LeetCode 3884: First Matching Character From Both Ends ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2026-03-29

Submissions: Accepted (first try)

Approach
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Find the smallest i with s[i] == s[n - 1 - i]. Check from i = 0 toward the middle; the first match is the answer.

Checking up to the middle is enough: past the middle, the pair (i, n - 1 - i) was already checked.

Observation: with an odd length, the middle character is compared with itself and always matches, so odd lengths always have an answer; only even lengths can return -1.

My Solution
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class Solution:
    def firstMatchingIndex(self, s: str) -> int:
        n = len(s)
        for i in range((n + 1) // 2):
            if s[i] == s[n - 1 - i]:
                return i
        return -1

My original looped to n, with the same result (when nothing matches, the second half just rechecks the same pairs); this version stops at the middle.

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

Mistakes to Avoid
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Looping only to n // 2: with an odd length, the middle character is missed.

Interview Explanation
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I check s[i] against s[n - 1 - i] from i = 0 toward the middle and return the first match; the middle is far enough, since later pairs were already seen. With an odd length the middle character always matches itself, so only even lengths can fail. Time O(n).