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Find the array element having minimum sum of absolute differences with all other array elements

  • Difficulty Level : Easy
  • Last Updated : 18 Jun, 2021

Given an array arr[] of size N, the task is to find the minimum sum of absolute differences of an array element with all elements of another array.

Input: arr[ ] = {1, 2, 3, 4, 5}, N = 5
Output: 3
Explanation: 
For arr[0](= 1): Sum = abs(2 – 1) + abs(3 – 1) + abs(4 – 1) + abs(5 – 1) = 10.
For arr[1](= 2): Sum = abs(2 – 1) + abs(3 – 2) + abs(4 – 2) + abs(5 – 2) = 7.
For arr[2](= 3): Sum = abs(3 – 1) + abs(3 – 2) + abs(4 – 3) + abs(5 – 3) = 6 (Minimum).
For arr[3](= 4): Sum = abs(4 – 1) + abs(4 – 2) + abs(4 – 3) + abs(5 – 4) = 7.
For arr[0](= 1): Sum = 10.

Input: arr[ ] = {1, 2, 3, 4}, N = 4
Output: 2

Approach: The problem can be solved based on the observation that the sum of absolute differences of all array elements is minimum for the median of the array. Follow the steps below to solve the problem:

  • Sort the array arr[].
  • Print the middle element of the sorted array as the required answer.

Below is the implementation of the above approach:



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the array element
// having minimum sum of absolute
// differences with other array elements
void minAbsDiff(int arr[], int n)
{
 
    // Sort the array
    sort(arr, arr + n);
 
    // Print the middle element
    cout << arr[n / 2] << endl;
}
 
// Driver Code
int main()
{
 
    int n = 5;
    int arr[] = { 1, 2, 3, 4, 5 };
 
    minAbsDiff(arr, n);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.Arrays;
public class GFG
{
   
  // Function to return the array element
  // having minimum sum of absolute
  // differences with other array elements
  static void minAbsDiff(int arr[], int n)
  {
 
    // Sort the array
    Arrays.sort(arr);
 
    // Print the middle element
    System.out.println(arr[n / 2]);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int n = 5;
    int arr[] = { 1, 2, 3, 4, 5 };
    minAbsDiff(arr, n);
  }
}
 
// This code is contributed by abhinavjain194

Python3




# Python3 program for the above approach
 
# Function to return the array element
# having minimum sum of absolute
# differences with other array elements
def minAbsDiff(arr, n):
 
    # Sort the array
    arr.sort()
 
    # Print the middle element
    print(arr[n // 2])
 
# Driver Code
if __name__ == '__main__':
 
    n = 5
    arr = [ 1, 2, 3, 4, 5 ]
     
    minAbsDiff(arr, n)
 
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to return the array element
// having minimum sum of absolute
// differences with other array elements
static void minAbsDiff(int []arr, int n)
{
     
    // Sort the array
    Array.Sort(arr);
     
    // Print the middle element
    Console.WriteLine(arr[n / 2]);
}
 
// Driver code
public static void Main(string[] args)
{
    int n = 5;
    int []arr = { 1, 2, 3, 4, 5 };
     
    minAbsDiff(arr, n);
}
}
 
// This code is contributed by ankThon

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to return the array element
// having minimum sum of absolute
// differences with other array elements
function minAbsDiff(arr, n){
 
    // Sort the array
    arr.sort(function(a, b){return a-b})
 
    // Print the middle element
    document.write(arr[Math.floor(n / 2)])
    }
     
     
// main code
let n = 5
let arr = [ 1, 2, 3, 4, 5 ]
     
minAbsDiff(arr, n)
 
// This code is contributed by AnkThon
 
</script>
Output: 
3

 

Time Complexity: O(NlogN) 
Auxiliary Space: O(1)
 

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