Make all array elements equal with minimum cost

3.5

Given an array which contains integer values, we need to make all values of this array equal to some integer value with minimum cost where the cost of changing an array value x to y is abs(x-y).
Examples:

Input  : arr[] = [1, 100, 101]
Output : 100
We can change all its values to 100 with minimum cost,
|1 - 100| + |100 - 100| + |101 - 100| = 100

Input  : arr[] = [4, 6]
Output : 2
We can change all its values to 5 with minimum cost,
|4 - 5| + |5 - 6| = 2

This problem can be solved by observing the cost while changing the target equal value, i.e. we will see the change in cost when target equal value is changed. It can be observed that, as we increase the target equal value the total cost decreases up to a limit and then starts increasing i.e. the cost graph with respect to target equal value is of U-shape and as cost graph is in U-shape, the ternary search can be applied to this search space and our goal is to get that bottom most point of the curve which will represent the smallest cost. We will make smallest and largest value of the array as the limit of our search space and then we will keep skipping 1/3 part of the search space until we reach to the bottom most point of our U-curve.
Please see below code for better understanding,

C++

// C/C++ program to find minimum cost to
// make all elements equal
#include <bits/stdc++.h>
using namespace std;

// Utility method to compute cost, when
// all values of array are made equal to X
int computeCost(int arr[], int N, int X)
{
    int cost = 0;
    for (int i = 0; i < N; i++)
        cost += abs(arr[i] - X);
    return cost;
}

//  Method to find minimum cost to make all
// elements equal
int minCostToMakeElementEqual(int arr[], int N)
{
    int low, high;
    low = high = arr[0];

    // setting limits for ternary search by
    // smallest and largest element
    for (int i = 0; i < N; i++)
    {
        if (low > arr[i])
            low = arr[i];
        if (high < arr[i])
            high = arr[i];
    }

    /* loop until difference between low and high
       become less than 3, because after that
       mid1 and mid2 will start repeating
    */
    while ((high - low) > 2)
    {
        // mid1 and mid2 are representative array
        // equal values of search space
        int mid1 = low + (high - low) / 3;
        int mid2 = high - (high - low) / 3;

        int cost1 = computeCost(arr, N, mid1);
        int cost2 = computeCost(arr, N, mid2);

        // if mid2 point gives more total cost,
        // skip third part
        if (cost1 < cost2)
            high = mid2;

        // if mid1 point gives more total cost,
        // skip first part
        else
            low = mid1;
    }

    // computeCost gets optimum cost by sending
    // average of low and high as X
    return computeCost(arr, N, (low + high) / 2);
}

//  Driver code to test above method
int main()
{
    int arr[] = {1, 100, 101};
    int N = sizeof(arr) / sizeof(int);
    cout << minCostToMakeElementEqual(arr, N);
    return 0;
}

Java

// JAVA Code for Make all array elements 
// equal with minimum cost
class GFG {
     
    // Utility method to compute cost, when
    // all values of array are made equal to X
    public static int computeCost(int arr[], int N, 
                                           int X)
    {
        int cost = 0;
        for (int i = 0; i < N; i++)
            cost += Math.abs(arr[i] - X);
        return cost;
    }
     
    //  Method to find minimum cost to make all
    // elements equal
    public static int minCostToMakeElementEqual(int arr[], 
                                                  int N)
    {
        int low, high;
        low = high = arr[0];
     
        // setting limits for ternary search by
        // smallest and largest element
        for (int i = 0; i < N; i++)
        {
            if (low > arr[i])
                low = arr[i];
            if (high < arr[i])
                high = arr[i];
        }
     
        /* loop until difference between low and high
           become less than 3, because after that
           mid1 and mid2 will start repeating
        */
        while ((high - low) > 2)
        {
            // mid1 and mid2 are representative array
            // equal values of search space
            int mid1 = low + (high - low) / 3;
            int mid2 = high - (high - low) / 3;
     
            int cost1 = computeCost(arr, N, mid1);
            int cost2 = computeCost(arr, N, mid2);
     
            // if mid2 point gives more total cost,
            // skip third part
            if (cost1 < cost2)
                high = mid2;
     
            // if mid1 point gives more total cost,
            // skip first part
            else
                low = mid1;
        }
     
        // computeCost gets optimum cost by sending
        // average of low and high as X
        return computeCost(arr, N, (low + high) / 2);
    }
    
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int arr[] = {1, 100, 101};
        int N = arr.length;
        System.out.println(minCostToMakeElementEqual(arr, N));
    }
  }
// This code is contributed by Arnav Kr. Mandal.


Output:

100

Time Complexity : O (n Log n)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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