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 quadrupletsint 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 codeint 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 quadrupletsclass 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 codepublic 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 codeif __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 approachusing 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 quadrupletsfunction 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 arrayfunction 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 codelet 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)); |
Output:
2
Time Complexity: O(n*log(n)+m*log(n)) where n=n1*n2 and m=n3*n4
Auxiliary Space: O(n)