通过重复删除给定操作获得的最大值来清空给定数组所需的成本

给定一个由N 个整数组成的数组arr[] ，任务是在以指定的顺序执行以下操作任意次数后，找到删除所有数组元素的成本：

• 将给定数组中存在的最大元素添加到成本中。
• 从给定数组中删除最大元素。
• 将给定数组中的所有整数减1 。
• 删除减少为 0 的元素。

例子：

Input: arr[] = {1, 6, 7, 4, 2, 5, 3}
Output: 16
Explanation: 
Step 1: Current maximum element = 7. Removing 7 and reducing each element by 1 modifies the array to {0, 5, 3, 1, 4, 2}. Cost = 7. 
Step 2: Current maximum element = 5. Removing 5 and reducing each element by 1 modifies the array to {2, 0, 3, 1}. Cost = 7 + 5 = 12. 
Step 3: Current maximum = 3. Removing 3 and reducing each element by 1 modifies the array to {1, 1}. Cost = 12 + 3 = 15 
Step 4: Final cost obtained = 15 + 1 = 16

Input: arr[] = {6, 4, 6, 1}
Output: 13

朴素的方法：解决问题的最简单的方法如下：

• 从给定数组中找到最大值
• 从数组中删除后将其添加到成本中。
• 删除最大元素后，将所有剩余的数组元素减少 1。删除所有减少为 0 的元素。
• 重复上述步骤，直到删除所有数组元素。
• 最后，打印cost的值。

时间复杂度： O(N ² )，其中 N 是给定数组的大小。
辅助空间： O(1)

有效的方法：为了优化上述方法，其思想是观察数组中每个^第i 大的元素都添加arr[i] – i次到成本中，其中 i 是每个元素arr[i]的索引。请按照以下步骤解决上述问题：

1. 按降序对给定数组进行排序。
2. 遍历数组，并添加值ARR [Ⅰ] -我对于每个阵列元素到成本，如果它是大于0。
3. 完成上述步骤后，打印cost的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the total cost
// of removing all array elements
int findCost(int* a, int n)
{
    // Sort the array in descending order
    sort(a, a + n, greater());
 
    // Stores the total cost
    int count = 0;
 
    for (int j = 0; j < n; j++) {
 
        // Contribution of i-th
        // greatest element to the cost
        int p = a[j] - j;
 
        // Remove the element
        a[j] = 0;
 
        // If negative
        if (p < 0) {
            p = 0;
            continue;
        }
 
        // Add to the final cost
        count += p;
    }
 
    // Return the cost
    return count;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 6, 7, 4, 2, 5, 3 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << findCost(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the total cost
// of removing all array elements
static int findCost(Integer [] a, int n)
{
    // Sort the array in descending order
     Arrays.sort(a, Collections.reverseOrder());
 
    // Stores the total cost
    int count = 0;
 
    for (int j = 0; j < n; j++)
    {
        // Contribution of i-th
        // greatest element to the cost
        int p = a[j] - j;
 
        // Remove the element
        a[j] = 0;
 
        // If negative
        if (p < 0)
        {
            p = 0;
            continue;
        }
 
        // Add to the final cost
        count += p;
    }
 
    // Return the cost
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    Integer arr[] = {1, 6, 7,
                     4, 2, 5, 3};
 
    int N = arr.length;
 
    // Function Call
    System.out.print(findCost(arr, N));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
 
# Function to find the total cost
# of removing all array elements
def findCost(a, n):
 
    # Sort the array in descending order
    a.sort(reverse = True)
 
    # Stores the total cost
    count = 0
 
    for j in range(n):
 
        # Contribution of i-th
        # greatest element to the cost
        p = a[j] - j
 
        # Remove the element
        a[j] = 0
 
        # If negative
        if(p < 0):
            p = 0
            continue
 
        # Add to the final cost
        count += p
 
    # Return the cost
    return count
 
# Driver Code
 
# Given array arr[]
arr = [ 1, 6, 7, 4, 2, 5, 3 ]
 
N = len(arr)
 
# Function call
print(findCost(arr, N))
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find the total cost
// of removing all array elements
static int findCost(int []a, int n)
{
    // Sort the array in descending order
    Array.Sort(a);
    Array.Reverse(a);
 
    // Stores the total cost
    int count = 0;
 
    for (int j = 0; j < n; j++)
    {
        // Contribution of i-th
        // greatest element to the cost
        int p = a[j] - j;
 
        // Remove the element
        a[j] = 0;
 
        // If negative
        if (p < 0)
        {
            p = 0;
            continue;
        }
 
        // Add to the readonly cost
        count += p;
    }
 
    // Return the cost
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int []arr = {1, 6, 7,
                 4, 2, 5, 3};
 
    int N = arr.Length;
 
    // Function Call
    Console.Write(findCost(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

      // JavaScript program for the above approach
      // Function to find the total cost
      // of removing all array elements
      function findCost(a, n) {
        // Sort the array in descending order
        a.sort((x, y) => y - x);
         
        // Stores the total cost
        var count = 0;
 
        for (var j = 0; j < n; j++) {
          // Contribution of i-th
          // greatest element to the cost
          var p = a[j] - j;
 
          // Remove the element
          a[j] = 0;
 
          // If negative
          if (p < 0) {
            p = 0;
            continue;
          }
          // Add to the final cost
          count += p;
        }
        // Return the cost
        return count;
      }
 
      // Driver Code
      // Given array arr[]
      var arr = [1, 6, 7, 4, 2, 5, 3];
      var N = arr.length;
      // Function Call
      document.write(findCost(arr, N));
    

输出：

16

时间复杂度： O(NlogN)，其中 N 是给定数组的大小。
辅助空间： O(1)

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