通过删除第一个或添加先前删除的元素来最大化 Array 的第一个元素

给定一个大小为N的数组arr[]和一个整数K ，任务是在K个操作中最大化数组的第一个元素，其中在每个操作中：

• 如果数组不为空，则删除数组的最顶部元素。
• 在数组的开头添加任何一个先前删除的元素。

例子：

Input: arr[] = [5, 2, 2, 4, 0, 6], K = 4
Output: 5
Explanation: The 4 operations are as:

• Remove the topmost element = 5. The arr becomes [2, 2, 4, 0, 6].
• Remove the topmost element = 2. The arr becomes [2, 4, 0, 6].
• Remove the topmost element = 2. The arr becomes [4, 0, 6].
• Add 5 back onto the arr. The arr becomes [5, 4, 0, 6].

Here 5 is the largest answer possible after 4 moves.

Input: arr[] = [2], K = 1
Output: -1
Explanation: Only one move can be applied and in the first move. 
The only option is to remove the first element of the arr[]. 
If that is done the array becomes empty. So answer is -1

方法：这个问题可以借助基于以下思想的贪心方法来解决：

In first K-1 operations, the K-1 value of the starting can be removed. So currently at Kth node. Now in the last operation there are two possible choice:

• Either remove the current starting node (optimal if the value of the (K+1)th node is greater than the largest amongst first K-1 already removed elements)
• Add the largest from the already removed K-1 elements (optimal when the (K+1)th node has less value than this largest one)

请按照下图所示进行更好的理解。

插图：

For example arr[] = {5, 2, 2, 4, 0, 6}, K = 4

1st Operation:
        => Remove 5. arr[] = {2, 2, 4, 0, 6}
        => maximum = 5, K = 4 – 1 = 3

2nd Operation:
        => Remove 2. arr[] = {2, 4, 0, 6}
        => maximum = max (5, 2) = 5, K = 3 – 1 = 2

3rd Operation:
        => Remove 2. arr[] = {4, 0, 6}
        => maximum = max (5, 2) = 5, K = 2 – 1 = 1

4th Operation:
        => Here the current 2nd element i.e. 0 is less than 5.
        => So add 5 back in the array. arr[] = {5, 4, 0, 6}
        => maximum = max (5, 0) = 5, K = 1 – 1 = 0

Therefore the maximum possible first element is 5.

请按照以下步骤解决问题：

• 如果K = 0 ，则返回第一个节点值。
• 如果K = 1 ，则返回第二个节点值（如果有），否则返回 -1 （因为在K操作之后列表不存在）。
• 如果链表的大小是1，那么在每个奇数操作（即1、3、5、...）中，返回-1，否则返回第一个节点值（因为如果执行奇数操作，则数组将变为空）。
• 如果K > 2 ，则：
  • 遍历前K-1个节点，找出最大值。
  • 将该最大值与第(K+1) 个节点值进行比较。
  • 如果第 (K+1) 个值大于之前的最大值，则使用第 (K+1) 个节点值对其进行更新。否则，不要更新最大值。
• 返回最大值。

以下是上述方法的实现：

C++

// C++ code to implement the approach
#include 
using namespace std;
 
int maximumTopMost(int arr[], int k,int N){
 
   // Checking if k is odd and
    // length of array is 1
    if (N == 1 and k % 2 != 0)
        return -1;
 
    // Initializing ans with -1
    int ans = -1;
 
    // If k is greater or equal to the
    // length of array
    for(int i  = 0; i <  min(N, k - 1); i++)
        ans = max(ans, arr[i]);
 
    // If k is less than length of array
    if (k < N)
        ans = max(ans, arr[k]);
 
    // Returning ans
    return ans;
}
 
// Driver code
int main() {
 
      int arr[] = {5, 2, 2, 4, 0, 6};
      int N = 6;
      int K = 4;
      cout <<(maximumTopMost(arr, K, N));
      return 0;
}
 
// This code is contributed by hrithikgarg03188.

Java

// Java code to implement the approach
class GFG {
 
  static int maximumTopMost(int[] arr, int k, int N)
  {
 
    // Checking if k is odd and
    // length of array is 1
    if (N == 1 && k % 2 != 0)
      return -1;
 
    // Initializing ans with -1
    int ans = -1;
 
    // If k is greater or equal to the
    // length of array
    for (int i = 0; i < Math.min(N, k - 1); i++)
      ans = Math.max(ans, arr[i]);
 
    // If k is less than length of array
    if (k < N)
      ans = Math.max(ans, arr[k]);
 
    // Returning ans
    return ans;
  }
  public static void main(String[] args)
  {
 
    int[] arr = { 5, 2, 2, 4, 0, 6 };
    int N = 6;
    int K = 4;
    System.out.println(maximumTopMost(arr, K, N));
  }
}
 
// This code is contributed by phasing17.

Python3

# Python code to implement the approach
 
def maximumTopMost(arr, k):
 
    # Checking if k is odd and
    # length of array is 1
    if len(arr) == 1 and k % 2 != 0:
        return -1
 
    # Initializing ans with -1
    ans = -1
 
    # If k is greater or equal to the
    # length of array
    for i in range(min(len(arr), k - 1)):
        ans = max(ans, arr[i])
 
    # If k is less than length of array
    if k < len(arr):
        ans = max(ans, arr[k])
 
    # Returning ans
    return ans
 
 
# Driver code
if __name__ == "__main__":
    arr = [5, 2, 2, 4, 0, 6]
    K = 4
    print(maximumTopMost(arr, K))

C#

// C# code to implement the approach
using System;
class GFG {
 
  static int maximumTopMost(int[] arr, int k, int N)
  {
 
    // Checking if k is odd and
    // length of array is 1
    if (N == 1 && k % 2 != 0)
      return -1;
 
    // Initializing ans with -1
    int ans = -1;
 
    // If k is greater or equal to the
    // length of array
    for (int i = 0; i < Math.Min(N, k - 1); i++)
      ans = Math.Max(ans, arr[i]);
 
    // If k is less than length of array
    if (k < N)
      ans = Math.Max(ans, arr[k]);
 
    // Returning ans
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
 
    int[] arr = { 5, 2, 2, 4, 0, 6 };
    int N = 6;
    int K = 4;
    Console.Write(maximumTopMost(arr, K, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code to implement the approach
 
    const maximumTopMost = (arr, k) => {
        // Checking if k is odd and
        // length of array is 1
        if (arr.length == 1 && k % 2 != 0)
            return -1;
 
        // Initializing ans with -1
        let ans = -1;
 
        // If k is greater or equal to the
        // length of array
        for (let i = 0; i < Math.min(arr.length, k - 1); ++i)
            ans = Math.max(ans, arr[i]);
 
        // If k is less than length of array
        if (k < arr.length)
            ans = Math.max(ans, arr[k]);
 
        // Returning ans
        return ans;
    }
 
 
    // Driver code
    let arr = [5, 2, 2, 4, 0, 6];
    let K = 4;
    document.write(maximumTopMost(arr, K));
 
// This code is contributed by rakeshsahni
 

输出

5

时间复杂度： O(N)
辅助空间： O(1)