Given an array, we need to find the sum of elements of an array after changing the element as arr[i] will become abs(arr[i]-x) where x is an array element.
Examples:
Input : {2, 5, 1, 7, 4}
Output : 9
We get minimum sum when we choose
x = 4. The minimum sum is
abs(2-4) + abs(5-4) + abs(1-4) + (7-4)
abs(4-4) = 9
Input : {5, 11, 14, 10, 17, 15}
Output : 20
We can either choose x = 14 or x = 11
The idea is based on fact that middle element would cause minimum sum of differences. When there are even number of elements, we can take any of the middle two elements. We can verify this fact by taking few examples.
Below is the implementation of above idea:
C++
// C++ program to find minimum sum of absolute // differences with an array element. #include<bits/stdc++.h> using namespace std; // function to find min sum after operation int absSumDidd(int a[],int n) { // Sort the array sort(a,a+n); // Pick middle value int midValue = a[(int)(n / 2)]; // Sum of absolute differences. int sum = 0; for (int i = 0; i < n; i++) { sum = sum + abs(a[i] - midValue); } return sum; } // Driver Code int main() { int arr[] = { 5, 11, 14, 10, 17, 15 }; int n=sizeof(arr)/sizeof(arr[0]); cout<< absSumDidd(arr,n); } // Contributed by mits |
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Java
// Java program to find minimum sum of absolute // differences with an array element. import java.lang.*; import java.util.*; public class GFG { // function to find min sum after operation static int absSumDidd(int a[]) { // Sort the array Arrays.sort(a); // Pick middle value int midValue = a[a.length / 2]; // Sum of absolute differences. int sum = 0; for (int i = 0; i < a.length; i++) { sum = sum + Math.abs(a[i] - midValue); } return sum; } // Driver Code public static void main(String[] args) { int arr[] = { 5, 11, 14, 10, 17, 15 }; System.out.print(absSumDidd(arr)); } // Contributed by Saurav Jain } |
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Python3
# Python3 program to find minimum sum of # absolute differences with an array element. # function to find min sum after operation def absSumDidd(a, n): # Sort the array a.sort() # Pick middle value midValue = a[(int)(n // 2)] # Sum of absolute differences. sum = 0 for i in range(n): sum = sum + abs(a[i] - midValue) return sum # Driver Code arr = [5, 11, 14, 10, 17, 15] n = len(arr) print(absSumDidd(arr, n)) # This code is contributed # by sahishelangia |
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C#
// C# program to find minimum sum of absolute // differences with an array element. using System; class GFG { // function to find min sum after operation static int absSumDidd(int []a) { // Sort the array Array.Sort(a); // Pick middle value int midValue = a[a.Length / 2]; // Sum of absolute differences. int sum = 0; for (int i = 0; i < a.Length; i++) { sum = sum + Math.Abs(a[i] - midValue); } return sum; } // Driver Code public static void Main() { int []arr = { 5, 11, 14, 10, 17, 15 }; Console.Write(absSumDidd(arr)); } // Contributed by Subhadeep } |
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PHP
<?php // PHP program to find minimum // sum of absolute differences // with an array element. // function to find min sum // after operation function absSumDidd($a, $n) { // Sort the array sort($a); // Pick middle value $midValue = $a[($n / 2)]; // Sum of absolute differences. $sum = 0; for ( $i = 0; $i < $n; $i++) { $sum = $sum + abs($a[$i] - $midValue); } return $sum; } // Driver Code $arr = array(5, 11, 14, 10, 17, 15 ); $n = count($arr); echo absSumDidd($arr, $n); // This code is contributed // by anuj_67 ?> |
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Output:
20
Time Complexity: O(n Log n)
We can further optimize above solution to O(n) by using linear time algorithm for median finding.
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