Intuitive solution for Pair sum to zero in language tools and time complexity module

[Nikhil-Patro]

/* Given a random integer array A of size N. Find and print the pair of elements in the array which sum to 0.
Array A can contain duplicate elements.
While printing a pair, print the smaller element first.
That is, if a valid pair is (6, -6) print "-6 6". There is no constraint that out of 5 pairs which have to be printed in 1st line. You can print pairs in any order, just be careful about the order of elements in a pair.
Input format :
Line 1 : Integer N (Array size)
Line 2 : Array elements (separated by space)
Output format :
Line 1 : Pair 1 elements (separated by space)
Line 2 : Pair 2 elements (separated by space)
Line 3 : and so on
Constraints :
0 <= N <= 10^4
Sample Input:
5
2 1 -2 2 3
Sample Output :
-2 2
-2 2 */

#include<bits/stdc++.h>
using namespace std;
int pairSum(int arr, int n) {
int count=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
m[arr[i]]+=1;
}
for(int i=0;i<n;i++){
if(arr[i]==0) continue; // we will handle 0 seperately
int pos = m[arr[i]];
int neg = m[-arr[i]];
if(pos >0 && neg >0){
count = count + (posneg);
m[arr[i]]=0;
m[-arr[i]]=0;
}
}
//handle 0 elements
int zeroes = m[0];
count = count + (zeroes*(zeroes-1))/2;
return count;
}

[parikshit223933]

If you follow Contribution guidelines It will be easier for me to have a look at your code. I appreciate your effort though.