Minimum Cost to make all array elements equal using given operations

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Given an array arr[] of positive integers and three integers A, R, M, where

The cost of adding 1 to an element of the array is A,
the cost of subtracting 1 from an element of the array is R and
the cost of adding 1 to an element and subtracting 1 from another element simultaneously is M.

The task is to find the minimum total cost to make all the elements of the array equal.

Examples:

Input: arr[] = {5, 5, 3, 6, 5}, A = 1, R = 2, M = 4
Output: 4
Explanation:
Operation 1: Add two times to element 3, Array – {5, 5, 5, 6, 5}, Cost = 2
Operation 2: Subtract one time to element 6, Array – {5, 5, 5, 5, 5}, Cost = 4
Therefore, minimum cost is 4.

Input: arr[] = {5, 5, 3, 6, 5}, A = 1, R = 2, M = 2
Output: 3

Approach: The idea is to:

find the minimum of the M and A + R as M can make both operations simultaneously.
Then, store prefix sum in an array to find the sum in constant time.
Now for each element calculate the cost of making equal to the current element and find the minimum of them.
The smallest answer can also exist when we make all elements equal to the average of the array.
Therefore, at the end also compute the cost for making all elements equal to the approximate average sum of elements.

Below is the implementation of above approach:

C++

// C++ implementation to find the
// minimum cost to make all array
// elements equal
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns the cost of
// making all elements equal to current element
int costCalculation(
    int current, int arr[],
    int n, int pref[],
    int a, int r,
    int minimum)
{
 
    // Compute the lower bound
    // of current element
    int index
        = lower_bound(
              arr, arr + n, current)
          - arr;
 
    // Calculate the requirement
    // of add operation
    int left
        = index * current - pref[index];
 
    // Calculate the requirement
    // of subtract operation
    int right
        = pref[n] - pref[index]
          - (n - index)
                * current;
 
    // Compute minimum of left and right
    int res = min(left, right);
    left -= res;
    right -= res;
 
    // Computing the total cost of add
    // and subtract operations
    int total = res * minimum;
    total += left * a;
    total += right * r;
 
    return total;
}
 
// Function that prints minimum cost
// of making all elements equal
void solve(int arr[], int n,
           int a, int r, int m)
{
    // Sort the given array
    sort(arr, arr + n);
 
    // Calculate minimum from a + r and m
    int minimum = min(a + r, m);
 
    int pref[n + 1] = { 0 };
 
    // Compute prefix sum
    // and store in pref
    // array
    for (int i = 0; i < n; i++)
        pref[i + 1] = pref[i] + arr[i];
 
    int ans = 10000;
 
    // Find the minimum cost
    // from the given elements
    for (int i = 0; i < n; i++)
        ans
            = min(
                ans,
                costCalculation(
                    arr[i], arr, n,
                    pref, a, r,
                    minimum));
 
    // Finding the minimum cost
    // from the other cases where
    // minimum cost can occur
    ans
        = min(
            ans,
            costCalculation(
                pref[n] / n, arr,
                n, pref, a,
                r, minimum));
    ans
        = min(
            ans,
            costCalculation(
                pref[n] / n + 1,
                arr, n, pref,
                a, r, minimum));
 
    // Printing the minimum cost of making
    // all elements equal
    cout << ans << "\n";
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 5, 3, 6, 5 };
 
    int A = 1, R = 2, M = 4;
 
    int size = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    solve(arr, size, A, R, M);
 
    return 0;
}

Java

// Java implementation to find the
// minimum cost to make all array
// elements equal
import java.lang.*;
import java.util.*;
 
class GFG{
 
public static int lowerBound(int[] array, int length, 
                                          int value)
{
    int low = 0;
    int high = length;
     
    while (low < high)
    {
        final int mid = (low + high) / 2;
         
        // Checks if the value is less 
        // than middle element of the array
        if (value <= array[mid])
        {
            high = mid;
        }
        else
        {
            low = mid + 1;
        }
    }
    return low;
}
 
// Function that returns the cost of making 
// all elements equal to current element
public static int costCalculation(int current, int arr[],
                                  int n, int pref[],
                                  int a, int r, 
                                  int minimum)
{
 
    // Compute the lower bound
    // of current element
    int index = lowerBound(arr, arr.length, current);
 
    // Calculate the requirement
    // of add operation
    int left = index * current - pref[index];
 
    // Calculate the requirement
    // of subtract operation
    int right = pref[n] - 
                pref[index]- (n - index)* current;
 
    // Compute minimum of left and right
    int res = Math.min(left, right);
    left -= res;
    right -= res;
 
