根据给定步骤将所有数组元素减少到 1 所需的最小步骤

给定一个大小为N的数组arr[] 。任务是找到将所有数组元素减少到 1 所需的最小步骤。在每个步骤中，执行以下给定操作：

• 选择任何起始索引，比如i ，然后跳转到第 ( arr[i] + i ⁾索引，将第 i和第 ( arr [i] + ⁱ ) 索引减 1，按照此过程直到到达数组
• 如果一个元素已经减少到 1，它不能再减少，它保持不变。

例子：

Input: arr[] = {4, 2, 3, 2, 2, 2, 1, 2}, N = 8
Output: 5
Explanation: Series of operations can be performed in the following way:

• {4, 2, 3, 2, 2, 2, 1, 2}, decrement values by 1, arr[] = {4, 2, 2, 2, 2, 1, 1, 1}
• {4, 2, 2, 2, 2, 1, 1, 1}, decrement values by 1, arr[] = {3, 2, 2, 2, 1, 1, 1, 1}
• {3, 2, 2, 2, 1, 1, 1, 1}, decrement values by 1, arr[] = {2, 2, 2, 1, 1, 1, 1, 1}
• {2, 2, 2, 1, 1, 1, 1, 1}, decrement values by 1, arr[] = {1, 2, 1, 1, 1, 1, 1, 1}
• {1, 2, 1, 1, 1, 1, 1, 1}, decrement values by 1, arr[] = {1, 1, 1, 1, 1, 1, 1, 1}

So, total steps required = 5

Input: arr[] = {1, 3, 1, 2, 2}, N = 5
Output: 2

方法：给定的问题可以通过将问题分为 4 个部分来解决：-（0 到 i-1） |我| (i + 1) | （i + 2 到 n – 1）。请按照以下步骤解决问题：

1. 拿一个向量说v ，它表示一个元素由于访问之前的元素而减少了多少次。
2. v 中的每个元素，即v[i]表示由于访问从0到 ( i-1 ) 的元素而导致的arr[i]中的递减计数。
3. 遍历数组arr[] ，并取一个变量k来存储由于元素arr[i]在使0到 ( i-1 ) 元素等于1之后必须添加到答案中的遍数。
4. 在循环内部，运行另一个循环（即第二个循环）来计算当前arr[i]元素对 ( i+2 ) 到N个元素的影响（访问第 i 个元素后减少）。

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 steps
// required to reduce all array
// elements to 1
int minSteps(int arr[], int N)
{
    // Variable to store the answer
    int steps = 0;
 
    // Vector with all elements initialized
    // with 0
    vector v(N + 1, 0);
 
    // Traverse the array
    for (int i = 0; i < N; ++i) {
        // Variable to store the numbers of
        // passes that have to add in answer
        // due to element arr[i] after making
        // 0 to (i-1) elements equal to 1
        int k = max(0ll, arr[i] - 1 - v[i]);
 
        // Increment steps by K
        steps += k;
        // Update element in v
        v[i] += k;
 
        // Loop to compute how much current element
        // effect the (i+2) to N elements
        for (int j = i + 2; j <= min(i + arr[i], N); j++) {
            v[j]++;
        }
        v[i + 1] += v[i] - arr[i] + 1;
    }
    // Return steps which is the answer
    return steps;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 2, 3, 2, 2, 2, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << minSteps(arr, N);
 
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG
{
   
    // Function to find minimum steps
    // required to reduce all array
    // elements to 1
    static int minSteps(int arr[], int N)
    {
       
        // Variable to store the answer
        int steps = 0;
 
        // Vector with all elements initialized
        // with 0
        int v[] = new int[N + 1];
 
        // Traverse the array
        for (int i = 0; i < N; ++i)
        {
           
            // Variable to store the numbers of
            // passes that have to add in answer
            // due to element arr[i] after making
            // 0 to (i-1) elements equal to 1
            int k = Math.max(0, arr[i] - 1 - v[i]);
 
            // Increment steps by K
            steps += k;
            // Update element in v
            v[i] += k;
 
