Minimum Cost to make all array elements equal using given operations

• Difficulty Level : Expert
• Last Updated : 29 May, 2021

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:
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:

1. find the minimum of the M and A + R as M can make both operations simultaneously.
2. Then, store prefix sum in an array to find the sum in constant time.
3. Now for each element calculate the cost of making equal to the current element and find the minimum of them.
4. The smallest answer can also exist when we make all elements equal to the average of the array.
5. 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 using namespace std; // Function that returns the cost of// making all elements equal to current elementint 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 equalvoid 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 Codeint main(){    int arr[] = { 5, 5, 3, 6, 5 };     int A = 1, R = 2, M = 4;     int size = sizeof(arr) / sizeof(arr);     // Function Call    solve(arr, size, A, R, M);     return 0;}

Java

 // Java implementation to find the// minimum cost to make all array// elements equalimport 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 elementpublic 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 equalpublic 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 Codepublic 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 equaldef 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 elementdef 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 equaldef 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 =  * (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 Codeif __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 equalusing 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 elementpublic 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 equalpublic 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 Codepublic 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


Output:
4

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