Count of quadruplets with given sum | Set 3

Last Updated : 27 Feb, 2023

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

Approach: Two different approaches to solve this problem has been discussed in Set 1 and Set 2 of this article. Here, an approach using the binary search will be discussed. Pick any two arrays and calculate all the possible pair sums and store them in a vector. Now, pick the other two arrays and calculate all possible sums and for every sum say tempSum, check whether sum – temp exists in the vector created earlier (after sorting it) using the binary search. Below is the implementation of the above approach:

CPP

// 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 value)
 
{
    vector<int> sum1;
    vector<int>::iterator it;
    vector<int>::iterator it2;
    int cnt = 0;
 
    // Take every possible pair sum
    // from the two arrays
    for (int i = 0; i < n1; i++) {
        for (int j = 0; j < n2; j++) {
 
            // Push the sum to a vector
            sum1.push_back(arr1[i] + arr2[j]);
        }
    }
 
    // Sort the sum vector
    sort(sum1.begin(), sum1.end());
 
    // Calculate the pair sums from
    // the other two arrays
    for (int i = 0; i < n3; i++) {
        for (int j = 0; j < n4; j++) {
 
            // Calculate the sum
            int temp = arr3[i] + arr4[j];
 
            // Check whether temp can be added to any
            // sum stored in the sum1 vector such that
            // the result is the required sum
            if (binary_search(sum1.begin(), sum1.end(), value - temp)) {
 
                // Add the count of such values from the sum1 vector
                it = lower_bound(sum1.begin(), sum1.end(), value - temp);
                it2 = upper_bound(sum1.begin(), sum1.end(), value - temp);
                cnt = cnt + ((it2 - sum1.begin()) - (it - sum1.begin()));
            }
        }
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    int arr1[] = { 0, 2 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
 
    int arr2[] = { -1, -2 };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
 
    int arr3[] = { 2, 1 };
    int n3 = sizeof(arr3) / sizeof(arr3[0]);
 
    int arr4[] = { 2, -1 };
    int n4 = sizeof(arr4) / sizeof(arr4[0]);
 
    int sum = 0;
 
    cout << countQuadruplets(arr1, n1, arr2, n2,
                             arr3, n3, arr4, n4, sum);
 
    return 0;
}

Java

import java.util.Arrays;
import java.util.Vector;
import java.util.*;
 
// Function to return the count
// of the required quadruplets
class Main{
public static int countQuadruplets(int[] arr1, int n1, int[] arr2,
                int n2, int[] arr3, int n3,
                 int[] arr4, int n4, int value)
{
    Vector<Integer> sum1 = new Vector<>();
    int cnt = 0;
 
    // Take every possible pair sum
    // from the two arrays
    for (int i = 0; i < n1; i++) {
        for (int j = 0; j < n2; j++) {
 
            // Push the sum to a vector
            sum1.add(arr1[i] + arr2[j]);
        }
    }
 
    // Sort the sum vector
    sum1.sort(null);
 
    // Calculate the pair sums from
    // the other two arrays
    for (int i = 0; i < n3; i++) {
        for (int j = 0; j < n4; j++) {
 
            // Calculate the sum
            int temp = arr3[i] + arr4[j];
 
            // Check whether temp can be added to any
            // sum stored in the sum1 vector such that
            // the result is the required sum
            if (Collections.binarySearch(sum1, value - temp) >= 0) {
 
                // Add the count of such values from the sum1 vector
                int it = Collections.binarySearch(sum1, value - temp);
                int it2 = it + 1;
                while (it2 < sum1.size() && sum1.get(it2).equals(value - temp)) {
                    it2++;
                }
                cnt = cnt + (it2 - it);
            }
        }
    }
 
    return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    int[] arr1 = { 0, 2 };
    int n1 = arr1.length;
 
    int[] arr2 = { -1, -2 };
    int n2 = arr2.length;
 
    int[] arr3 = { 2, 1 };
    int n3 = arr3.length;
 
    int[] arr4 = { 2, -1 };
    int n4 = arr4.length;
 
    int sum = 0;
 
    System.out.println(countQuadruplets(arr1, n1, arr2, n2,
                     arr3, n3, arr4, n4, sum));
}
}

Python3

from typing import List
import bisect
 
def countQuadruplets(arr1: List[int], n1: int, arr2: List[int], n2: int,
                arr3: List[int], n3: int, arr4: List[int], n4: int, value: int) -> int:
 
    sum1 = []
 
    # Take every possible pair sum from the two arrays
    for i in range(n1):
        for j in range(n2):
            # Append the sum to the list
            sum1.append(arr1[i] + arr2[j])
 
    # Sort the sum list
    sum1.sort()
 
    cnt = 0
 
    # Calculate the pair sums from the other two arrays
    for i in range(n3):
        for j in range(n4):
            # Calculate the sum
            temp = arr3[i] + arr4[j]
 
