Maximize first element of Array by deleting first or adding a previously deleted element

Given an array arr[] of size N, and an integer K, the task is to maximize the first element of the array in K operations where in each operation:

• If the array is not empty, remove the topmost element of the array.
• Add any one of the previously removed element back at the starting of the array.

Examples:

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

Approach: This problem can be solved with the help of the Greedy approach based on the following idea:

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)

Follow the illustration shown below for a better understanding.

Illustration:

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.

Follow the steps to solve the problem:

• If K = 0, then return first node value.
• If K = 1, then return the second node value(if any) else return -1 (because after K operations the list does not exist).
• If the size of the linked list is one then in every odd operation (i.e. 1, 3, 5, . . . ), return -1, else return the first node value (because if performed odd operation then array will become empty).
• If K > 2, then:
  • Traverse first K-1 nodes and find out the maximum value.
  • Compare that maximum value with the (K+1)th node value.
  • If (K+1)th value is greater than the previous maximum value, update it with (K+1)th Node value. Otherwise, don't update the maximum value.
• Return the maximum value.

Below is the implementation of the above approach :

// C++ code to implement the approach
#include <iostream>
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 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.

# 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# 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.

    <script>
        // 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

    </script>

5

Time Complexity: O(N)
Auxiliary Space: O(1)