使用给定操作使所有数组元素相等的最低成本

给定一个正整数和三个整数A ， R ， M的数组arr [] ，其中

• 将1加到数组元素的成本为A ，
• 从数组的元素中减去1的成本为R和
• 将一个元素加1并同时从另一个元素减1的成本为M。

我们的任务是找到使阵列中所有元素相等的最小总成本。
例子：

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: Substract 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

方法：想法是：

1. 找到M和A + R的最小值，因为M可以同时进行两个操作。
2. 然后，将前缀和存储在数组中以找到恒定时间的和。
3. 现在，为每个元素计算与当前元素相等的成本，并找到它们中的最小值。
4. 当我们使所有元素等于数组的平均值时，最小的答案也可能存在。
5. 因此，最后还要计算使所有元素等于元素的近似平均和的成本。

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum cost to make all array
// elements equal
  
#include 
using namespace std;
  
// Function that returns the cost of
// makeing 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];
  
    // Calcuate 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);
  
    // Calcuate 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 makeing 
// 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];
  
    // Calcuate 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);
  
    // Calcuate 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 makeing 
# 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]
  
    # Calcuate 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()
  
    # Calcuate 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 makeing 
// 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];
  
    // Calcuate 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);
  
    // Calcuate 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

输出：

4