How do you solve Maximize the Topmost Element After K Moves on LeetCode?

Maximize the Topmost Element After K Moves (LeetCode #2202) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: You can only consider elements that are within the first k moves or the last element if k is larger than the number of elements.

What is the brute force solution for Maximize the Topmost Element After K Moves?

The brute force approach for Maximize the Topmost Element After K Moves is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves simulating all possible moves to see which elements can be the topmost after k moves. This method is straightforward but inefficient as it checks every possibility. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize the Topmost Element After K Moves?

Maximize the Topmost Element After K Moves uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Maximize the Topmost Element After K Moves?

Always clarify the constraints and edge cases before diving into coding. Think about the problem in terms of what you can remove and add back; visualize the pile. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Maximize the Topmost Element After K Moves?

Forgetting to check if k is greater than or equal to the length of nums when considering the last element. Not properly handling the case where k is less than the number of elements.

#2202

Maximize the Topmost Element After K Moves

Medium

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach leverages the fact that we only need to consider elements that can potentially be the topmost after k moves. We check the first k elements and the last element if k is less than the length of nums.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to track the maximum top element found.
2Step 2: Loop through the first min(k, len(nums)) elements and update the maximum.
3Step 3: If k is greater than or equal to the length of nums, also consider the last element in nums.
4Step 4: Return the maximum found or -1 if no valid topmost element exists.

solution.py11 lines

1# Full working Python code
2
3def maxTop(nums, k):
4    max_top = -1
5    limit = min(k, len(nums))
6    for i in range(limit):
7        max_top = max(max_top, nums[i])
8    if k >= len(nums):
9        max_top = max(max_top, nums[-1])
10    return max_top if max_top != -1 else -1
11

ℹ

Complexity note: The time complexity is O(n) because we only loop through the first k elements and possibly the last element. The space complexity is O(1) as we only use a few variables.

1You can only consider elements that are within the first k moves or the last element if k is larger than the number of elements.
2If k is less than the length of nums, you cannot add back any removed elements beyond the first k.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.