Make all array elements equal with minimum cost

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++ 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
import java.io.*;
public 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.

Python3

# Python3 program to find minimum
# cost to make all elements equal

# Utility method to compute cost, when
# all values of array are made equal to X
def computeCost(arr, N, X):

    cost = 0
    for i in range(N):
        cost += abs(arr[i] - X)
    return cost


# Method to find minimum cost to
# make all elements equal
def minCostToMakeElementEqual(arr, N):

    low = high = arr[0]

    # Setting limits for ternary search
    # by smallest and largest element
    for i in range(N):

        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
        mid1 = low + (high - low) // 3
        mid2 = high - (high - low) // 3

        cost1 = computeCost(arr, N, mid1)
        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
arr = [1, 100, 101]
N = len(arr)
print(minCostToMakeElementEqual(arr, N))

# This code is contributed by Anant Agarwal.

C#

// C# Code to Make all array elements
// equal with minimum cost
using System;

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()
    {
        int[] arr = { 1, 100, 101 };
        int N = arr.Length;
        Console.Write(minCostToMakeElementEqual(arr, N));
    }
}

// This code is contributed by nitin mittal

Javascript

<script>

// Javascript program to find minimum cost to
// make all elements equal

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

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

        // setting limits for ternary search by
        // smallest and largest element
        for (let 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
            let mid1 = low + (high - low) / 3;
            let mid2 = high - (high - low) / 3;

            let cost1 = computeCost(arr, N, mid1);
            let 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 Math.round(computeCost(arr, N, (low + high) / 2));
    }


// Driver Code

    let arr = [ 1, 100, 101 ];
    let N = arr.length;
    document.write(minCostToMakeElementEqual(arr, N));

</script>

PHP

<?php
// PHP program to find minimum cost
// to make all elements equal

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

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

    // setting limits for ternary
    // search by smallest and
    // largest element
    for ($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
        $mid1 = $low + (floor($high - $low) / 3);
        $mid2 = $high - ($high - $low) / 3;

        $cost1 = computeCost($arr, $N, $mid1);
        $cost2 = computeCost($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
$arr = array( 1, 100, 101 );
$N = sizeof($arr) / sizeof($arr[0]);
echo minCostToMakeElementEqual($arr, $N);

// This code is contributed by nitin mittal.
?>

Output

100

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

Alternate Solution 
Think geometrically. Assume that array elements are co-ordinates on x axis. The problem reduces to finding another co-ordinate such that the sum of distances between this choice and other co-ordinates is minimized. 
Observe that: If number of coordinates are odd then y = middle element. If even then y is any number in between middle 2 co-ordinates. Say Input = [a, b, c, d]. Output is any number between b and c including both. Hence the cost is sum which can be computed easily now that we have chosen y. sum|(y-ai)| for all i. 

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// This function assumes that a[] is
// sorted. If a[] is not sorted, we need
// to sort it first.
int minCostToMakeElementEqual(int a[], int n)
{

    // If there are odd elements, we choose
    // middle element
    int y;
    if (n % 2 == 1)
        y = a[n / 2];

    // If there are even elements, then we choose
    // the average of middle two.
    else
        y = (a[n / 2] + a[(n - 2) / 2]) / 2;

    // After deciding the final value, find the
    // result.
    int s = 0;
    for(int i = 0; i < n; i++)
        s += abs(a[i] - y);

    return s;
}

// Driver code
int main()
{
    int a[] = { 1, 100, 101 };
    int n = sizeof(a) / sizeof(a[0]);
    sort(a, a + n);

    cout << (minCostToMakeElementEqual(a, n));
}

// This code is contributed by chitranayal

Java

import java.io.*;
import java.util.*;

class GFG{

// This function assumes that a[] is
// sorted. If a[] is not sorted, we need
// to sort it first.
public static int minCostToMakeElementEqual(int a[],
                                            int n)
{

    // If there are odd elements, we choose
    // middle element
    int y;

    if (n % 2 == 1)
        y = a[n / 2];

    // If there are even elements, then we
    // choose the average of middle two.
    else
        y = (a[n / 2] + a[(n - 2) / 2]) / 2;

    // After deciding the final value,
    // find the result.
    int s = 0;

    for(int i = 0; i < n; i++)
        s += Math.abs(a[i] - y);

    return s;
}

// Driver code
public static void main (String[] args)
{
    int a[] = { 1, 100, 101 };

      Arrays.sort(a);
    int n = a.length;

    System.out.println(minCostToMakeElementEqual(a, n));
}
}

// This code is contributed by parascoding

Python3

# This function assumes that a[] is
# sorted. If a[] is not sorted, we need
# to sort it first.
def minCostToMakeElementEqual(a):
    l = len(a)

    # If there are odd elements, we choose
    # middle element
    if (l%2 == 1):
        y = a[l//2]

    # If there are even elements, then we choose
    # the average of middle two.
    else:
        y = (a[l//2] + a[(l-2)//2])//2

    # After deciding the final value, find the
    # result.
    s = 0
    for i in range(l):
        s += abs(a[i]-y)
    return s

# Driver code
a = [1, 1700, 101]
a1 = sorted(a)
print(minCostToMakeElementEqual(a1))

C#

using System;
using System.Collections.Generic;

class GFG{

// This function assumes that a[] is
// sorted. If a[] is not sorted, we need
// to sort it first.
static int minCostToMakeElementEqual(int[] a,
                                     int n)
{

    // If there are odd elements, we choose
    // middle element
    int y;

    if (n % 2 == 1)
        y = a[n / 2];

    // If there are even elements, then we
    // choose the average of middle two.
    else
        y = (a[n / 2] + a[(n - 2) / 2]) / 2;

    // After deciding the final value,
    // find the result.
    int s = 0;

    for(int i = 0; i < n; i++)
        s += Math.Abs(a[i] - y);

    return s;
}

// Driver code
static void Main()
{
    int[] a = { 1, 100, 101 };
    int n = a.Length;
    Array.Sort(a);

    Console.WriteLine(
        minCostToMakeElementEqual(a, n));
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>


// This function assumes that a[] is
// sorted. If a[] is not sorted, we need
// to sort it first.
function minCostToMakeElementEqual( a, n)
{

    // If there are odd elements, we choose
    // middle element
    let y;
    if (n % 2 == 1)
        y = a[(Math.trunc(n / 2))];

    // If there are even elements, then we choose
    // the average of middle two.
    else
        y = Math.trunc((a[n / 2] + a[(n - 2) / 2]) / 2);

    // After deciding the final value, find the
    // result.
    let s = 0;
    for(let i = 0; i < n; i++)
        s += Math.abs(a[i] - y);

    return s;
}



    // Driver program

    let a = [ 1, 1030, 1101 ];
    let n = a.length;
    a.sort((a1, b1) => a1 - b1);

    document.write(minCostToMakeElementEqual(a, n));


</script>

Output

100

Time Complexity: O(nlogn) – Since we are using sorting
Auxiliary Space: O(1)

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