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
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++
#include <bits/stdc++.h>
using namespace std;
int countQuadruplets( int arr1[], int n1, int arr2[], int n2,
int arr3[], int n3, int arr4[], int n4, int sum)
{
int cnt = 0;
for ( int i = 0; i < n1; i++) {
for ( int j = 0; j < n2; j++) {
for ( int k = 0; k < n3; k++) {
for ( int l = 0; l < n4; l++) {
if (arr1[i] + arr2[j] + arr3[k] + arr4[l] == sum) {
cnt++;
}
}
}
}
}
return cnt;
}
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
class GFG
{
static int countQuadruplets( int arr1[], int n1, int arr2[], int n2,
int arr3[], int n3, int arr4[], int n4, int sum)
{
int cnt = 0 ;
for ( int i = 0 ; i < n1; i++)
{
for ( int j = 0 ; j < n2; j++)
{
for ( int k = 0 ; k < n3; k++)
{
for ( int l = 0 ; l < n4; l++)
{
if (arr1[i] + arr2[j] + arr3[k] + arr4[l] == sum)
{
cnt++;
}
}
}
}
}
return cnt;
}
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));
}
}
|
Python3
def countQuadruplets(P, Q, R, S, sum ):
cnt = 0
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
P = [ 0 , 2 ]
Q = [ - 1 , - 2 ]
R = [ 2 , 1 ]
S = [ 2 , - 1 ]
sum = 0
print (countQuadruplets(P, Q, R, S, sum ))
|
C#
using System;
class GFG
{
static int countQuadruplets( int []arr1, int n1, int []arr2, int n2,
int []arr3, int n3, int []arr4, int n4, int sum)
{
int cnt = 0;
for ( int i = 0; i < n1; i++)
{
for ( int j = 0; j < n2; j++)
{
for ( int k = 0; k < n3; k++)
{
for ( int l = 0; l < n4; l++)
{
if (arr1[i] + arr2[j] + arr3[k] + arr4[l] == sum)
{
cnt++;
}
}
}
}
}
return cnt;
}
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));
}
}
|
PHP
<?php
function countQuadruplets( $arr1 , $n1 , $arr2 , $n2 ,
$arr3 , $n3 , $arr4 , $n4 , $sum )
{
$cnt = 0;
for ( $i = 0; $i < $n1 ; $i ++)
{
for ( $j = 0; $j < $n2 ; $j ++)
{
for ( $k = 0; $k < $n3 ; $k ++)
{
for ( $l = 0; $l < $n4 ; $l ++)
{
if ( $arr1 [ $i ] + $arr2 [ $j ] + $arr3 [ $k ] +
$arr4 [ $l ] == $sum )
{
$cnt ++;
}
}
}
}
}
return $cnt ;
}
$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 );
?>
|
Javascript
<script>
function countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, sum)
{
let cnt = 0;
for (let i = 0; i < n1; i++)
{
for (let j = 0; j < n2; j++)
{
for (let k = 0; k < n3; k++)
{
for (let l = 0; l < n4; l++)
{
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>
|
Time Complexity: O(n4)
Space Complexity: O(1)
Efficient Approach: Store frequency of all possible sum of two elements from two different arrays in a map. Iterate over other two arrays and find the sum of any two elements in these two arrays,lets it be cur_sum. If sum – cur_sum is present in the map, this means that there exists four elements in four different arrays whose sum is equal to sum.
Implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countQuadruplets( int arr1[], int n1, int arr2[], int n2,
int arr3[], int n3, int arr4[], int n4, int sum)
{
int cnt = 0;
unordered_map< int , int > freq;
for ( int i = 0; i < n1; i++) {
for ( int j = 0; j < n2; j++) {
freq[arr1[i]+arr2[j]]++;
}
}
for ( int i = 0; i < n3; i++) {
for ( int j = 0; j < n4; j++) {
int cur_sum = arr3[i]+arr4[j];
cnt+=freq[sum-cur_sum];
}
}
return cnt;
}
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
import java.util.*;
class GFG {
static int countQuadruplets( int arr1[], int n1,
int arr2[], int n2,
int arr3[], int n3,
int arr4[], int n4, int sum)
{
int cnt = 0 ;
HashMap<Integer, Integer> freq = new HashMap<>();
for ( int i = 0 ; i < n1; i++) {
for ( int j = 0 ; j < n2; j++) {
freq.put(arr1[i] + arr2[j],freq.getOrDefault(arr1[i] + arr2[j], 0 )+ 1 );
}
}
for ( int i = 0 ; i < n3; i++) {
for ( int j = 0 ; j < n4; j++) {
int cur_sum = arr3[i] + arr4[j];
cnt += freq.getOrDefault(sum - cur_sum, 0 );
}
}
return cnt;
}
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));
}
}
|
Python3
from collections import defaultdict
def countQuadruplets(arr1, n1, arr2, n2, arr, n3, arr4, n4, S):
cnt = 0
freq = defaultdict( int )
for i in range (n1):
for j in range (n2):
freq[arr1[i] + arr2[j]] + = 1
for i in range (n3):
for j in range (n4):
cur_S = arr3[i] + arr4[j]
cnt + = freq[S - cur_S]
return cnt
if __name__ = = "__main__" :
arr1 = [ 0 , 2 ]
arr2 = [ - 1 , - 2 ]
arr3 = [ 2 , 1 ]
arr4 = [ 2 , - 1 ]
S = 0
n1 = len (arr1)
n2 = len (arr2)
n3 = len (arr3)
n4 = len (arr4)
print (countQuadruplets(arr1, n1, arr2, n2, arr3, n3, arr4, n4, S))
|
C#
using System;
using System.Collections.Generic;
class GFG {
static int countQuadruplets( int [] arr1, int n1,
int [] arr2, int n2,
int [] arr3, int n3,
int [] arr4, int n4, int sum)
{
int cnt = 0;
Dictionary< int , int > freq
= new Dictionary< int , int >();
for ( int i = 0; i < n1; i++) {
for ( int j = 0; j < n2; j++) {
freq.Add(arr1[i] + arr2[j],
freq.GetValueOrDefault(
arr1[i] + arr2[j], 0)
+ 1);
}
}
for ( int i = 0; i < n3; i++) {
for ( int j = 0; j < n4; j++) {
int cur_sum = arr3[i] + arr4[j];
cnt += freq.GetValueOrDefault(sum - cur_sum,
0);
}
}
return cnt;
}
public static 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));
}
}
|
Javascript
<script>
function countQuadruplets(arr1, n1, arr2, n2,
arr3, n3, arr4, n4, sum)
{
let cnt = 0;
let freq = new Map();
for (let i = 0; i < n1; i++) {
for (let j = 0; j < n2; j++) {
if (freq.has(arr1[i]+arr2[j])){
freq.set(arr1[i]+arr2[j],freq.get(arr1[i]+arr2[j])+1);
}
else
freq.set(arr1[i]+arr2[j],1);
}
}
for (let i = 0; i < n3; i++) {
for (let j = 0; j < n4; j++) {
let cur_sum = arr3[i]+arr4[j];
cnt+= freq.has(sum-cur_sum) == false ? 0 : freq.get(sum-cur_sum);
}
}
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>
|
Time Complexity: O(n*n)
Auxiliary Space: O(n*n)
Last Updated :
16 May, 2023
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