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📜  将所有数组元素减少为零所需的子数组的最小减量

📅  最后修改于: 2021-09-03 03:24:42             🧑  作者: Mango

给定一个由N 个非负整数组成的数组arr[] ,任务是找到需要减少 1 以使所有数组元素等于 0 的最小子数组数。

例子:

方法:
这可以通过从索引0遍历给定数组,找到直到索引i的答案来最佳地完成,其中0 ≤ i < N 。如果arr[i] ≥ arr[i+1] ,则第(i + 1)元素可以包含在i元素的每个子数组操作中,因此不需要额外的操作。如果arr[i] < arr[i + 1] ,则第(i + 1)元素可以包含在i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arr[i+1]-arr[i]额外的操作来将其归零。

请按照以下步骤解决问题:

  • 添加第一个元素arr[0]回答,因为我们至少需要arr[0]来使给定的数组0
  • 遍历索引[1, N-1]并且对于每个元素,检查它是否大于前一个元素。如果发现是真的,请将它们的差异添加到答案中

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


输出:

3

时间复杂度: O(N)
辅助空间: O(N)

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