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Count of quadruplets with given Sum

  • Difficulty Level : Basic
  • Last Updated : 11 May, 2021
Geek Week

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 <bits/stdc++.h>
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




<?php
// PHP implementation of the approach
 
// Function to return the count of the required quadruplets
function countQuadruplets($arr1, $n1, $arr2,$n2,
                $arr3, $n3, $arr4, $n4, $sum)
{
 
    // To store the count of required quadruplets
    $cnt = 0;
 
    // For arr1[]
    for ($i = 0; $i < $n1; $i++)
    {
 
        // For arr2[]
        for ($j = 0; $j < $n2; $j++)
        {
 
            // For arr3[]
            for ($k = 0; $k < $n3; $k++)
            {
 
                // For arr4[]
                for ( $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
$arr1 = array (0, 2 );
$arr2 = array( -1, -2 );
$arr3 = array( 2, 1 );
$arr4 =array( 2, -1 );
$sum = 0;
$n1 = count($arr1);
$n2 =count($arr2);
$n3 = count($arr3);
$n4 = count($arr4);
 
echo countQuadruplets($arr1, $n1, $arr2, $n2,
                    $arr3, $n3, $arr4, $n4, $sum);
 
 
// This code is contributed by ajit
?>

Javascript




<script>
    // Javascript program to implement the above approach
     
    // Function to return the count of the required quadruplets
    function countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, sum)
    {
 
        // To store the count of required quadruplets
        let cnt = 0;
 
        // For arr1[]
        for (let i = 0; i < n1; i++)
        {
 
            // For arr2[]
            for (let j = 0; j < n2; j++)
            {
 
                // For arr3[]
                for (let k = 0; k < n3; k++)
                {
 
                    // For arr4[]
                    for (let 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;
    }
     
    let arr1 = [ 0, 2 ];
    let arr2 = [ -1, -2 ];
    let arr3 = [ 2, 1 ];
    let arr4 = [ 2, -1 ];
    let sum = 0;
    let n1 = arr1.length;
    let n2 = arr2.length;
    let n3 = arr3.length;
    let n4 = arr4.length;
    document.write(countQuadruplets(arr1, n1, arr2, n2,
                                    arr3, n3, arr4, n4, sum));
 
</script>
Output: 
2

 

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

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