Leetcode #2202: Maximize the Topmost Element After K Moves
In this guide, we solve Leetcode #2202 Maximize the Topmost Element After K Moves in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
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.
Python Solution
class Solution:
def maximumTop(self, nums: List[int], k: int) -> int:
if k == 0:
return nums[0]
n = len(nums)
if n == 1:
if k % 2:
return -1
return nums[0]
ans = max(nums[: k - 1], default=-1)
if k < n:
ans = max(ans, nums[k])
return ans
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.