使 Array 最多为 0 的最小递减量，使得所有数组元素在一个数字减少到 0 后循环递减

给定一个包含N个整数的循环数组arr[]和一个数组cost[] ，任务是计算使数组的所有元素都等于0所需的最小操作数，其中在每个操作中递减索引i的值由1 。如果索引的值变为 0，则将arr[i+1]的值减少cost[i] ，将arr[i + 2]的值减少cost[i + 1]等等。

注意：如果一个元素变得小于 0，则认为它是 0。

例子：

Input: arr[] = {7, 2, 5}, cost[] = {8, 9, 3}
Output: 6
Explanation: Decrements can be made in the following way:

Decrement the value of arr[1] twice. Hence, the value of arr[2] will be decremented by cost[1] and the value of arr[0] will be decremented by cost[2]. Therefore, the final array will be arr[] = {4, 0, 0}.
Now, decrement the value of arr[0], 4 times to make it 0. Therefore, the array becomes arr[] = {0, 0, 0}.

Hence, the number of required operations to make all the elements of the array equal to zero is 6 which is the minimum possible.

Input: arr[] = {6, 7, 7, 10, 8, 2}, cost[] = {5, 10, 1, 4, 7, 7}
Output: 16

编程需要懂一点英语

方法：给定的问题可以通过以下观察使用贪心方法来解决：

数组arr[]的最后一个剩余非零元素，假设x将需要x减量操作。
假设在减量操作后， arr[i]的值变为0 ，那么对于arr[i] ，所需的减量数为arr[i] ，对于arr[i+1] ，所需的减量数为arr [i+1] – max(0, arr[i+1] – cost[i])等等。

以下是要遵循的步骤：

使用变量i遍历[0, N)范围内的给定数组arr[] 。
使数组的第 i^个索引等于 0 所需的操作数是arr[i] – min(arr[i], cost[i-1]) 。因此，在变量ans中的所有索引上保持该值的总和。
在[0, N)范围内的所有索引上，将ans的值增加 arr[i] 或 cost[i] 的最小值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find minimum decrements
// required to make all array elements 0
int minDecrements(int arr[], int powr[], int N)
{
    // Variable to store the answer
    int ans = 0;
    int mn = INT_MAX;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int idx = (i + 1) % N;
        int val = min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 7, 7, 10, 8, 2 };
    int powr[] = { 5, 10, 1, 4, 7, 7 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << minDecrements(arr, powr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
public class GFG
{
// Function to find minimum decrements
// required to make all array elements 0
static int minDecrements(int []arr, int []powr, int N)
{
    // Variable to store the answer
    int ans = 0;
    int mn = Integer.MAX_VALUE;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int idx = (i + 1) % N;
        int val = Math.min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = Math.min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int []arr = { 6, 7, 7, 10, 8, 2 };
    int []powr = { 5, 10, 1, 4, 7, 7 };
    int N = arr.length;
 
    System.out.println(minDecrements(arr, powr, N));
}
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
 
# Function to find minimum decrements
# required to make all array elements 0
def minDecrements(arr, powr, N):
 
    # Variable to store the answer
    ans = 0
    mn = 99999999
 
    # Traverse the array
    for i in range(N):
        idx = (i + 1) % N
        val = min(arr[idx], powr[i])
 
        ans += arr[idx] - val
 
       # Store the minimum one
        mn = min(mn, val)
    ans += mn
 
    # Return the ans
    return ans
 
# Driver Code
if __name__ == "__main__":
    arr = [6, 7, 7, 10, 8, 2]
    powr = [5, 10, 1, 4, 7, 7]
    N = len(arr)
    print(minDecrements(arr, powr, N))
 
# This code is contributed by Potta Lokesh

C#

// C# program for the above approach
using System;
using System.Collections;
 
class GFG
{
// Function to find minimum decrements
// required to make all array elements 0
static int minDecrements(int []arr, int []powr, int N)
{
    // Variable to store the answer
    int ans = 0;
    int mn = Int32.MaxValue;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int idx = (i + 1) % N;
        int val = Math.Min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = Math.Min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 6, 7, 7, 10, 8, 2 };
    int []powr = { 5, 10, 1, 4, 7, 7 };
    int N = arr.Length;
 
    Console.Write(minDecrements(arr, powr, N));
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

时间复杂度： O(N)
辅助空间： O(1)