# Count of quadruplets with given Sum

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: 2
(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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Generate all possible quadruplets and calculate the sum of every quadruplets. 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); ` `    ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2); ` `    ``int` `n3 = ``sizeof``(arr3) / ``sizeof``(arr3); ` `    ``int` `n4 = ``sizeof``(arr4) / ``sizeof``(arr4); ` ` `  `    ``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

 ` `

Output:

```2
```

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

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Improved By : Rajput-Ji, jit_t