Skip to content
Related Articles

Related Articles

Count pairs with given sum
  • Difficulty Level : Medium
  • Last Updated : 19 Mar, 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 – 
  1. Create a map to store frequency of each number in the array. (Single traversal is required)
  2. 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.
  3. 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. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes
arrow_drop_up
Next
last_page
Given an array A[] and a number x, check for pair in A[] with sum as x
Recommended Articles
Page :
Count of pairs {X, Y} from an array such that sum of count of set bits in X ⊕ Y and twice the count of set bits in X & Y is M
12, Mar 21
Maximize count of pairs whose bitwise XOR is even by replacing such pairs with their Bitwise XOR
22, Feb 21
Maximize count of pairs whose Bitwise AND exceeds Bitwise XOR by replacing such pairs with their Bitwise AND
22, Feb 21
Given an array of pairs, find all symmetric pairs in it
17, Sep 15
Count of unique pairs (i, j) in an array such that sum of A[i] and reverse of A[j] is equal to sum of reverse of A[i] and A[j]
12, Feb 21
Count number of distinct pairs whose sum exists in the given array
06, Sep 17
Maximum Count of pairs having equal Sum based on the given conditions
18, Aug 20
Count of replacements required to make the sum of all Pairs of given type from the Array equal
21, Aug 20
Count pairs from two linked lists whose sum is equal to a given value
03, Jul 17
Count pairs from two sorted matrices with given sum
29, Jan 18
Given two arrays count all pairs whose sum is an odd number
26, Apr 19
Count pairs from two sorted arrays whose sum is equal to a given value x
23, Jun 17
Count pairs with given sum | Set 2
12, Jul 19
Count distinct pairs with given sum
11, Dec 20
Sum of Bitwise AND of sum of pairs and their Bitwise AND from a given array
23, Feb 21
Sum of absolute differences of pairs from the given array that satisfy the given condition
06, Sep 19
Find the maximum cost of an array of pairs choosing at most K pairs
23, Apr 19
Count pairs from an array with even product of count of distinct prime factors
22, Feb 21
Count pairs in a sorted array whose sum is less than x
31, Mar 17
Count pairs in array whose sum is divisible by 4
08, Dec 17
Count number of pairs in array having sum divisible by K | SET 2
14, Aug 19
Count distinct pairs from two arrays having same sum of digits
23, Nov 18
Count of pairs in an array whose sum is a perfect square
27, Nov 18
Maximum count of pairs which generate the same sum
13, Mar 19
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox