Maximum difference between two subsets of m elements
Last Updated :
03 Aug, 2022
Given an array of n integers and a number m, find the maximum possible difference between two sets of m elements chosen from given array.
Examples:
Input : arr[] = 1 2 3 4 5
m = 4
Output : 4
The maximum four elements are 2, 3,
4 and 5. The minimum four elements are
1, 2, 3 and 4. The difference between
two sums is (2 + 3 + 4 + 5) - (1 + 2
+ 3 + 4) = 4
Input : arr[] = 5 8 11 40 15
m = 2
Output : 42
The difference is (40 + 15) - (5 + 8)
The idea is to first sort the array, then find sum of first m elements and sum of last m elements. Finally return difference between two sums.
Implementation:
CPP
#include <bits/stdc++.h>
using namespace std;
int find_difference(int arr[], int n, int m)
{
int max = 0, min = 0;
sort(arr, arr + n);
for (int i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int m = 4;
cout << find_difference(arr, n, m);
return 0;
}
|
Java
import java.util.Arrays;
class GFG {
static int find_difference(int arr[], int n,
int m)
{
int max = 0, min = 0;
Arrays.sort(arr);
for (int i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
public static void main(String arg[])
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
int m = 4;
System.out.print(find_difference(arr, n, m));
}
}
|
Python3
def find_difference(arr, n, m):
max = 0; min = 0
arr.sort();
j = n-1
for i in range(m):
min += arr[i]
max += arr[j]
j = j - 1
return (max - min)
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5]
n = len(arr)
m = 4
print(find_difference(arr, n, m))
|
C#
using System;
class GFG {
static int find_difference(int[] arr, int n,
int m)
{
int max = 0, min = 0;
Array.Sort(arr);
for (int i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
int m = 4;
Console.Write(find_difference(arr, n, m));
}
}
|
PHP
<?php
function find_difference($arr, $n, $m)
{
$max = 0; $min = 0;
sort($arr);
sort( $arr,$n);
for($i = 0, $j = $n - 1; $i <$m; $i++, $j--)
{
$min += $arr[$i];
$max += $arr[$j];
}
return ($max - $min);
}
{
$arr = array(1, 2, 3, 4, 5);
$n = sizeof($arr) / sizeof($arr[0]);
$m = 4;
echo find_difference($arr, $n, $m);
return 0;
}
?>
|
Javascript
<script>
function find_difference(arr, n,
m)
{
let max = 0, min = 0;
arr.sort();
for (let i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
let arr = [ 1, 2, 3, 4, 5 ];
let n = arr.length;
let m = 4;
document.write(find_difference(arr, n, m));
</script>
|
Time Complexity : O(n Log n)
Auxiliary Space : O(1)
We can optimize the above solution using more efficient approaches discussed in below post.
k largest(or smallest) elements in an array | added Min Heap method
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