通过递减 arr[0] 并反复移动到 end 将 arr[K] 更改为 0 的最小步骤

给定一个大小为N的数组arr[]和一个表示索引K 的整数，任务是找到arr[K]变为0的最小操作次数。在一次操作中，第一个数组元素的值减1并走到数组的末尾。如果在任何时候， arr[i]变为0，则将其从数组中删除，并对剩余元素执行操作。

例子：

Input:  arr[] = {2, 3, 2}, K = 2
Output: 6
Explanation: For the first input,  
After iteration-1, the array changes to [1, 2, 1]. Steps taken = 3 steps
After iteration-2, the array changes to [0, 1, 0]. Steps taken = 3 steps
Hence, for the element at index 2, it took 6 steps to become 0.

Input:  arr[] = {5, 1, 1, 1}, K = 0
Output: 8

方法：思路是不断遍历数组，当arr[i]大于0时，减小arr[i]的值，计算答案。请按照以下步骤解决问题：

• 将变量time初始化为0以存储答案。
• 在while循环中遍历直到arr[k]不为0并执行以下任务：
  • 使用变量i遍历范围[0, N)并执行以下任务：
    • 如果arr[i]大于0，则将arr[i]的值减1，然后将time的值加1。
    • 如果arr[k]变为0，则中断。
• 执行上述步骤后，打印时间值作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum number
// of steps
void findMinimumNumberOfSteps(vector arr,
                              int K)
{
 
    // Variable to store the answer
    int time = 0;
 
    // Traverse in the while loop
    while (arr[K] != 0) {
 
        // Iterate over the loop
        for (int i = 0; i < arr.size(); i++) {
 
            // Check the condition and
            // decrease the value
            if (arr[i] > 0) {
                arr[i] -= 1;
                time++;
            }
 
            // Break the loop
            if (arr[K] == 0)
                break;
        }
    }
 
    // Print the result
    cout << time;
}
 
// Driver Code
int main()
{
    vector arr = { 2, 3, 2 };
    int K = 2;
    findMinimumNumberOfSteps(arr, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the minimum number
  // of steps
  static void findMinimumNumberOfSteps(int arr[],
                                       int K)
  {
 
    // Variable to store the answer
    int time = 0;
 
    // Traverse in the while loop
    while (arr[K] != 0) {
 
      // Iterate over the loop
      for (int i = 0; i < arr.length; i++) {
 
        // Check the condition and
        // decrease the value
        if (arr[i] > 0) {
          arr[i] -= 1;
          time++;
        }
 
        // Break the loop
        if (arr[K] == 0)
          break;
      }
    }
 
    // Print the result
    System.out.println(time);
  }
 
  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 2, 3, 2 };
    int K = 2;
    findMinimumNumberOfSteps(arr, K);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python program to implement
# the above approach
 
# Function to find the minimum number
# of steps
def findMinimumNumberOfSteps(arr, K) :
 
    # Variable to store the answer
    time = 0
 
    # Traverse in the while loop
    while (arr[K] != 0) :
 
        # Iterate over the loop
        for i in range(0, len(arr)) :
             
            # Check the condition and
            # decrease the value
            if (arr[i] > 0) :
                arr[i] -= 1
                time += 1
             
 
            # Break the loop
            if (arr[K] == 0):
                break
 
    # Print the result
    print(time)
 
# Driver Code
arr = [ 2, 3, 2 ]
K = 2
findMinimumNumberOfSteps(arr, K)
 
# This code is contributed by sanjoy_62.

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to find the minimum number
  // of steps
  static void findMinimumNumberOfSteps(int []arr,
                                       int K)
  {
 
    // Variable to store the answer
    int time = 0;
 
    // Traverse in the while loop
    while (arr[K] != 0) {
 
      // Iterate over the loop
      for (int i = 0; i < arr.Length; i++) {
 
        // Check the condition and
        // decrease the value
        if (arr[i] > 0) {
          arr[i] -= 1;
          time++;
        }
 
        // Break the loop
        if (arr[K] == 0)
          break;
      }
    }
 
    // Print the result
    Console.WriteLine(time);
  }
 
  // Driver Code
  public static void Main () {
    int []arr = { 2, 3, 2 };
    int K = 2;
    findMinimumNumberOfSteps(arr, K);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

       // JavaScript code for the above approach
 
       // Function to find the minimum number
       // of steps
       function findMinimumNumberOfSteps(arr,
           K) {
 
           // Variable to store the answer
           let time = 0;
 
           // Traverse in the while loop
           while (arr[K] != 0) {
 
               // Iterate over the loop
               for (let i = 0; i < arr.length; i++) {
 
                   // Check the condition and
                   // decrease the value
                   if (arr[i] > 0) {
                       arr[i] -= 1;
                       time++;
                   }
 
                   // Break the loop
                   if (arr[K] == 0)
                       break;
               }
           }
 
           // Print the result
           document.write(time);
       }
 
       // Driver Code
 
       let arr = [2, 3, 2];
       let K = 2;
       findMinimumNumberOfSteps(arr, K);
 
 
      // This code is contributed by Potta Lokesh
   

输出

6

时间复杂度： O(N*X)，其中 X 是arr[K]的值
辅助空间： O(1)