Maximum difference between group of k-elements and rest of the array.
Last Updated :
19 Sep, 2023
You are given an array of n elements. You have to divide the given array into two group such that one group consists exactly k elements and second group consists rest of elements. Your result must be maximum possible difference of sum of elements of these two group.
Examples:
Input : arr[n] = {1, 5, 2, 6, 3} , k = 2
Output : Maximum Difference = 11
Explanation : group1 = 1+2 , group2 = 3+5+6
Maximum Difference = 14 - 3 = 11
Input : arr[n] = {1, -1, 3, -2, -3} , k = 2
Output : Maximum Difference = 10
Explanation : group1 = -1-2-3 , group2 = 1+3
Maximum Difference = 4 - (-6) = 10
For finding the maximum group difference we have two possibilities. For the first case k-smallest elements belongs to one group and rest elements to other group. For second case k-largest elements belongs to one group and rest elements to other group.
So, first of all sort the whole array and find group difference for both cases explained as above and then finally find the maximum difference among them.
Algorithm :
sort the arrayfind sum of whole array
case-1 -> find sum of first k-smallest elements
-> differece1 = abs( arraySum - 2*k_Smallest)
case-2 -> find sum of first k-largest elements
-> differece2 = abs( arraySum - 2*k_largest)
print max(difference1, difference2)
Implementation:
C++
// CPP to find maximum group difference
#include<bits/stdc++.h>
using namespace std;
// utility function for array sum
long long int arraySum(int arr[], int n)
{
long long int sum = 0;
for (int i=0; i<n; i++)
sum = sum + arr[i];
return sum;
}
// function for finding
// maximum group difference of array
long long int maxDiff (int arr[], int n, int k)
{
// sort the array
sort(arr, arr+n);
// find array sum
long long int arraysum = arraySum(arr, n);
// difference for k-smallest
// diff1 = (arraysum-k_smallest)-k_smallest
long long int diff1 = abs(arraysum - 2*arraySum(arr, k));
// reverse array for finding sum of 1st k-largest
reverse(arr, arr+n);
// difference for k-largest
// diff2 = (arraysum-k_largest)-k_largest
long long int diff2 = abs(arraysum - 2*arraySum(arr, k));
// return maximum difference value
return(max(diff1,diff2));
}
// driver program
int main()
{
int arr[] = {1, 7, 4, 8, -1, 5, 2, 1};
int n = sizeof(arr)/sizeof(arr[0]);
int k = 3;
cout << "Maximum Difference = " << maxDiff(arr,n,k);
return 0;
}
// java to find maximum group difference
import java.util.Arrays;
public class GFG {
// utility function for array sum
static long arraySum(int arr[], int n)
{
long sum = 0;
for (int i = 0; i < n; i++)
sum = sum + arr[i];
return sum;
}
// function for finding maximum group
// difference of array
static long maxDiff (int arr[], int n, int k)
{
// sort the array
Arrays.sort(arr);
// find array sum
long arraysum = arraySum(arr, n);
// difference for k-smallest
// diff1 = (arraysum-k_smallest)-k_smallest
long diff1 = Math.abs(arraysum -
2 * arraySum(arr, k));
// reverse array for finding sum of
// 1st k-largest
int end = arr.length - 1;
int start = 0;
while (start < end)
{
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
// difference for k-largest
// diff2 = (arraysum-k_largest)-k_largest
long diff2 = Math.abs(arraysum -
2 * arraySum(arr, k));
// return maximum difference value
return(Math.max(diff1, diff2));
}
public static void main(String args[]) {
int arr[] = {1, 7, 4, 8, -1, 5, 2, 1};
int n = arr.length;
int k = 3;
System.out.println("Maximum Difference = "
+ maxDiff(arr, n, k));
}
}
// This code is contributed by Sam007.