            // Loop to compute how much current element
            // effect the (i+2) to N elements
            for (int j = i + 2;
                 j <= Math.min(i + arr[i], N); j++) {
                v[j]++;
            }
            v[i + 1] += v[i] - arr[i] + 1;
        }
        // Return steps which is the answer
        return steps;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 4, 2, 3, 2, 2, 2, 1, 2 };
        int N = arr.length;
 
        System.out.println(minSteps(arr, N));
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python program for the above approach
 
# Function to find minimum steps
# required to reduce all array
# elements to 1
def minSteps(arr, N):
   
    # Variable to store the answer
    steps = 0
 
    # Vector with all elements initialized
    # with 0
    v = [0] * (N + 1)
 
    # Traverse the array
    for i in range(N):
       
        # Variable to store the numbers of
        # passes that have to add in answer
        # due to element arr[i] after making
        # 0 to (i-1) elements equal to 1
        k = max(0, arr[i] - 1 - v[i])
 
        # Increment steps by K
        steps += k
        # Update element in v
        v[i] += k
 
        # Loop to compute how much current element
        # effect the (i+2) to N elements
        for j in range(i + 2, min(i + arr[i], N) + 1):
            v[j] += 1
        v[i + 1] += v[i] - arr[i] + 1
 
    # Return steps which is the answer
    return steps
 
# Driver Code
arr = [4, 2, 3, 2, 2, 2, 1, 2]
N = len(arr)
 
print(minSteps(arr, N))
 
 # This code is contributed by gfgking.

C#

// C# code for the above approach
using System;
class GFG
{
   
    // Function to find minimum steps
    // required to reduce all array
    // elements to 1
    static int minSteps(int[] arr, int N)
    {
       
        // Variable to store the answer
        int steps = 0;
 
        // Vector with all elements initialized
        // with 0
        int[] v = new int[N + 1];
 
        // Traverse the array
        for (int i = 0; i < N; ++i)
        {
           
            // Variable to store the numbers of
            // passes that have to add in answer
            // due to element arr[i] after making
            // 0 to (i-1) elements equal to 1
            int k = Math.Max(0, arr[i] - 1 - v[i]);
 
            // Increment steps by K
            steps += k;
           
            // Update element in v
            v[i] += k;
 
            // Loop to compute how much current element
            // effect the (i+2) to N elements
            for (int j = i + 2;
                 j <= Math.Min(i + arr[i], N); j++) {
                v[j]++;
            }
            v[i + 1] += v[i] - arr[i] + 1;
        }
       
        // Return steps which is the answer
        return steps;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 4, 2, 3, 2, 2, 2, 1, 2 };
        int N = arr.Length;
 
        Console.Write(minSteps(arr, N));
    }
}
 
// This code is contributed by gfgking

Javascript

    // JavaScript program for the above approach
 
    // Function to find minimum steps
    // required to reduce all array
    // elements to 1
    const minSteps = (arr, N) => {
        // Variable to store the answer
        let steps = 0;
 
        // Vector with all elements initialized
        // with 0
        let v = new Array(N + 1).fill(0);
 
        // Traverse the array
        for (let i = 0; i < N; ++i) {
            // Variable to store the numbers of
            // passes that have to add in answer
            // due to element arr[i] after making
            // 0 to (i-1) elements equal to 1
            let k = Math.max(0, arr[i] - 1 - v[i]);
 
            // Increment steps by K
            steps += k;
            // Update element in v
            v[i] += k;
 
            // Loop to compute how much current element
            // effect the (i+2) to N elements
            for (let j = i + 2; j <= Math.min(i + arr[i], N); j++) {
                v[j]++;
            }
            v[i + 1] += v[i] - arr[i] + 1;
        }
        // Return steps which is the answer
        return steps;
    }
 
    // Driver Code
    let arr = [4, 2, 3, 2, 2, 2, 1, 2];
    let N = arr.length
 
    document.write(minSteps(arr, N));
 
    // This code is contributed by rakeshsahni
 

输出：

5

时间复杂度： O(N ² )
辅助空间： O(N)