Open In App

Count number of distinct pairs whose sum exists in the given array

Last Updated : 09 Feb, 2023
Comments
Improve
Suggest changes
Like Article
Like
Report

Given an array of N positive integers. Count the number of pairs whose sum exists in the given array. While repeating pairs will not be counted again. And we can't make a pair using same position element. Eg : (2, 1) and (1, 2) will be considered as only one pair. Please read all examples carefully. 

Examples:

Input : arr[] = {1, 2, 3, 5, 10}
Output : 2
Explanation : Here there are two such pairs:
(1 + 2) = 3, (2 + 3) = 5.
Note : Here we can't take pair (5, 5) as 
we can see 5 is not coming twice

Input : arr[] = {1, 5, 6, 4, -1, 5} 
Output : 4
Explanation : (1 + 5) = 6, (1 + 4) = 5, 
(5 + -1) = 4, (6 + -1) = 5
Note : Here (1, 5) comes twice will be 
considered as only one pair. 

Input : arr[] = {5, 5, 5, 5, 10} 
Output : 1
Explanation : (5 + 5) = 10
Note : Here (5, 5) comes twice will be
considered as only one pair. 

The idea is to map of pairs to find unique elements. We first store elements and their counts in a map. Then we traverse array elements, for every pair of elements (arr[i], arr[j]), we check if (arr[i] + arr[j]) exists in array. If exists, then we check if it is already counted using map of pairs. If not already counted, then we increment count. 

Implementation:

C++
// C++ implementation to find count of unique pairs
// whose sum exists in given array
#include <bits/stdc++.h>
using namespace std;

// Returns number of pairs in arr[0..n-1] with
// sum equal to 'sum'
int getPairsCount(int arr[], int n)
{
    // Store counts of all elements in map m
    // to find pair (arr[i], sum-arr[i])
    // because (arr[i]) + (sum - arr[i]) = sum
    map<int, int> m;
    for (int i = 0; i < n; i++)
        m[arr[i]]++;

    // To remove duplicate items we use result map
    map<pair<int, int>, int> pairs;

    int count = 0; // Initialize result

    // Consider all pairs
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            // If sum of current pair exists
            if (m[arr[i] + arr[j]] > 0 && 
                pairs[{ arr[i], arr[j] }] == 0) {
                count++;
            }

            // Insert current pair both ways to avoid
            // duplicates.
            pairs[{ arr[i], arr[j] }]++;
            pairs[{ arr[j], arr[i] }]++;
        }
    }
    return count;
}

// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 6, 4, -1, 5, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << getPairsCount(arr, n);
    return 0;
}
Java Python3 C# JavaScript

Output
6

Time complexity: O(n^2)
Space complexity: O(n)


Article Tags :
Practice Tags :

Similar Reads