# Count pairs with given sum | Set 2

• Difficulty Level : Medium
• Last Updated : 19 Mar, 2021

Given an array arr[] and an integer sum, the task is to 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:
Pairs with sum 6 are (1, 5) and (7, -1)
Input: arr[] = {1, 5, 7, -1, 5}, sum = 6
Output:
Pairs with sum 6 are (1, 5), (7, -1) & (1, 5)
Input: arr[] = {1, 1, 1, 1}, sum = 2
Output:

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Approach: Two Different methods have already been discussed here. Here, a method based on sorting will be discussed.

1. Sort the array and take two pointers i and j, one pointer pointing to the start of the array i.e. i = 0 and another pointer pointing to the end of the array i.e. j = n – 1.
• Greater than the sum then decrement j.
• Lesser than the sum then increment i.
• Equals to the sum then count such pairs.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of pairs``// from arr[] with the given sum``int` `pairs_count(``int` `arr[], ``int` `n, ``int` `sum)``{``    ``// To store the count of pairs``    ``int` `ans = 0;` `    ``// Sort the given array``    ``sort(arr, arr + n);` `    ``// Take two pointers``    ``int` `i = 0, j = n - 1;` `    ``while` `(i < j) {``        ``// If sum is greater``        ``if` `(arr[i] + arr[j] < sum)``            ``i++;` `        ``// If sum is lesser``        ``else` `if` `(arr[i] + arr[j] > sum)``            ``j--;` `        ``// If sum is equal``        ``else` `{``            ``// Find the frequency of arr[i]``            ``int` `x = arr[i], xx = i;``            ``while` `(i < j and arr[i] == x)``                ``i++;` `            ``// Find the frequency of arr[j]``            ``int` `y = arr[j], yy = j;``            ``while` `(j >= i and arr[j] == y)``                ``j--;` `            ``// If arr[i] and arr[j] are same``            ``// then remove the extra number counted``            ``if` `(x == y) {``                ``int` `temp = i - xx + yy - j - 1;``                ``ans += (temp * (temp + 1)) / 2;``            ``}``            ``else``                ``ans += (i - xx) * (yy - j);``        ``}``    ``}` `    ``// Return the required answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 5, 7, 5, -1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `sum = 6;` `    ``cout << pairs_count(arr, n, sum);` `    ``return` `0;``}`

