Count pairs with given sum

Difficulty Level : Medium
Last Updated : 02 Aug, 2021

Given an array of integers, and a number ‘sum’, find the number of pairs of integers in the array whose sum is equal to ‘sum’.

Examples:

Input  :  arr[] = {1, 5, 7, -1}, 
          sum = 6
Output :  2
Pairs with sum 6 are (1, 5) and (7, -1)

Input  :  arr[] = {1, 5, 7, -1, 5}, 
          sum = 6
Output :  3
Pairs with sum 6 are (1, 5), (7, -1) &
                     (1, 5)         

Input  :  arr[] = {1, 1, 1, 1}, 
          sum = 2
Output :  6
There are 3! pairs with sum 2.

Input  :  arr[] = {10, 12, 10, 15, -1, 7, 6, 
                   5, 4, 2, 1, 1, 1}, 
          sum = 11
Output :  9

Expected time complexity O(n)

Naive Solution – A simple solution is to traverse each element and check if there’s another number in the array which can be added to it to give sum.

C++

// C++ implementation of simple method to find count of
// pairs with given sum.
#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, int sum)
{
    int count = 0; // Initialize result
 
    // Consider all possible pairs and check their sums
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (arr[i] + arr[j] == sum)
                count++;
 
    return count;
}
 
// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;
    cout << "Count of pairs is "
         << getPairsCount(arr, n, sum);
    return 0;
}

Java

// Java implementation of simple method to find count of
// pairs with given sum.
public class find {
    public static void main(String args[])
    {
        int[] arr = { 1, 5, 7, -1, 5 };
        int sum = 6;
        getPairsCount(arr, sum);
    }
 
    // Prints number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    public static void getPairsCount(int[] arr, int sum)
    {
 
        int count = 0; // Initialize result
 
        // Consider all possible pairs and check their sums
        for (int i = 0; i < arr.length; i++)
            for (int j = i + 1; j < arr.length; j++)
                if ((arr[i] + arr[j]) == sum)
                    count++;
 
        System.out.printf("Count of pairs is %d", count);
    }
}
// This program is contributed by Jyotsna

Python3

# Python3 implementation of simple method
# to find count of pairs with given sum.
 
# Returns number of pairs in arr[0..n-1]
# with sum equal to 'sum'
 
 
def getPairsCount(arr, n, sum):
 
    count = 0  # Initialize result
 
    # Consider all possible pairs
    # and check their sums
    for i in range(0, n):
        for j in range(i + 1, n):
            if arr[i] + arr[j] == sum:
                count += 1
 
    return count
 
 
# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
print("Count of pairs is",
      getPairsCount(arr, n, sum))
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# implementation of simple
// method to find count of
// pairs with given sum.
using System;
 
class GFG {
    public static void getPairsCount(int[] arr, int sum)
    {
 
        int count = 0; // Initialize result
 
        // Consider all possible pairs
        // and check their sums
        for (int i = 0; i < arr.Length; i++)
            for (int j = i + 1; j < arr.Length; j++)
                if ((arr[i] + arr[j]) == sum)
                    count++;
 
        Console.WriteLine("Count of pairs is " + count);
    }
 
    // Driver Code
    static public void Main()
    {
        int[] arr = { 1, 5, 7, -1, 5 };
        int sum = 6;
        getPairsCount(arr, sum);
    }
}
 
// This code is contributed
// by Sach_Code

PHP

<?php
// PHP implementation of simple
// method to find count of
// pairs with given sum.
 
// Returns number of pairs in
// arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount($arr, $n, $sum)
{
    // Initialize result
    $count = 0;
 
    // Consider all possible pairs
    // and check their sums
    for ($i = 0; $i < $n; $i++)
        for ($j = $i + 1; $j < $n; $j++)
            if ($arr[$i] + $arr[$j] == $sum)
                $count++;
 
    return $count;
}
 
    // Driver Code
    $arr = array(1, 5, 7, -1, 5) ;
    $n = sizeof($arr);
    $sum = 6;
    echo "Count of pairs is "
         , getPairsCount($arr, $n, $sum);
          
// This code is contributed by nitin mittal.
?>

Javascript

<script>
 
// Javascript implementation of simple method to find count of
// pairs with given sum.
 
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(arr, n, sum)
{
    let count = 0; // Initialize result
 
    // Consider all possible pairs and check their sums
    for (let i = 0; i < n; i++)
        for (let j = i + 1; j < n; j++)
            if (arr[i] + arr[j] == sum)
                count++;
 
    return count;
}
 
// Driver function to test the above function
    let arr = [ 1, 5, 7, -1, 5 ];
    let n = arr.length;
    let sum = 6;
    document.write("Count of pairs is "
        + getPairsCount(arr, n, sum));
     
// This code is contributed by Mayank Tyagi
 
</script>

Output

Count of pairs is 3

Time Complexity: O(n²)
Auxiliary Space: O(1)

Efficient solution –
A better solution is possible in O(n) time. Below is the Algorithm –

Create a map to store frequency of each number in the array. (Single traversal is required)
In the next traversal, for every element check if it can be combined with any other element (other than itself!) to give the desired sum. Increment the counter accordingly.
After completion of second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide count by 2 and return.