    // Computing the total cost of add
    // and subtract operations
    int total = res * minimum;
    total += left * a;
    total += right * r;
 
    return total;
}
 
// Function that prints minimum cost
// of making all elements equal
public static void solve(int arr[], int n,
                         int a, int r, int m)
{
     
    // Sort the given array
    Arrays.sort(arr);
 
    // Calculate minimum from a + r and m
    int minimum = Math.min(a + r, m);
 
    int []pref = new int [n + 1]; 
    Arrays.fill(pref, 0);
     
    // Compute prefix sum and 
    // store in pref array
    for(int i = 0; i < n; i++)
       pref[i + 1] = pref[i] + arr[i];
 
    int ans = 10000;
 
    // Find the minimum cost
    // from the given elements
    for(int i = 0; i < n; i++)
       ans = Math.min(ans, costCalculation(arr[i], arr, 
                                           n, pref, 
                                           a, r, minimum));
 
    // Finding the minimum cost
    // from the other cases where
    // minimum cost can occur
    ans = Math.min(ans, costCalculation(pref[n] / n, arr,
                                        n, pref, a, r,
                                        minimum));
    ans = Math.min(ans, costCalculation(pref[n] / n + 1, 
                                        arr, n, pref,
                                        a, r, minimum));
 
    // Printing the minimum cost of making
    // all elements equal
    System.out.println(ans);
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 5, 5, 3, 6, 5 };
    int A = 1, R = 2, M = 4;
    int size = arr.length ;
 
    // Function Call
    solve(arr, size, A, R, M);
}
}
 
// This code is contributed by SoumikMondal

Python3

# Python3 implementation to find the
# minimum cost to make all array
# elements equal
def lowerBound(array, length, value):
 
    low = 0
    high = length
     
    while (low < high):
        mid = (low + high) // 2
         
        # Checks if the value is less 
        # than middle element of the array
        if (value <= array[mid]):
            high = mid
        else:
            low = mid + 1
     
    return low
 
# Function that returns the cost of making 
# all elements equal to current element
def costCalculation(current, arr,
                    n, pref, a, r, minimum):
 
    # Compute the lower bound
    # of current element
    index = lowerBound(arr, len(arr), current)
 
    # Calculate the requirement
    # of add operation
    left = index * current - pref[index]
 
    # Calculate the requirement
    # of subtract operation
    right = (pref[n] - pref[index] -
            (n - index) * current)
 
    # Compute minimum of left and right
    res = min(left, right)
    left -= res
    right -= res
 
    # Computing the total cost of add
    # and subtract operations
    total = res * minimum
    total += left * a
    total += right * r
 
    return total
 
# Function that prints minimum cost
# of making all elements equal
def solve(arr, n, a, r, m):
     
    # Sort the given array
    arr.sort()
 
    # Calculate minimum from a + r and m
    minimum = min(a + r, m)
 
    pref = [0] * (n + 1)
     
    # Compute prefix sum and 
    # store in pref array
    for i in range(n):
        pref[i + 1] = pref[i] + arr[i]
 
    ans = 10000
 
    # Find the minimum cost
    # from the given elements
    for i in range(n):
        ans = min(ans, costCalculation(arr[i], arr, 
                                       n, pref, a, 
                                       r, minimum))
 
    # Finding the minimum cost
    # from the other cases where
    # minimum cost can occur
    ans = min(ans, costCalculation(pref[n] // n, 
                                   arr, n, pref, 
                                   a, r, minimum))
    ans = min(ans, costCalculation(pref[n] // n + 1, 
                                   arr, n, pref, a, 
                                   r, minimum))
 
    # Printing the minimum cost of making
    # all elements equal
    print(ans)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 5, 5, 3, 6, 5 ]
    A = 1
    R = 2
    M = 4
    size = len(arr)
 
    # Function call
    solve(arr, size, A, R, M)
 
# This code is contributed by chitranayal

C#

// C# implementation to find the
// minimum cost to make all array
// elements equal
using System;
 
class GFG{
 
public static int lowerBound(int[] array,
                             int length, 
                             int value)
{
    int low = 0;
    int high = length;
     
    while (low < high)
    {
        int mid = (low + high) / 2;
         
        // Checks if the value is less 
        // than middle element of the array
        if (value <= array[mid])
        {
            high = mid;
        }
        else
        {
            low = mid + 1;
        }
    }
    return low;
}
 
// Function that returns the cost of making 
// all elements equal to current element
public static int costCalculation(int current, 
                                  int []arr, int n,
                                  int []pref, int a,
                                  int r, int minimum)
{
 
    // Compute the lower bound
    // of current element
    int index = lowerBound(arr, arr.Length, current);
 
