将所有数组元素减少到零所需的最小子数组减量

给定由N个非负整数组成的数组arr [] ，任务是找到需要减少1的最小子数组数，以使所有数组元素均等于0。

例子：

Input: arr[] = {1, 2, 3, 2, 1}
Output: 3
Explanation: 
Operation 1: {1, 2, 3, 2, 1} -> {0, 1, 2, 1, 0} 
Operation 2: {0, 1, 2, 1, 0} -> {0, 0, 1, 0, 0} 
Operation 3: {0, 0, 1, 0, 0} -> {0, 0, 0, 0, 0}

Input: arr[] = {5, 4, 3, 4, 4}
Output: 6
Explanation: 
{5, 4, 3, 4, 4} -> {4, 3, 2, 3, 3} -> {3, 2, 1, 2, 2} -> {2, 1, 0, 1, 1} -> {2, 1, 0, 0, 0} -> {1, 0, 0, 0, 0} -> {0, 0, 0, 0, 0} 
 

方法：
可以通过从索引0遍历给定数组，找到直到索引i的答案来最佳地做到这一点，其中0≤i 。如果arr [i]≥arr [i + 1] ，则第(i +1)^个元素可以包含在^第i^个元素的每个子数组运算中，因此不需要额外的运算。如果arr [i] ，则第(i +1)^个元素可以包含在^第i^个元素的每个子数组操作中，并且在所有操作之后， arr [i + 1]变为arr [i + 1] -arr [i] 。因此，我们需要额外的arr [i + 1] -arr [i]运算以将其减小为零。

请按照以下步骤解决问题：

• 添加第一个元素arr [0]来回答，因为我们至少需要arr [0]才能使给定数组0 。
• 遍历索引[1，N-1] ，对于每个元素，检查它是否大于前一个元素。如果发现是正确的，则将它们的差添加到答案中。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to count the minimum
// number of subarrays that are
// required to be decremented by 1
int min_operations(vector& A)
{
    // Base Case
    if (A.size() == 0)
        return 0;
 
    // Initialize ans to first element
    int ans = A[0];
 
    for (int i = 1; i < A.size(); i++) {
 
        // For A[i] > A[i-1], operation
        // (A[i] - A[i - 1]) is required
        ans += max(A[i] - A[i - 1], 0);
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
int main()
{
    vector A{ 1, 2, 3, 2, 1 };
 
    cout << min_operations(A) << "\n";
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.io.*;
 
class GFG {
 
    // Function to count the minimum
    // number of subarrays that are
    // required to be decremented by 1
    static int min_operations(int A[], int n)
    {
        // Base Case
        if (n == 0)
            return 0;
 
        // Initializing ans to first element
        int ans = A[0];
        for (int i = 1; i < n; i++) {
 
            // For A[i] > A[i-1], operation
            // (A[i] - A[i - 1]) is required
            if (A[i] > A[i - 1]) {
                ans += A[i] - A[i - 1];
            }
        }
 
        // Return the count
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
        int A[] = { 1, 2, 3, 2, 1 };
        System.out.println(min_operations(A, n));
    }
}

Python

# Python Program to implement
# the above approach
 
# Function to count the minimum
# number of subarrays that are
# required to be decremented by 1
def min_operations(A):
 
    # Base case
    if len(A) == 0:
        return 0
 
    # Initializing ans to first element
    ans = A[0]
    for i in range(1, len(A)):
 
        if A[i] > A[i-1]:
            ans += A[i]-A[i-1]
 
    return ans
 
 
# Driver Code
A = [1, 2, 3, 2, 1]
print(min_operations(A))

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to count the minimum
// number of subarrays that are
// required to be decremented by 1
static int min_operations(int[] A, int n)
{
     
    // Base Case
    if (n == 0)
        return 0;
 
    // Initializing ans to first element
    int ans = A[0];
     
    for(int i = 1; i < n; i++)
    {
         
        // For A[i] > A[i-1], operation
        // (A[i] - A[i - 1]) is required
        if (A[i] > A[i - 1])
        {
            ans += A[i] - A[i - 1];
        }
    }
     
    // Return the count
    return ans;
}
 
// Driver Code
public static void Main()
{
    int n = 5;
    int[] A = { 1, 2, 3, 2, 1 };
     
    Console.WriteLine(min_operations(A, n));
}
}
 
// This code is contributed by bolliranadheer

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to count the minimum
// number of subarrays that are
// required to be decremented by 1
function min_operations(A)
{
     
    // Base Case
    if (A.length == 0)
        return 0;
 
    // Initialize ans to first element
    let ans = A[0];
 
    for(let i = 1; i < A.length; i++)
    {
         
        // For A[i] > A[i-1], operation
        // (A[i] - A[i - 1]) is required
        ans += Math.max(A[i] - A[i - 1], 0);
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
let A = [ 1, 2, 3, 2, 1 ];
document.write(min_operations(A));
 
// This code is contributed by subhammahato348
 

输出：

3

时间复杂度： O(N)
辅助空间： O(N)