# Python3 to find maximum group difference
# utility function for array sum
def arraySum(arr, n):
sum = 0
for i in range(n):
sum = sum + arr[i]
return sum
# function for finding
# maximum group difference of array
def maxDiff (arr, n, k):
# sort the array
arr.sort()
# find array sum
arraysum = arraySum(arr, n)
# difference for k-smallest
# diff1 = (arraysum-k_smallest)-k_smallest
diff1 = abs(arraysum - 2 * arraySum(arr, k))
# reverse array for finding sum
# of 1st k-largest
arr.reverse()
# difference for k-largest
# diff2 = (arraysum-k_largest)-k_largest
diff2 = abs(arraysum - 2 * arraySum(arr, k))
# return maximum difference value
return(max(diff1, diff2))
# Driver Code
if __name__ == "__main__":
arr = [1, 7, 4, 8, -1, 5, 2, 1]
n = len(arr)
k = 3
print ("Maximum Difference =",
maxDiff(arr, n, k))
# This code is contributed by ita_c
// C# to find maximum group difference
using System;
public class GFG {
// utility function for array sum
static long arraySum(int []arr, int n)
{
long sum = 0;
for (int i = 0; i < n; i++)
sum = sum + arr[i];
return sum;
}
// function for finding maximum group
// difference of array
static long maxDiff (int []arr, int n, int k)
{
// sort the array
Array.Sort(arr);
// find array sum
long arraysum = arraySum(arr, n);
// difference for k-smallest
// diff1 = (arraysum-k_smallest)-k_smallest
long diff1 = Math.Abs(arraysum -
2 * arraySum(arr, k));
// reverse array for finding sum of
// 1st k-largest
Array.Reverse(arr);
// difference for k-largest
// diff2 = (arraysum-k_largest)-k_largest
long diff2 = Math.Abs(arraysum -
2 * arraySum(arr, k));
// return maximum difference value
return(Math.Max(diff1, diff2));
}
// Driver program
static public void Main ()
{
int []arr = {1, 7, 4, 8, -1, 5, 2, 1};
int n = arr.Length;
int k = 3;
Console.WriteLine("Maximum Difference = "
+ maxDiff(arr, n, k));
}
}
// This Code is contributed by vt_m.
<?php
// PHP to find maximum group difference
// utility function for array sum
function arraySum($arr, $n)
{
$sum = 0;
for ($i = 0; $i < $n; $i++)
$sum = $sum + $arr[$i];
return $sum;
}
// function for finding
// maximum group difference
// of array
function maxDiff ($arr, $n, $k)
{
// sort the array
sort($arr);
// find array sum
$arraysum = arraySum($arr, $n);
// difference for k-smallest
// diff1 = (arraysum - k_smallest)
// - k_smallest
$diff1 = abs($arraysum - 2 *
arraySum($arr, $k));
// reverse array for finding
// sum of 1st k-largest
array_reverse($arr);
// difference for k-largest
// diff2 = (arraysum - k_largest)
// - k_largest
$diff2 = abs($arraysum - 2 *
arraySum($arr, $k));
// return maximum difference value
return(max($diff1,$diff2));
}
// Driver Code
$arr = array(1, 7, 4, 8, -1, 5, 2, 1);
$n = count($arr);
$k = 3;
echo "Maximum Difference = " ,
maxDiff($arr, $n, $k);
// This Code is contributed by vt_m.
?>
<script>
// Javascript to find maximum group difference
// utility function for array sum
function arraySum(arr, n)
{
var sum = 0;
for (var i=0; i<n; i++)
sum = sum + arr[i];
return sum;
}
// function for finding
// maximum group difference of array
function maxDiff (arr, n, k)
{
// sort the array
arr.sort((a,b)=>a-b);
// find array sum
var arraysum = arraySum(arr, n);
// difference for k-smallest
// diff1 = (arraysum-k_smallest)-k_smallest
var diff1 = Math.abs(arraysum - 2*arraySum(arr, k));
// reverse array for finding sum of 1st k-largest
arr.reverse();
// difference for k-largest
// diff2 = (arraysum-k_largest)-k_largest
var diff2 = Math.abs(arraysum - 2*arraySum(arr, k));
// return maximum difference value
return(Math.max(diff1,diff2));
}
// driver program
var arr = [1, 7, 4, 8, -1, 5, 2, 1];
var n = arr.length;
var k = 3;
document.write( "Maximum Difference = " + maxDiff(arr,n,k));
</script>
OutputMaximum Difference = 25
Time Complexity: O(n log n)
Auxiliary Space: O(1)
Optimizations to above solution :
- We can avoid reversing the array and find sum k elements from end using a different loop.
- We can also find sum of k largest and smallest elements more efficiently using methods discussed in below posts.
K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)
K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst Case Linear Time)
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