LeetCode 2000: Reverse Prefix of Word

Meta Data

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Difficulty: easy First Attempt: 2024-05-01

• Submissions: Accepted (first try)

Approach

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i = word.find(ch) is -1 when absent; then word[:0] is empty and appending all of word returns it unchanged, so no special case is needed. By hand, swap [0, i] with converging pointers (Two Pointers Template 1).

My Solution

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class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        i = word.find(ch)
        return word[:i + 1][::-1] + word[i + 1:]

Complexity

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

Mistakes to Avoid

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• index raises when not found; find returns -1.

Interview Explanation

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I find ch’s first index, reverse the prefix through it and append the rest; when absent, find returns -1, the prefix is empty, and the word comes back unchanged. Time O(n).