            # Check whether temp can be added to any sum stored in the sum1
            # list such that the result is the required sum
            index = bisect.bisect_left(sum1, value - temp)
 
            if index != len(sum1) and sum1[index] == value - temp:
                # Add the count of such values from the sum1 list
                it = index
                it2 = it + 1
                while it2 < len(sum1) and sum1[it2] == value - temp:
                    it2 += 1
                cnt += (it2 - it)
 
    return cnt
 
# Driver code
if __name__ == "__main__":
    arr1 = [0, 2]
    n1 = len(arr1)
 
    arr2 = [-1, -2]
    n2 = len(arr2)
 
    arr3 = [2, 1]
    n3 = len(arr3)
 
    arr4 = [2, -1]
    n4 = len(arr4)
 
    sum = 0
 
    print(countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, sum))
     
 # This code is contributed by Prince Kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
public class GFG
{
  // Function to return the count
  // of the required quadruplets
  public static int CountQuadruplets(int[] arr1, int n1, int[] arr2,
                                     int n2, int[] arr3, int n3,
                                     int[] arr4, int n4, int value)
  {
    List<int> sum1 = new List<int>();
    int cnt = 0;
 
    // Take every possible pair sum
    // from the two arrays
    for (int i = 0; i < n1; i++)
    {
      for (int j = 0; j < n2; j++)
      {
        // Push the sum to a vector
        sum1.Add(arr1[i] + arr2[j]);
      }
    }
 
    // Sort the sum vector
    sum1.Sort();
 
    // Calculate the pair sums from
    // the other two arrays
    for (int i = 0; i < n3; i++)
    {
      for (int j = 0; j < n4; j++)
      {
        // Calculate the sum
        int temp = arr3[i] + arr4[j];
 
        // Check whether temp can be added to any
        // sum stored in the sum1 vector such that
        // the result is the required sum
        if (sum1.BinarySearch(value - temp) >= 0)
        {
          // Add the count of such values from the sum1 vector
          int it = sum1.BinarySearch(value - temp);
          int it2 = it + 1;
          while (it2 < sum1.Count && sum1[it2] == (value - temp))
          {
            it2++;
          }
          cnt = cnt + (it2 - it);
        }
      }
    }
 
    return cnt;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int[] arr1 = { 0, 2 };
    int n1 = arr1.Length;
 
    int[] arr2 = { -1, -2 };
    int n2 = arr2.Length;
 
    int[] arr3 = { 2, 1 };
    int n3 = arr3.Length;
 
    int[] arr4 = { 2, -1 };
    int n4 = arr4.Length;
 
    int sum = 0;
 
    Console.WriteLine(CountQuadruplets(arr1, n1, arr2, n2,
                                       arr3, n3, arr4, n4, sum));
  }
}
 
// This code is contributed by Aman Kumar.

Javascript

// Function to return the count
// of the required quadruplets
function countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, value) {
  let sum1 = [];
 
  // Take every possible pair sum
  // from the two arrays
  for (let i = 0; i < n1; i++) {
    for (let j = 0; j < n2; j++) {
 
      // Push the sum to an array
      sum1.push(arr1[i] + arr2[j]);
    }
  }
 
  // Sort the sum array
  sum1.sort((a, b) => a - b);
 
  let cnt = 0;
 
  // Calculate the pair sums from
  // the other two arrays
  for (let i = 0; i < n3; i++) {
    for (let j = 0; j < n4; j++) {
 
      // Calculate the sum
      let temp = arr3[i] + arr4[j];
 
      // Check whether temp can be added to any
      // sum stored in the sum1 array such that
      // the result is the required sum
      if (binarySearch(sum1, value - temp) >= 0) {
 
        // Add the count of such values from the sum1 array
        let it = binarySearch(sum1, value - temp);
        let it2 = it + 1;
        while (it2 < sum1.length && sum1[it2] === value - temp) {
          it2++;
        }
        cnt = cnt + (it2 - it);
      }
    }
  }
 
  return cnt;
}
 
// Binary search function to search for an element in the array
function binarySearch(arr, x) {
  let l = 0, r = arr.length - 1;
  while (l <= r) {
    let m = Math.floor((l + r) / 2);
    if (arr[m] === x) {
      return m;
    } else if (arr[m] < x) {
      l = m + 1;
    } else {
      r = m - 1;
    }
  }
  return -1;
}
 
// Driver code
let arr1 = [0, 2];
let n1 = arr1.length;
 
let arr2 = [-1, -2];
let n2 = arr2.length;
 
let arr3 = [2, 1];
let n3 = arr3.length;
 
let arr4 = [2, -1];
let n4 = arr4.length;
 
let sum = 0;
 
console.log(countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, sum));