LeetCode 2401: Longest Nice Subarray

Meta Data

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Difficulty: medium First Attempt: 2026-01-22

• Submissions: Accepted (first try)

Approach

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Every pair in the window ANDs to 0 ⇔ each bit is used by at most one number. So one integer used holds the OR of the window:

• If the new number x overlaps used (used & x != 0), shrink the left until it does not.
• When nums[left] leaves, its bits belong to it alone, so used ^= nums[left] (or &= ~) removes them.
• Add x: used |= x.

Numbers have at most 30 bits, so a valid window has at most 30 elements.

My Solution

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class Solution:
    def longestNiceSubarray(self, nums: List[int]) -> int:
        used = left = best = 0
        for right, x in enumerate(nums):
            while used & x:                        # bit conflict
                used ^= nums[left]                 # its bits are its own, so XOR removes them
                left += 1
            used |= x
            best = max(best, right - left + 1)
        return best

Complexity

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

Mistakes to Avoid

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• Removing bits with -= or ^= is only correct because the window’s bits never overlap; with overlaps it would not work.

Interview Explanation

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Pairwise AND 0 means each bit is used by at most one number, so one integer records the bits the window uses. If a new number overlaps it, I shrink the left; the leaving number’s bits are its own, so XOR removes them. Tracking the longest length gives O(n).