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

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

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++

#include <bits/stdc++.h> using namespace std;    void minAbsDiff(int arr[], int n) {             sort(arr, arr + n);             cout << arr[n / 2] << endl; }    int main() {        int n = 5;     int arr[] = { 1, 2, 3, 4, 5 };        minAbsDiff(arr, n);        return 0; }

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

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