Count of quadruplets with given Sum | Set 2
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: We pick any two arrays and calculate all possible sums and and keep their counts in a map. Using the remaining two arrays, we calculate all possible sums and check how many times their additive inverse exists in the map which will be the count of required quadruplets.
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 value)
{
int cnt = 0;
unordered_map< int , int > sum;
for ( int i = 0; i < n1; i++)
for ( int j = 0; j < n2; j++)
sum[arr1[i] + arr2[j]]++;
for ( int i = 0; i < n3; i++)
for ( int j = 0; j < n4; j++)
cnt += sum[value - (arr3[i] + arr4[j])];
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 value)
{
int cnt = 0 ;
Map<Integer, Integer> sum = new HashMap<>();
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 );
}
}
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;
}
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(arr1, n1, arr2, n2,
arr3, n3, arr4, n4, value):
cnt = 0
sum = {i: 0 for i in range ( - 4 , 10 , 1 )}
for i in range (n1):
for j in range (n2):
sum [arr1[i] + arr2[j]] + = 1
for i in range (n3):
for j in range (n4):
cnt + = sum [value - (arr3[i] + arr4[j])]
return cnt
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 ))
|
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 value)
{
int cnt = 0;
Dictionary< int , int > sum = new Dictionary< int , int >();
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);
}
}
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;
}
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));
}
}
|
PHP
<?php
function countQuadruplets( $arr1 , $n1 , $arr2 , $n2 ,
$arr3 , $n3 , $arr4 , $n4 , $value )
{
$cnt = 0;
$sum = array ();
for ( $i = 0; $i < $n1 ; $i ++)
for ( $j = 0; $j < $n2 ; $j ++)
$sum [ $arr1 [ $i ] + $arr2 [ $j ]] = 0 ;
for ( $i = 0; $i < $n1 ; $i ++)
for ( $j = 0; $j < $n2 ; $j ++)
$sum [ $arr1 [ $i ] + $arr2 [ $j ]]++;
for ( $i = 0; $i < $n3 ; $i ++)
for ( $j = 0; $j < $n4 ; $j ++)
$cnt += $sum [ $value - ( $arr3 [ $i ] + $arr4 [ $j ])];
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, value)
{
var cnt = 0;
var sum = new Map();
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);
}
}
}
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;
}
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));
</script>
|
Time Complexity : O(n1*n2+n3*n4) where, n1, n2, n3, n4 are size of arrays arr1, arr2, arr3 and arr4 respectively.
Auxiliary Space : O(n) , to store the elements in map.
Last Updated :
16 Jun, 2022
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