Below is the implementation of above idea :

C++

// C++ implementation of simple method to find count of
// pairs with given sum.
#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, int sum)
{
    unordered_map<int, int> m;
 
    // Store counts of all elements in map m
    for (int i = 0; i < n; i++)
        m[arr[i]]++;
 
    int twice_count = 0;
 
    // iterate through each element and increment the
    // count (Notice that every pair is counted twice)
    for (int i = 0; i < n; i++) {
        twice_count += m[sum - arr[i]];
 
        // if (arr[i], arr[i]) pair satisfies the condition,
        // then we need to ensure that the count is
        // decreased by one such that the (arr[i], arr[i])
        // pair is not considered
        if (sum - arr[i] == arr[i])
            twice_count--;
    }
 
    // return the half of twice_count
    return twice_count / 2;
}
 
// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;
    cout << "Count of pairs is "
         << getPairsCount(arr, n, sum);
    return 0;
}

Java

/* Java implementation of simple method to find count of
pairs with given sum*/
 
import java.util.HashMap;
 
class Test {
    static int arr[] = new int[] { 1, 5, 7, -1, 5 };
 
    // Returns number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    static int getPairsCount(int n, int sum)
    {
        HashMap<Integer, Integer> hm = new HashMap<>();
 
        // Store counts of all elements in map hm
        for (int i = 0; i < n; i++) {
 
            // initializing value to 0, if key not found
            if (!hm.containsKey(arr[i]))
                hm.put(arr[i], 0);
 
            hm.put(arr[i], hm.get(arr[i]) + 1);
        }
        int twice_count = 0;
 
        // iterate through each element and increment the
        // count (Notice that every pair is counted twice)
        for (int i = 0; i < n; i++) {
            if (hm.get(sum - arr[i]) != null)
                twice_count += hm.get(sum - arr[i]);
 
            // if (arr[i], arr[i]) pair satisfies the
            // condition, then we need to ensure that the
            // count is decreased by one such that the
            // (arr[i], arr[i]) pair is not considered
            if (sum - arr[i] == arr[i])
                twice_count--;
        }
 
        // return the half of twice_count
        return twice_count / 2;
    }
 
    // Driver method to test the above function
    public static void main(String[] args)
    {
 
        int sum = 6;
        System.out.println(
            "Count of pairs is "
            + getPairsCount(arr.length, sum));
    }
}
// This code is contributed by Gaurav Miglani

Python3

# Python 3 implementation of simple method
# to find count of pairs with given sum.
import sys
 
# Returns number of pairs in arr[0..n-1]
# with sum equal to 'sum'
 
 
def getPairsCount(arr, n, sum):
 
    m = [0] * 1000
 
    # Store counts of all elements in map m
    for i in range(0, n):
        m[arr[i]] += 1
 
    twice_count = 0
 
    # Iterate through each element and increment
    # the count (Notice that every pair is
    # counted twice)
    for i in range(0, n):
 
        twice_count += m[sum - arr[i]]
 
        # if (arr[i], arr[i]) pair satisfies the
        # condition, then we need to ensure that
        # the count is  decreased by one such
        # that the (arr[i], arr[i]) pair is not
        # considered
        if (sum - arr[i] == arr[i]):
            twice_count -= 1
 
    # return the half of twice_count
    return int(twice_count / 2)
 
 
# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
 
print("Count of pairs is", getPairsCount(arr,
                                         n, sum))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# implementation of simple method to
// find count of pairs with given sum
using System;
using System.Collections.Generic;
 
class GFG {
    public static int[] arr = new int[] { 1, 5, 7, -1, 5 };
 
    // Returns number of pairs in arr[0..n-1]
    // with sum equal to 'sum'
    public static int getPairsCount(int n, int sum)
    {
        Dictionary<int, int> hm
            = new Dictionary<int, int>();
 
        // Store counts of all elements
        // in map hm
        for (int i = 0; i < n; i++) {
 
            // initializing value to 0,
            // if key not found
            if (!hm.ContainsKey(arr[i])) {
                hm[arr[i]] = 0;
            }
 
            hm[arr[i]] = hm[arr[i]] + 1;
        }
        int twice_count = 0;
 
        // iterate through each element and
        // increment the count (Notice that
        // every pair is counted twice)
        for (int i = 0; i < n; i++) {
            if (hm[sum - arr[i]] != 0) {
                twice_count += hm[sum - arr[i]];
            }
 
            // if (arr[i], arr[i]) pair satisfies
            // the condition, then we need to ensure
            // that the count is decreased by one
            // such that the (arr[i], arr[i])
            // pair is not considered
            if (sum - arr[i] == arr[i]) {
                twice_count--;
            }
        }
 
        // return the half of twice_count
        return twice_count / 2;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int sum = 6;
        Console.WriteLine("Count of pairs is "
                          + getPairsCount(arr.Length, sum));
    }
}
 
// This code is contributed by Shrikant13

Output

Count of pairs is 3

This article is contributed by Ashutosh Kumar.

C++

// C++ implementation of simple method to find count of
// pairs with given sum.
#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, int k)
{
    unordered_map<int, int> m;
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (m.find(k - arr[i]) != m.end()) {
            count += m[k - arr[i]];
        }
        m[arr[i]]++;
    }
    return count;
}
 
// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;
    cout << "Count of pairs is "
         << getPairsCount(arr, n, sum);
    return 0;
}

Output

Count of pairs is 3

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