    // Calculate the requirement
    // of add operation
    int left = index * current - pref[index];
 
    // Calculate the requirement
    // of subtract operation
    int right = pref[n] - pref[index] -
                          (n - index) * 
                           current;
 
    // Compute minimum of left and right
    int res = Math.Min(left, right);
    left -= res;
    right -= res;
 
    // Computing the total cost of add
    // and subtract operations
    int total = res * minimum;
    total += left * a;
    total += right * r;
 
    return total;
}
 
// Function that prints minimum cost
// of making all elements equal
public static void solve(int []arr, int n,
                         int a, int r, int m)
{
     
    // Sort the given array
    Array.Sort(arr);
 
    // Calculate minimum from a + r and m
    int minimum = Math.Min(a + r, m);
 
    int []pref = new int [n + 1]; 
    Array.Fill(pref, 0);
     
    // Compute prefix sum and 
    // store in pref array
    for(int i = 0; i < n; i++)
        pref[i + 1] = pref[i] + arr[i];
 
    int ans = 10000;
 
    // Find the minimum cost
    // from the given elements
    for(int i = 0; i < n; i++)
        ans = Math.Min(ans, costCalculation(arr[i], arr, 
                                            n, pref, a, 
                                            r, minimum));
 
    // Finding the minimum cost
    // from the other cases where
    // minimum cost can occur
    ans = Math.Min(ans, costCalculation(pref[n] / n, arr,
                                        n, pref, a, r,
                                        minimum));
    ans = Math.Min(ans, costCalculation(pref[n] / n + 1, 
                                        arr, n, pref,
                                        a, r, minimum));
 
    // Printing the minimum cost of making
    // all elements equal
    Console.WriteLine(ans);
}
 
// Driver Code
public static void Main(string []args)
{
    int []arr = { 5, 5, 3, 6, 5 };
    int A = 1, R = 2, M = 4;
    int size = arr.Length ;
 
    // Function Call
    solve(arr, size, A, R, M);
}
}
 
// This code is contributed by SoumikMondal

Javascript

<script>
 
// javascript implementation to find the
// minimum cost to make all array
// elements equal    
function lowerBound(array, length, value) 
{
        var low = 0;
        var high = length;
 
        while (low < high) 
        {
             var mid = parseInt((low + high) / 2);
 
            // Checks if the value is less
            // than middle element of the array
            if (value <= array[mid])
            {
                high = mid;
            }
            else
            {
                low = mid + 1;
            }
        }
        return low;
    }
 
    // Function that returns the cost of making
    // all elements equal to current element
    function costCalculation(current , arr , n , pref , a , r , minimum) 
    {
 
        // Compute the lower bound
        // of current element
        var index = lowerBound(arr, arr.length, current);
 
        // Calculate the requirement
        // of add operation
        var left = index * current - pref[index];
 
        // Calculate the requirement
        // of subtract operation
        var right = pref[n] - pref[index] - (n - index) * current;
 
        // Compute minimum of left and right
        var res = Math.min(left, right);
        left -= res;
        right -= res;
 
        // Computing the total cost of add
        // and subtract operations
        var total = res * minimum;
        total += left * a;
        total += right * r;
 
        return total;
    }
 
    // Function that prints minimum cost
    // of making all elements equal
    function solve(arr , n , a , r , m) {
 
        // Sort the given array
        arr.sort();
 
        // Calculate minimum from a + r and m
        var minimum = Math.min(a + r, m);
 
        var pref = Array(n+1).fill(0);
 
 
        // Compute prefix sum and
        // store in pref array
        for (i = 0; i < n; i++)
            pref[i + 1] = pref[i] + arr[i];
 
        var ans = 10000;
 
        // Find the minimum cost
        // from the given elements
        for (i = 0; i < n; i++)
            ans = Math.min(ans, costCalculation(arr[i], arr, n, pref, a, r, minimum));
 
        // Finding the minimum cost
        // from the other cases where
        // minimum cost can occur
        ans = Math.min(ans, costCalculation(pref[n] / n, arr, n, pref, a, r, minimum));
        ans = Math.min(ans, costCalculation(pref[n] / n + 1, arr, n, pref, a, r, minimum));
 
        // Printing the minimum cost of making
        // all elements equal
        document.write(ans);
    }
 
    // Driver Code
        var arr = [ 5, 5, 3, 6, 5 ];
        var A = 1, R = 2, M = 4;
        var size = arr.length;
 
        // Function Call
        solve(arr, size, A, R, M);
 
// This code is contributed by Rajput-Ji.
</script>

Output:

Time Complexity: O(n log n), used for sorting
Auxiliary Space: O(n), as extra space is used of size n to create prefix array

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