Zero Pair Sum

PROBLEM STATEMENT

You are given an array ‘arr’ of ‘N’ integers. Your task is to find the count of pairs with a sum equal to zero.

Specifically, find the count of all pairs ( i , j ) such that i < j and arr[i] + arr[j] = 0

Input Format :

The first line contains a single integer ‘T’ denoting the number of test cases, then each test case follows

The first line of each testcase contains a single integers ‘N’ denoting the length of the array.

The second line of each testcase contains ‘N’ integers denoting the array elements.

Output Format :

For each test case print a single integer denoting the count of pairs with zero-sum.

Output for each test case will be printed in a separate line.

Note :

You are not required to print anything; it has already been taken care of. Just implement the function.

Constraints :

1 <= T <= 10      
1 <= N <= 10^4
10^-9 <= arr[i] <= 10^9

Time limit: 1 sec

Sample Input 1 :

Sample Output 1 :

2
0

Explanation Of Sample Output 1 :

For test case 1 :
Two pairs exist which have their sums equal to zero:
Pair formed by the first two elements has sum equal to zero, ie: 1 + (-1) = 0
Also, pair formed by the last two elements has sum qual to zero, ie: 2 + (-2) = 0 

For test case 2 :
No pair can be formed with sum equal to zero.

Sample Input 2 :

Sample Output 2 :

0
0

Approach 1

We can generate all pairs possible and check whether their sum is equal to zero.

To generate all the pairs: for each x in the range [0, N-2] iterate through all y in the range [x+1, N-1].

The steps are as follows :

Initialize count to 0.
Run outer for loop for x from 0 to N-2
Run inner for loop for y from x+1 to N-1
For each generated pair of (x,y) increment the count if the arr[x]+arr[y] =0
Return the final value of count.

Try Problem

Approach 2

Create a hash-map to store the count of the frequency of each array element. Now simply iterate on the created hash-map, if the current key is equal to x then search if -x exists in the hash-map or not, if it exists then add the frequency of x multiplied by the frequency of -x to the count. Notice that we are double counting things here, so remember to return count/2.

Remember to separately handle the border case when array element is equal to 0.

The steps are as follows :

Initialize count to 0.
Create a hash-map to store the frequency of each array element.
Iterate the hash-map, if the current key is equal to curKey and (curKey !=0), search if -curKey exists.
Multiply frequency of curKey and frequency of -curKey, and add it to count
If the frequency of elements equal to 0 is f, add f * (f-1) to count
Return the final value of count / 2