给定总和的四元组计数 | 2套

给定包含整数元素和整数sum 的四个数组，任务是对四元组进行计数，以便从不同的数组中选择每个元素，并且所有四个元素的总和等于给定的总和。
例子：

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 
 

方法：我们选择任意两个数组并计算所有可能的总和，并将它们的计数保存在一个映射中。使用剩余的两个数组，我们计算所有可能的和并检查它们的加法逆存在于映射中的次数，这将是所需四元组的计数。
下面是上述方法的实现：

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 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 sum = new Dictionary();
 
        // 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 */

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count of the required quadruplets
function countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, value)
{
    var cnt = 0;
    var sum = new Map();
 
    // All possible sums from arr1[] and arr2[]
    for (var i = 0; i < n1; i++)
    {
        for (var j = 0; j < n2; j++)
        {
            if(sum.has(arr1[i] + arr2[j]))
            {
                sum.set(arr1[i] + arr2[j], sum.get(arr1[i] + arr2[j])+1);
            }
            else
            {
                sum.set(arr1[i] + arr2[j], 1);
            }
        }
    }
 
    // Find the count of quadruplets
    for (var i = 0; i < n3; i++)
        for (var j = 0; j < n4; j++)
            if(sum.has((value - (arr3[i] + arr4[j]))))
            {
                cnt += sum.get((value - (arr3[i] + arr4[j])));
            }
             
    return cnt;
}
 
// Driver code
var arr1 = [0, 2];
var arr2 = [-1, -2];
var arr3 = [2, 1];
var arr4 = [2, -1];
var sum = 0;
var n1 = arr1.length;
var n2 = arr2.length;
var n3 = arr3.length;
var n4 = arr4.length;
document.write( countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, sum));
 
// This code is contributed by noob2000.

输出：

2

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