# Count of quadruplets with given Sum

• Difficulty Level : Basic
• Last Updated : 23 Nov, 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: Generate all possible quadruplets and calculate the sum of every quadruplet. Count all such quadruplets whose sum is equal to the given sum.

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` `sum)``{` `    ``// To store the count of required quadruplets``    ``int` `cnt = 0;` `    ``// For arr1[]``    ``for` `(``int` `i = 0; i < n1; i++) {` `        ``// For arr2[]``        ``for` `(``int` `j = 0; j < n2; j++) {` `            ``// For arr3[]``            ``for` `(``int` `k = 0; k < n3; k++) {` `                ``// For arr4[]``                ``for` `(``int` `l = 0; l < n4; l++) {` `                    ``// If current quadruplet has the required sum``                    ``if` `(arr1[i] + arr2[j] + arr3[k] + arr4[l] == sum) {``                        ``cnt++;``                    ``}``                ``}``            ``}``        ``}``    ``}` `    ``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 program to implement``// the above approach` `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` `sum)``{` `    ``// To store the count of required quadruplets``    ``int` `cnt = ``0``;` `    ``// For arr1[]``    ``for` `(``int` `i = ``0``; i < n1; i++)``    ``{` `        ``// For arr2[]``        ``for` `(``int` `j = ``0``; j < n2; j++)``        ``{` `            ``// For arr3[]``            ``for` `(``int` `k = ``0``; k < n3; k++)``            ``{` `                ``// For arr4[]``                ``for` `(``int` `l = ``0``; l < n4; l++)``                ``{` `                    ``// If current quadruplet has the required sum``                    ``if` `(arr1[i] + arr2[j] + arr3[k] + arr4[l] == sum)``                    ``{``                        ``cnt++;``                    ``}``                ``}``            ``}``        ``}``    ``}` `    ``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 implementation of the approach` `# Function to return the count of the required quadruplets``def` `countQuadruplets(P, Q, R, S, ``sum``):``    ` `    ``# To store the count of required quadruplets``    ``cnt ``=` `0``    ` `    ``# Using four loops generate all possible quadruplets``    ``for` `elem1 ``in` `P:``        ``for` `elem2 ``in` `Q:``            ``for` `elem3 ``in` `R:``                ``for` `elem4 ``in` `S:``                    ``if` `elem1 ``+` `elem2 ``+` `elem3 ``+` `elem4 ``=``=` `sum``:``                        ``cnt ``=` `cnt ``+` `1``    ``return` `cnt` `# Driver code``P ``=` `[ ``0``, ``2``]``Q ``=` `[``-``1``, ``-``2``]``R ``=` `[``2``, ``1``]``S ``=` `[ ``2``, ``-``1``]``sum` `=` `0` `print``(countQuadruplets(P, Q, R, S, ``sum``))`

## C#

 `// C# program to implement``// the above approach``using` `System;` `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` `sum)``{` `    ``// To store the count of required quadruplets``    ``int` `cnt = 0;` `    ``// For arr1[]``    ``for` `(``int` `i = 0; i < n1; i++)``    ``{` `        ``// For arr2[]``        ``for` `(``int` `j = 0; j < n2; j++)``        ``{` `            ``// For arr3[]``            ``for` `(``int` `k = 0; k < n3; k++)``            ``{` `                ``// For arr4[]``                ``for` `(``int` `l = 0; l < n4; l++)``                ``{` `                    ``// If current quadruplet has the required sum``                    ``if` `(arr1[i] + arr2[j] + arr3[k] + arr4[l] == sum)``                    ``{``                        ``cnt++;``                    ``}``                ``}``            ``}``        ``}``    ``}` `    ``return` `cnt;``}` `// Driver code``static` `public` `void` `Main ()``{``    ` `    ``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 akt_mit`

## PHP

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## Javascript

 ``

Output:

`2`

Time Complexity: O(n4
Space Complexity: O(1)

Efficient Approach: Store frequency of all possible sum of two elements from two different arrays in a map. Iterate over other two arrays and find the sum of any two elements in these two arrays,lets it be cur_sum. If sum – cur_sum is present in the map, this means that there exists four elements in four different arrays whose sum is equal to sum.

