# Count of quadruplets with given Sum | Set 2

• Last Updated : 16 Jun, 2022

Given four arrays containing integer elements and an integer sum, the task is to count the quadruplets such that each element is chosen from a different array and the sum of all the four elements is equal to the given sum.
Examples:

Input: P[] = {0, 2}, Q[] = {-1, -2}, R[] = {2, 1}, S[] = {2, -1}, sum = 0
Output:
(0, -1, 2, -1) and (2, -2, 1, -1) are the required quadruplets.
Input: P[] = {1, -1, 2, 3, 4}, Q[] = {3, 2, 4}, R[] = {-2, -1, 2, 1}, S[] = {4, -1}, sum = 3
Output: 10

Approach: We pick any two arrays and calculate all possible sums and and keep their counts in a map. Using the remaining two arrays, we calculate all possible sums and check how many times their additive inverse exists in the map which will be the count of required quadruplets.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of the required quadruplets``int` `countQuadruplets(``int` `arr1[], ``int` `n1, ``int` `arr2[], ``int` `n2,``                     ``int` `arr3[], ``int` `n3, ``int` `arr4[], ``int` `n4, ``int` `value)``{``    ``int` `cnt = 0;``    ``unordered_map<``int``, ``int``> sum;` `    ``// All possible sums from arr1[] and arr2[]``    ``for` `(``int` `i = 0; i < n1; i++)``        ``for` `(``int` `j = 0; j < n2; j++)``            ``sum[arr1[i] + arr2[j]]++;` `    ``// Find the count of quadruplets``    ``for` `(``int` `i = 0; i < n3; i++)``        ``for` `(``int` `j = 0; j < n4; j++)``            ``cnt += sum[value - (arr3[i] + arr4[j])];` `    ``return` `cnt;``}` `// Driver code``int` `main()``{` `    ``int` `arr1[] = { 0, 2 };``    ``int` `arr2[] = { -1, -2 };``    ``int` `arr3[] = { 2, 1 };``    ``int` `arr4[] = { 2, -1 };``    ``int` `sum = 0;``    ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]);``    ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);``    ``int` `n3 = ``sizeof``(arr3) / ``sizeof``(arr3[0]);``    ``int` `n4 = ``sizeof``(arr4) / ``sizeof``(arr4[0]);` `    ``cout << countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, sum);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to return the count of the required quadruplets``    ``static` `int` `countQuadruplets(``int` `arr1[], ``int` `n1, ``int` `arr2[], ``int` `n2,``                                ``int` `arr3[], ``int` `n3, ``int` `arr4[], ``int` `n4, ``int` `value)``    ``{``        ``int` `cnt = ``0``;``        ``Map sum = ``new` `HashMap<>();` `        ``// All possible sums from arr1[] and arr2[]``        ``for` `(``int` `i = ``0``; i < n1; i++)``            ``for` `(``int` `j = ``0``; j < n2; j++) {``                ``if` `(sum.containsKey(arr1[i] + arr2[j])) {``                    ``sum.put(arr1[i] + arr2[j], sum.get(arr1[i] + arr2[j]) + ``1``);``                ``}``                ``else` `{``                    ``sum.put(arr1[i] + arr2[j], ``1``);``                ``}``            ``}` `        ``// Find the count of quadruplets``        ``for` `(``int` `i = ``0``; i < n3; i++)``            ``for` `(``int` `j = ``0``; j < n4; j++)``                ``if` `(sum.containsKey(value - (arr3[i] + arr4[j])))``                    ``cnt += sum.get(value - (arr3[i] + arr4[j]));` `        ``return` `cnt;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr1[] = { ``0``, ``2` `};``        ``int` `arr2[] = { -``1``, -``2` `};``        ``int` `arr3[] = { ``2``, ``1` `};``        ``int` `arr4[] = { ``2``, -``1` `};``        ``int` `sum = ``0``;``        ``int` `n1 = arr1.length;``        ``int` `n2 = arr2.length;``        ``int` `n3 = arr3.length;``        ``int` `n4 = arr4.length;` `        ``System.out.println(countQuadruplets(arr1, n1, arr2, n2,``                                            ``arr3, n3, arr4, n4, sum));``    ``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Python 3 implementation of the approach` `# Function to return the count``# of the required quadruplets``def` `countQuadruplets(arr1, n1, arr2, n2,``                     ``arr3, n3, arr4, n4, value):``    ``cnt ``=` `0``    ``sum` `=` `{i:``0` `for` `i ``in` `range``(``-``4``, ``10``, ``1``)}` `    ``# All possible sums from arr1[] and arr2[]``    ``for` `i ``in` `range``(n1):``        ``for` `j ``in` `range``(n2):``            ``sum``[arr1[i] ``+` `arr2[j]] ``+``=` `1` `    ``# Find the count of quadruplets``    ``for` `i ``in` `range``(n3):``        ``for` `j ``in` `range``(n4):``            ``cnt ``+``=` `sum``[value ``-` `(arr3[i] ``+` `arr4[j])]` `    ``return` `cnt` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr1 ``=` `[``0``, ``2``]``    ``arr2 ``=` `[``-``1``, ``-``2``]``    ``arr3 ``=` `[``2``, ``1``]``    ``arr4 ``=` `[``2``, ``-``1``]``    ``sum` `=` `0``    ``n1 ``=` `len``(arr1)``    ``n2 ``=` `len``(arr2)``    ``n3 ``=` `len``(arr3)``    ``n4 ``=` `len``(arr4)` `    ``print``(countQuadruplets(arr1, n1, arr2, n2,``                           ``arr3, n3, arr4, n4, ``sum``))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``// Function to return the count of the required quadruplets``    ``static` `int` `countQuadruplets(``int``[] arr1, ``int` `n1,``                                ``int``[] arr2, ``int` `n2,``                                ``int``[] arr3, ``int` `n3,``                                ``int``[] arr4, ``int` `n4, ``int` `value)``    ``{``        ``int` `cnt = 0;``        ``Dictionary<``int``, ``int``> sum = ``new` `Dictionary<``int``, ``int``>();` `        ``// All possible sums from arr1[] and arr2[]``        ``for` `(``int` `i = 0; i < n1; i++)``            ``for` `(``int` `j = 0; j < n2; j++) {``                ``if` `(sum.ContainsKey(arr1[i] + arr2[j])) {``                    ``var` `obj = sum[arr1[i] + arr2[j]] + 1;``                    ``sum.Remove(arr1[i] + arr2[j]);``                    ``sum.Add(arr1[i] + arr2[j], obj);``                ``}``                ``else` `{``                    ``sum.Add(arr1[i] + arr2[j], 1);``                ``}``            ``}` `        ``// Find the count of quadruplets``        ``for` `(``int` `i = 0; i < n3; i++)``            ``for` `(``int` `j = 0; j < n4; j++)``                ``if` `(sum.ContainsKey(value - (arr3[i] + arr4[j])))``                    ``cnt += sum[value - (arr3[i] + arr4[j])];` `        ``return` `cnt;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr1 = { 0, 2 };``        ``int``[] arr2 = { -1, -2 };``        ``int``[] arr3 = { 2, 1 };``        ``int``[] arr4 = { 2, -1 };``        ``int` `sum = 0;``        ``int` `n1 = arr1.Length;``        ``int` `n2 = arr2.Length;``        ``int` `n3 = arr3.Length;``        ``int` `n4 = arr4.Length;` `        ``Console.WriteLine(countQuadruplets(arr1, n1, arr2, n2,``                                           ``arr3, n3, arr4, n4, sum));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## PHP

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## Javascript

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Output:

`2`

Time Complexity : O(n1*n2+n3*n4) where, n1, n2, n3, n4 are size of arrays arr1, arr2, arr3 and arr4 respectively.

Auxiliary Space : O(n) , to store the elements in map.

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