n);
for(int i = 0; i < n; ++i){
scanf("%d",&elements[i]);
}

unordered_map<int,int> frecuency; /// a hash table default c++11 , expected time 0(1)
for(int i = 0; i < n; ++i){
if(i > 0){
elements[i] += elements[i-1];
}
frecuency[elements[i]]++; /// update the frecuency
}

int goal = 0;
int ways = 0;
for(int i = 0; i < n; ++i){
ways += frecuency[goal];
goal = elements[i];
frecuency[elements[i]]--;
}
return ways;
}

int main(){


//give a array of just -1 or 1 , how many subarray there are , so that the sum = 0 ?
freopen("in.c","r",stdin);


cout << solve() << endl;

return 0;
}

- Elvis Capia February 06, 2017 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>

using namespace std;


/// Expected Time O(n)
int solve(){

/// read the lenght of array
int n;
scanf("%d",&n);

/// read the elements
vector<int> elements(n);
for(int i = 0; i < n; ++i){
scanf("%d",&elements[i]);
}

unordered_map<int,int> frecuency; /// a hash table default c++11 , expected time 0(1)
for(int i = 0; i < n; ++i){
if(i > 0){
elements[i] += elements[i-1];
}
frecuency[elements[i]]++; /// update the frecuency
}

int goal = 0;
int ways = 0;
for(int i = 0; i < n; ++i){
ways += frecuency[goal];
goal = elements[i];
frecuency[elements[i]]--;
}
return ways;
}

int main(){


//give a array of just -1 or 1 , how many subarray there are , so that the sum = 0 ?
freopen("in.c","r",stdin);


cout << solve() << endl;

return 0;
}

- Elvis Capia February 06, 2017 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>

using namespace std;


/// Expected Time O(n)
int solve(){

    /// read the lenght of array
    int n;
    scanf("%d",&n);

    /// read the elements
    vector<int> elements(n);
    for(int i = 0; i < n; ++i){
        scanf("%d",&elements[i]);
    }

    unordered_map<int,int> frecuency; /// a hash table default c++11 , expected time 0(1)
    for(int i = 0; i < n; ++i){
        if(i > 0){
        elements[i] += elements[i-1];
        }
        frecuency[elements[i]]++; /// update the frecuency
    }

    int goal = 0;
    int ways = 0;
    for(int i = 0; i < n; ++i){
        ways += frecuency[goal];
        goal = elements[i];
        frecuency[elements[i]]--;
    }
    return ways;
}

int main(){


    //give a array of just -1 or 1 , how many subarray there are , so that the sum = 0 ?
    freopen("in.c","r",stdin);


    cout << solve() << endl;

    return 0;
}

- Elvis Capia February 06, 2017 | Flag Reply
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0
of 0 vote

/* shows how declarative paradigm is powerful */
def count_zero_sum_sub_arrays( arr ){
  len = size(arr)
  // generate sum from combinations of [0,1,2,...len-1] taking 2 at a time 
  sum ( comb( [0:len] , 2 ) ) ->{
    // when the sum is 0 add 1, else 0 
    ( 0 == sum( [$.0:$.1+1] ) -> { arr[$.o] } ) ? 1 : 0 
  }
}
arr = [-1,1,-1,1]  
println( count_zero_sum_sub_arrays(arr) )

- NoOne February 07, 2017 | Flag Reply
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of 0 vote

We can solve this in linear time by traversing through the array and keeping the current sum of all the values. When we encounter a sum that we have already found previously that means that there is a subarray with sum 0.

For example, for input: [-1, 1, -1, 1]
As we traverse and keep the sum we get the following values:
0, -1, 0, -1, 0

three zeros indicate there are three subarrays that sum to zero. Two -1s indicate there is one subarray that sums to zero.

Java implementation:

public int countSubArraysSumZero(int[] nums) {
      Map<Integer, Integer> count = new HashMap<Integer, Integer>();
      count.put(0, 1);

      int sum = 0, res = 0;
      for(int i : nums) {
          sum += i;
          if(count.containsKey(sum)) {
              res += count.get(sum);
              count.put(sum, count.get(sum)+1);
          }else{
              count.put(sum, 1);
          }
      }

      return res;
  }

- Martin Copes February 08, 2017 | Flag Reply
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0
of 0 vote

def count_zero(a):
	n = len(a)
	d = {}
	temp_sum = 0

	for i in xrange(n):
		temp_sum = temp_sum + a[i]
		if temp_sum in d:
			d[temp_sum] = d[temp_sum] + 1
		else:
			d[temp_sum] = 1
			
	count = 0
	for i in d:
		count = count + (d[i]*(d[i]-1))/2

	if 0 in d:
		count = count + d[0]
	return count

a = map(int, raw_input().strip().split(' '))
print count_zero(a)

- itsvks February 08, 2017 | Flag Reply
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of 0 vote

Looks like it should be possible to do this in O(N) time. I just maintain a list of sums of all subarrays ending at each index.

def find_sums(arr):
	l = len(arr);
	if l == 1:
		res = [{}];
		res[0][arr[0]] = 1;
		return res;
	temp = {};
	prev_res = find_sums(arr[1:]);
	if arr[0] == -1:
		last_res = prev_res[-1];
		for key in last_res.keys():
			temp[key-1] = last_res[key];
		temp[-1] = 1 if -1 not in temp.keys() else (temp[-1] + 1);
	elif arr[0] == 1:
		last_res = prev_res[-1];
		for key in last_res.keys():
			temp[key+1] = last_res[key];
		temp[1] = 1 if 1 not in temp.keys() else (temp[1] + 1);
	
	prev_res.append(temp);
	return prev_res;


print find_sums([-1,1,1,-1,-1,1,1,1,-1,-1,1,-1,1]);

Then we just take a sum of 0 indexed elements at each index.

- Aayush February 18, 2017 | Flag Reply


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