LeetCode 2831: Find the Longest Equal Subarray

Meta Data

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Difficulty: medium First Attempt: 2026-01-11

• Submissions: Wrong Answer → Time Limit Exceeded ×2 → Accepted

Approach

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What remains is copies of one value v. Handle each value separately:

• List the positions where v appears, pos = [p0, p1, ...].
• Keeping the copies at pos[l..r] means deleting everything between them: (pos[r] - pos[l] + 1) - (r - l + 1) deletions.
• Slide a window over pos: shrink the left while deletions exceed k; the answer is the largest r - l + 1.

All the position lists together have length n, so this is O(n).

My Attempts

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Solution

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class Solution:
    def longestEqualSubarray(self, nums: List[int], k: int) -> int:
        positions = defaultdict(list)
        for i, v in enumerate(nums):
            positions[v].append(i)

        best = 0
        for pos in positions.values():
            left = 0
            for right in range(len(pos)):
                while (pos[right] - pos[left] + 1) - (right - left + 1) > k:
                    left += 1                      # too many deletions in between
                best = max(best, right - left + 1)
        return best

Complexity

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Time O(n), space O(n).

Mistakes to Avoid

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• Using window length - k as the answer: the window does not necessarily delete exactly k.
• Taking max over the count table every step: O(n · distinct).

Interview Explanation

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What remains is one value, so I handle each value separately: list its positions, and keeping a consecutive run of them requires deleting the span of positions minus the count. A sliding window over the position list shrinks when deletions exceed k, and the number of copies in the window is a candidate answer. All lists total n, so time is O(n).