使数组不递减所需的非递减子数组的最小增量

给定一个由N个整数组成的数组arr [] ，任务是找到使数组不递减所需的最少操作数，其中，每个操作都涉及将给定数组中非递减子数组的所有元素增加1 。

例子：

Input: arr[] = {1, 3, 1, 2, 4} 
Output: 2 
Explanation: 
Operation 1: Incrementing arr[2] modifies array to {1, 3, 2, 2, 4} 
Operation 2: Incrementing subarray {arr[2], arr[3]} modifies array to {1, 3, 3, 3, 4} 
Therefore, the final array is non-decreasing.

Input: arr[] = {1, 3, 5, 10} 
Output: 0 
Explanation: The array is already non-decreasing.

方法：请按照以下步骤解决问题：

• 如果该数组已经是一个非递减数组，则无需进行任何更改。
• 否则，对于0≤i 任何索引i ，如果arr [i]> arr [i + 1] ，则将差值添加到ans 。
• 最后打印答的答案。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
  
// Function to return to the minimum
// number of operations required to
// make the array non-decreasing
int getMinOps(int arr[], int n)
{
      
    // Stores the count of operations
    int ans = 0;
    for(int i = 0; i < n - 1; i++)
    {
  
        // If arr[i] > arr[i + 1], add
        // arr[i] - arr[i + 1] to the answer
        // Otherwise, add 0
        ans += max(arr[i] - arr[i + 1], 0);
    }
    return ans;
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 3, 1, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
      
    cout << (getMinOps(arr, n));
}
  
// This code is contributed by PrinciRaj1992

Java

// Java Program to implement the
// above approach
import java.io.*;
import java.util.*;
  
class GFG {
  
    // Function to return to the minimum
    // number of operations required to
    // make the array non-decreasing
    public static int getMinOps(int[] arr)
    {
        // Stores the count of operations
        int ans = 0;
        for (int i = 0; i < arr.length - 1; i++) {
  
            // If arr[i] > arr[i + 1], add
            // arr[i] - arr[i + 1] to the answer
            // Otherwise, add 0
            ans += Math.max(arr[i] - arr[i + 1], 0);
        }
  
        return ans;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 3, 1, 2, 4 };
  
        System.out.println(getMinOps(arr));
    }
}

Python3

# Python3 program to implement
# the above approach
  
# Function to return to the minimum 
# number of operations required to
# make the array non-decreasing
def getMinOps(arr):
  
    # Stores the count of operations
    ans = 0
      
    for i in range(len(arr) - 1):
  
        # If arr[i] > arr[i + 1], add
        # arr[i] - arr[i + 1] to the answer
        # Otherwise, add 0
        ans += max(arr[i] - arr[i + 1], 0)
  
    return ans
  
# Driver Code
  
# Given array arr[]
arr = [ 1, 3, 1, 2, 4 ]
  
# Function call
print(getMinOps(arr))
  
# This code is contributed by Shivam Singh

C#

// C# Program to implement the
// above approach
using System;
class GFG
{
  
  // Function to return to the minimum
  // number of operations required to
  // make the array non-decreasing
  public static int getMinOps(int[] arr)
  {
    // Stores the count of operations
    int ans = 0;
    for (int i = 0; i < arr.Length - 1; i++) 
    {
  
      // If arr[i] > arr[i + 1], add
      // arr[i] - arr[i + 1] to the answer
      // Otherwise, add 0
      ans += Math.Max(arr[i] - arr[i + 1], 0);
    }
    return ans;
  }
  
  // Driver Code
  public static void Main(String[] args)
  {
    int[] arr = { 1, 3, 1, 2, 4 };
  
    Console.WriteLine(getMinOps(arr));
  }
}
  
// This code is contributed by Amit Katiyar

输出：

2

时间复杂度： O(N)
辅助空间： O(1)