Minimize sum of absolute difference between all pairs of array elements by decrementing and incrementing pairs by 1
Given an array arr[] ( 1-based indexing ) consisting of N integers, the task is to find the minimum sum of the absolute difference between all pairs of array elements by decrementing and incrementing any pair of elements by 1 any number of times.
Examples:
Input: arr[] = {1, 2, 3}
Output: 0
Explanation:
Modify the array elements by performing the following operations:
- Choose the pairs of element (arr[1], arr[3]) and incrementing and decrementing the pairs modifies the array to {2, 2, 2}.
After the above operations, the sum of the absolute differences is |2 – 2| + |2 – 2| + |2 – 2| = 0. Therefore, print 0.
Input: arr[] = {0, 1, 0, 1}
Output: 4
Approach: The given problem can be solved by using a Greedy Approach. It can be observed that to minimize the sum of the absolute difference between every pair of array elements arr[], the idea to make every array element closed to each other. Follow the steps below to solve the problem:
- Find the sum of the array elements arr[] and store it in a variable, say sum.
- Now, if the value of sum % N is 0, then print 0 as all the array elements can be made equal and the resultant value of the expression is always 0. Otherwise, find the value of sum % N and store it in a variable, say R.
- Now, if all the array elements are sum/N, then we can make R number of array elements as 1 and the rest of the array elements as 0 to minimize the resultant value.
- After the above steps, the minimum sum of the absolute difference is given by R*(N – R).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int minSumDifference( int ar[], int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += ar[i];
int rem = sum % n;
return rem * (n - rem);
}
int main()
{
int arr[] = { 3, 6, 8, 5, 2,
1, 11, 7, 10, 4 };
int N = sizeof (arr) / sizeof ( int );
cout << minSumDifference(arr, N);
return 0;
}
|
Java
class GFG {
public static int minSumDifference( int ar[], int n) {
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += ar[i];
int rem = sum % n;
return rem * (n - rem);
}
public static void main(String args[]) {
int [] arr = { 3 , 6 , 8 , 5 , 2 , 1 , 11 , 7 , 10 , 4 };
int N = arr.length;
System.out.println(minSumDifference(arr, N));
}
}
|
Python3
def minSumDifference(ar, n):
sum = 0
for i in range (n):
sum + = ar[i]
rem = sum % n
return rem * (n - rem)
if __name__ = = '__main__' :
arr = [ 3 , 6 , 8 , 5 , 2 , 1 , 11 , 7 , 10 , 4 ]
N = len (arr)
print (minSumDifference(arr, N))
|
C#
using System;
class GFG{
public static int minSumDifference( int [] ar, int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += ar[i];
int rem = sum % n;
return rem * (n - rem);
}
public static void Main()
{
int [] arr = { 3, 6, 8, 5, 2,
1, 11, 7, 10, 4 };
int N = arr.Length;
Console.Write(minSumDifference(arr, N));
}
}
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Javascript
<script>
function minSumDifference(ar, n) {
let sum = 0;
for (let i = 0; i < n; i++) sum += ar[i];
let rem = sum % n;
return rem * (n - rem);
}
let arr = [3, 6, 8, 5, 2, 1, 11, 7, 10, 4];
let N = arr.length;
document.write(minSumDifference(arr, N));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
06 Aug, 2021
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