## Java

 `//Java implementation of the approach``import` `java.util.Arrays;``import` `java.io.*;` `class` `GFG``{``    ` `// Function to return the count of pairs``// from arr[] with the given sum``static` `int` `pairs_count(``int` `arr[], ``int` `n, ``int` `sum)``{``    ``// To store the count of pairs``    ``int` `ans = ``0``;` `    ``// Sort the given array``    ``Arrays.sort(arr);` `    ``// Take two pointers``    ``int` `i = ``0``, j = n - ``1``;` `    ``while` `(i < j)``    ``{``        ` `        ``// If sum is greater``        ``if` `(arr[i] + arr[j] < sum)``            ``i++;` `        ``// If sum is lesser``        ``else` `if` `(arr[i] + arr[j] > sum)``            ``j--;` `        ``// If sum is equal``        ``else``        ``{``            ` `            ``// Find the frequency of arr[i]``            ``int` `x = arr[i], xx = i;``            ``while` `((i < j ) && (arr[i] == x))``                ``i++;` `            ``// Find the frequency of arr[j]``            ``int` `y = arr[j], yy = j;``            ``while` `((j >= i )&& (arr[j] == y))``                ``j--;` `            ``// If arr[i] and arr[j] are same``            ``// then remove the extra number counted``            ``if` `(x == y)``            ``{``                ``int` `temp = i - xx + yy - j - ``1``;``                ``ans += (temp * (temp + ``1``)) / ``2``;``            ``}``            ``else``                ``ans += (i - xx) * (yy - j);``        ``}``    ``}` `    ``// Return the required answer``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `arr[] = { ``1``, ``5``, ``7``, ``5``, -``1` `};``    ``int` `n = arr.length;``    ``int` `sum = ``6``;``    ` `    ``System.out.println (pairs_count(arr, n, sum));``}``}` `// The code is contributed by ajit..`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of pairs``# from arr with the given sum``def` `pairs_count(arr, n, ``sum``):``    ` `    ``# To store the count of pairs``    ``ans ``=` `0` `    ``# Sort the given array``    ``arr ``=` `sorted``(arr)` `    ``# Take two pointers``    ``i, j ``=` `0``, n ``-` `1` `    ``while` `(i < j):``        ` `        ``# If sum is greater``        ``if` `(arr[i] ``+` `arr[j] < ``sum``):``            ``i ``+``=` `1` `        ``# If sum is lesser``        ``elif` `(arr[i] ``+` `arr[j] > ``sum``):``            ``j ``-``=` `1``            ` `        ``# If sum is equal``        ``else``:``            ` `            ``# Find the frequency of arr[i]``            ``x ``=` `arr[i]``            ``xx ``=` `i``            ``while` `(i < j ``and` `arr[i] ``=``=` `x):``                ``i ``+``=` `1` `            ``# Find the frequency of arr[j]``            ``y ``=` `arr[j]``            ``yy ``=` `j``            ``while` `(j >``=` `i ``and` `arr[j] ``=``=` `y):``                ``j ``-``=` `1` `            ``# If arr[i] and arr[j] are same``            ``# then remove the extra number counted``            ``if` `(x ``=``=` `y):``                ``temp ``=` `i ``-` `xx ``+` `yy ``-` `j ``-` `1``                ``ans ``+``=` `(temp ``*` `(temp ``+` `1``)) ``/``/` `2``            ``else``:``                ``ans ``+``=` `(i ``-` `xx) ``*` `(yy ``-` `j)` `    ``# Return the required answer``    ``return` `ans` `# Driver code``arr ``=` `[``1``, ``5``, ``7``, ``5``, ``-``1``]``n ``=` `len``(arr)``sum` `=` `6` `print``(pairs_count(arr, n, ``sum``))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{``    ` `// Function to return the count of pairs``// from arr[] with the given sum``static` `int` `pairs_count(``int` `[]arr,``                       ``int` `n, ``int` `sum)``{``    ``// To store the count of pairs``    ``int` `ans = 0;` `    ``// Sort the given array``    ``Array.Sort(arr);` `    ``// Take two pointers``    ``int` `i = 0, j = n - 1;` `    ``while` `(i < j)``    ``{``        ` `        ``// If sum is greater``        ``if` `(arr[i] + arr[j] < sum)``            ``i++;` `        ``// If sum is lesser``        ``else` `if` `(arr[i] + arr[j] > sum)``            ``j--;` `        ``// If sum is equal``        ``else``        ``{``            ` `            ``// Find the frequency of arr[i]``            ``int` `x = arr[i], xx = i;``            ``while` `((i < j) && (arr[i] == x))``                ``i++;` `            ``// Find the frequency of arr[j]``            ``int` `y = arr[j], yy = j;``            ``while` `((j >= i) && (arr[j] == y))``                ``j--;` `            ``// If arr[i] and arr[j] are same``            ``// then remove the extra number counted``            ``if` `(x == y)``            ``{``                ``int` `temp = i - xx + yy - j - 1;``                ``ans += (temp * (temp + 1)) / 2;``            ``}``            ``else``                ``ans += (i - xx) * (yy - j);``        ``}``    ``}` `    ``// Return the required answer``    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main (String[] args)``{``    ``int` `[]arr = { 1, 5, 7, 5, -1 };``    ``int` `n = arr.Length;``    ``int` `sum = 6;``    ` `    ``Console.WriteLine (pairs_count(arr, n, sum));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`3`

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