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> |
3
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
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