# Minimum Increment / decrement to make array elements equal

• Difficulty Level : Easy
• Last Updated : 26 Apr, 2021

Given an array of integers where . In one operation you can either Increment/Decrement any element by 1. The task is to find the minimum operations needed to be performed on the array elements to make all array elements equal.
Examples

Input : A[] = { 1, 5, 7, 10 }
Output : 11
Increment 1 by 4, 5 by 0.
Decrement 7 by 2, 10 by 5.
New array A = { 5, 5, 5, 5 } with
cost of operations = 4 + 0 + 2 + 5 = 11.

Input : A = { 10, 2, 20 }
Output : 18

Approach:

1. Sort the array of Integers in increasing order.
2. Now, to make all elements equal with min cost. We will have to make the elements equal to the middle element of this sorted array. So, select the middle value, Let it be K.
Note: In case of even numbers of element, we will have to check for the costs of both middle elements and take minimum.
3. If A[i] < K, Increment the element by K – A[i].
4. If A[i] > K, Decrement the element by A[i] – K.
5. Update cost of each operation performed.

Below is the implementation of above approach:

## C++

 // C++ program to find minimum Increment or // decrement to make array elements equal #include  using namespace std;   // Function to return minimum operations need // to be make each element of array equal int minCost(int A[], int n) {     // Initialize cost to 0     int cost = 0;       // Sort the array     sort(A, A + n);       // Middle element     int K = A[n / 2];       // Find Cost     for (int i = 0; i < n; ++i)         cost += abs(A[i] - K);       // If n, is even. Take minimum of the     // Cost obtained by considering both     // middle elements     if (n % 2 == 0) {         int tempCost = 0;           K = A[(n / 2) - 1];           // Find cost again         for (int i = 0; i < n; ++i)             tempCost += abs(A[i] - K);           // Take minimum of two cost         cost = min(cost, tempCost);     }       // Return total cost     return cost; }   // Driver Code int main() {     int A[] = { 1, 6, 7, 10 };       int n = sizeof(A) / sizeof(A);       cout << minCost(A, n);       return 0; }

## Java

 // Java program to find minimum Increment or  // decrement to make array elements equal  import java.util.*; class GfG {    // Function to return minimum operations need  // to be make each element of array equal  static int minCost(int A[], int n)  {      // Initialize cost to 0      int cost = 0;        // Sort the array      Arrays.sort(A);        // Middle element      int K = A[n / 2];        // Find Cost      for (int i = 0; i < n; ++i)          cost += Math.abs(A[i] - K);        // If n, is even. Take minimum of the      // Cost obtained by considering both      // middle elements      if (n % 2 == 0) {          int tempCost = 0;            K = A[(n / 2) - 1];            // Find cost again          for (int i = 0; i < n; ++i)              tempCost += Math.abs(A[i] - K);            // Take minimum of two cost          cost = Math.min(cost, tempCost);      }        // Return total cost      return cost;  }    // Driver Code  public static void main(String[] args)  {      int A[] = { 1, 6, 7, 10 };        int n = A.length;        System.out.println(minCost(A, n));  } }

## Python3

 # Python3 program to find minimum Increment or # decrement to make array elements equal       # Function to return minimum operations need # to be make each element of array equal def minCost(A, n):           # Initialize cost to 0     cost = 0           # Sort the array     A.sort();           # Middle element     K = A[int(n / 2)]           #Find Cost     for i in range(0, n):         cost = cost + abs(A[i] - K)           # If n, is even. Take minimum of the     # Cost obtained by considering both     # middle elements     if n % 2 == 0:         tempCost = 0         K = A[int(n / 2) - 1]                   # FInd cost again         for i in range(0, n):             tempCost = tempCost + abs(A[i] - K)                   # Take minimum of two cost         cost = min(cost, tempCost)               # Return total cost     return cost       # Driver code A = [1, 6, 7, 10] n = len(A)   print(minCost(A, n))           # This code is contributed  # by Shashank_Sharma

## C#

 // C# program to find minimum Increment or // decrement to make array elements equal using System;   class GFG {       // Function to return minimum operations need // to be make each element of array equal static int minCost(int []A, int n) {     // Initialize cost to 0     int cost = 0;       // Sort the array     Array.Sort(A);       // Middle element     int K = A[n / 2];       // Find Cost     for (int i = 0; i < n; ++i)         cost += Math.Abs(A[i] - K);       // If n, is even. Take minimum of the     // Cost obtained by considering both     // middle elements     if (n % 2 == 0) {         int tempCost = 0;           K = A[(n / 2) - 1];           // Find cost again         for (int i = 0; i < n; ++i)             tempCost += Math.Abs(A[i] - K);           // Take minimum of two cost         cost = Math.Min(cost, tempCost);     }       // Return total cost     return cost; }   // Driver Code public static void Main(String[] args) {     int []A = new int []{ 1, 6, 7, 10 };       int n = A.Length;       Console.WriteLine(minCost(A, n)); } }

## PHP

 

## Javascript

 

Output:

10

Time Complexity: O(N*log(N))
Further Optimization We can find median in linear time and reduce time complexity to O(N)

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