Package g2201_2300.s2202_maximize_the_topmost_element_after_k_moves

Class Solution

java.lang.Object
g2201_2300.s2202_maximize_the_topmost_element_after_k_moves.Solution
public class Solution extends Object
2202 - Maximize the Topmost Element After K Moves\. Medium You are given a **0-indexed** integer array `nums` representing the contents of a **pile** , where `nums[0]` is the topmost element of the pile. In one move, you can perform **either** of the following: * If the pile is not empty, **remove** the topmost element of the pile. * If there are one or more removed elements, **add** any one of them back onto the pile. This element becomes the new topmost element. You are also given an integer `k`, which denotes the total number of moves to be made. Return _the **maximum value** of the topmost element of the pile possible after **exactly**_ `k` _moves_. In case it is not possible to obtain a non-empty pile after `k` moves, return `-1`. **Example 1:** **Input:** nums = [5,2,2,4,0,6], k = 4 **Output:** 5 **Explanation:** One of the ways we can end with 5 at the top of the pile after 4 moves is as follows: - Step 1: Remove the topmost element = 5. The pile becomes [2,2,4,0,6]. - Step 2: Remove the topmost element = 2. The pile becomes [2,4,0,6]. - Step 3: Remove the topmost element = 2. The pile becomes [4,0,6]. - Step 4: Add 5 back onto the pile. The pile becomes [5,4,0,6]. Note that this is not the only way to end with 5 at the top of the pile. It can be shown that 5 is the largest answer possible after 4 moves. **Example 2:** **Input:** nums = [2], k = 1 **Output:** -1 **Explanation:** In the first move, our only option is to pop the topmost element of the pile. Since it is not possible to obtain a non-empty pile after one move, we return -1. **Constraints:** * 1 <= nums.length <= 105 * 0 <= nums[i], k <= 109

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maximumTop

      public int maximumTop(int[] nums, int k)