LeetCode 392: Is Subsequence

Meta Data

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Difficulty: easy First Attempt: 2022-08-23

• Submissions: Runtime Error ×2 → Accepted ×2 → Runtime Error → Accepted ×2 → Runtime Error → Wrong Answer → Accepted ×3 → Runtime Error → Wrong Answer → Runtime Error ×2 → Accepted

Approach

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One pointer per string (Two Pointers Template 2): j scans t, and i advances when t[j] == s[i]. Loop while i < len(s) and j < len(t), and succeed if i == len(s). Empty s or t then needs no special case. For the follow-up with many s, store the positions of each character of t and binary search for the next one per s.

My Attempts

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I wrote this many times (2022, 2025, 2026), and almost every error was about empty strings and bounds:

• 2022-08 used a queue with pop(0); with an empty s, queue[0] was out of range.
• 2022-11 wrote if not t or not s: return False, but an empty s should be True: ('', '') returned False.
• In 2025 the inner while t[r] != s[l] raised IndexError for an empty t; adding if not t: return False broke ('', '') again, and r += 1 after a match was never bound-checked: ('bb', 'b') raised IndexError.
• The 2026 version checked right == m only on a mismatch, so an empty t overran on the first comparison; putting both bounds in the while condition passed.

Solution

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class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        i = 0
        for c in t:
            if i < len(s) and s[i] == c:
                i += 1
        return i == len(s)

Complexity

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

Mistakes to Avoid

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• An empty s is always a subsequence, even of an empty t.
• Bound-check j before reading t[j], including right after a match advances it.

Interview Explanation

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With two pointers I scan t and advance s’s pointer on each match; s is a subsequence if its pointer reaches the end. Time O(n + m), space O(1).