3 sum problem

The problem is defined as given a array find a set of elements such that c=a+b.

Approach
1. Sort the array
2. for each element c=arr[i] in the array
3. left=0; right=i;
4. while left<right
5. check if c==arr[left]+arr[right]
6. then add c to output
7. else if c<arr[left]+arr[right] right--
8. else left++
9. end
10.end

#include <iostream>

using namespace std;

int compare (const void * a, const void * b);
void sum3(int arr[], int n);
void main(){
 int arr[]={2,3,5,1,4,9};
 
 qsort(arr,6,sizeof(int),compare);

 sum3(arr,6);

}
void sum3(int arr[], int n){
 int l,r;
 int count=0; 
 int index[6];
 int curr=0;
 for(int i=n-1;i>=0;i--)
 { 
  l=0; r=i;
  while(l<r){
   curr=arr[l] +arr[r];  
   if(arr[i] == curr){
    index[count++]=i; 
    cout<<arr[i]<<" ";
    break;
   }
   else if(arr[i] < curr)
    r--;
   else l++;
  }
 }
}
int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

Complexity of the above procedure is O(n^2). Apparently a better solution for the problem doesn't exist. Also, a number of problems can be reduced to the 3 sum problem and are called 3sum hard.