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` `sum)``{` `    ``// To store the count of required quadruplets``    ``int` `cnt = 0;``      ``// To store the frequency of sum of``    ``// two elements of two different arrays``    ``unordered_map<``int``,``int``> freq;``    ``// For arr1[]``    ``for` `(``int` `i = 0; i < n1; i++) {` `        ``// For arr2[]``        ``for` `(``int` `j = 0; j < n2; j++) {` `            ``freq[arr1[i]+arr2[j]]++;``        ``}``    ``}``    ` `      ``// for arr3[]``   ``for` `(``int` `i = 0; i < n3; i++) {` `        ``// For arr4[]``        ``for` `(``int` `j = 0; j < n4; j++) {``            ``int` `cur_sum = arr3[i]+arr4[j];``            ``cnt+=freq[sum-cur_sum];``        ``}``    ``}``    ``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

 `/*package whatever //do not write package name here */``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` `sum)``  ``{` `    ``// To store the count of required quadruplets``    ``int` `cnt = ``0``;``    ` `    ``// To store the frequency of sum of``    ``// two elements of two different arrays``    ``HashMap freq = ``new` `HashMap<>();``    ` `    ``// For arr1[]``    ``for` `(``int` `i = ``0``; i < n1; i++) {` `      ``// For arr2[]``      ``for` `(``int` `j = ``0``; j < n2; j++) {``        ``freq.put(arr1[i] + arr2[j],freq.getOrDefault(arr1[i] + arr2[j], ``0``)+ ``1``);``      ``}``    ``}` `    ``// for arr3[]``    ``for` `(``int` `i = ``0``; i < n3; i++) {` `      ``// For arr4[]``      ``for` `(``int` `j = ``0``; j < n4; j++) {``        ``int` `cur_sum = arr3[i] + arr4[j];``        ``cnt += freq.getOrDefault(sum - cur_sum,``0``);``      ``}``    ``}``    ``return` `cnt;``  ``}` `  ``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 is contributed by utkarshshirode02`

## Python3

 `# Python3 implementation of the approach`  `# Function to return the count of the required quadruplets``from` `collections ``import` `defaultdict`  `def` `countQuadruplets(arr1, n1, arr2, n2, arr, n3, arr4, n4, S):` `    ``# To store the count of required quadruplets``    ``cnt ``=` `0``    ``# To store the frequency of S of``    ``# two elements of two different arrays``    ``freq ``=` `defaultdict(``int``)``    ``# For arr1[]``    ``for` `i ``in` `range``(n1):` `        ``# For arr2[]``        ``for` `j ``in` `range``(n2):` `            ``freq[arr1[i] ``+` `arr2[j]] ``+``=` `1` `    ``# for arr3[]``    ``for` `i ``in` `range``(n3):` `        ``# For arr4[]``        ``for` `j ``in` `range``(n4):``            ``cur_S ``=` `arr3[i] ``+` `arr4[j]``            ``cnt ``+``=` `freq[S ``-` `cur_S]` `    ``return` `cnt`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr1 ``=` `[``0``, ``2``]``    ``arr2 ``=` `[``-``1``, ``-``2``]``    ``arr3 ``=` `[``2``, ``1``]``    ``arr4 ``=` `[``2``, ``-``1``]``    ``S ``=` `0``    ``n1 ``=` `len``(arr1)``    ``n2 ``=` `len``(arr2)``    ``n3 ``=` `len``(arr3)``    ``n4 ``=` `len``(arr4)` `    ``print``(countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, S))`

## C#

 `/*package whatever //do not write package name here */``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` `sum)``  ``{` `    ``// To store the count of required quadruplets``    ``int` `cnt = 0;` `    ``// To store the frequency of sum of``    ``// two elements of two different arrays``    ``Dictionary<``int``, ``int``> freq``      ``= ``new` `Dictionary<``int``, ``int``>();` `    ``// For arr1[]``    ``for` `(``int` `i = 0; i < n1; i++) {` `      ``// For arr2[]``      ``for` `(``int` `j = 0; j < n2; j++) {``        ``freq.Add(arr1[i] + arr2[j],``                 ``freq.GetValueOrDefault(``                   ``arr1[i] + arr2[j], 0)``                 ``+ 1);``      ``}``    ``}` `    ``// for arr3[]``    ``for` `(``int` `i = 0; i < n3; i++) {` `      ``// For arr4[]``      ``for` `(``int` `j = 0; j < n4; j++) {``        ``int` `cur_sum = arr3[i] + arr4[j];``        ``cnt += freq.GetValueOrDefault(sum - cur_sum,``                                      ``0);``      ``}``    ``}``    ``return` `cnt;``  ``}` `  ``public` `static` `void` `Main()``  ``{``    ``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 is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output

`2`

Time Complexity: O(n*n)

Auxiliary Space: O(n)

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