LeetCode 3828: Final Element After Subarray Deletions ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-02-01

Submissions: Accepted (first try)

Approach
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A move may delete any subarray of length < m (the current length), so one move can leave a single element, as long as it is the first or the last (delete a subarray of length m - 1).

Lower bound: Alice’s first move leaves only nums[0] or nums[-1], whichever is larger, ending the game. The answer is at least max(first, last).
Upper bound: if Alice’s first move does not end the game, she deleted one contiguous subarray and cannot remove both ends, so the new array keeps at least one original end. Bob then deletes down to the smaller of the new array’s ends, which is <= that original end <= max(first, last).

So the answer is exactly max(nums[0], nums[-1]).

My Solution
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class Solution:
    def finalElement(self, nums: List[int]) -> int:
        return max(nums[0], nums[-1])

Complexity
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Time O(1), space O(1).

Mistakes to Avoid
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Simulating the whole game: once “one move can leave an end” is seen, it is constant time.

Interview Explanation
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Every move can delete down to just the first or last element. Alice does that immediately and gets the larger end; if she does not, she only removed one block, so an original end survives, and Bob deletes down to the smaller of the new ends, which cannot exceed the original larger end. So the answer is max(first, last).