LeetCode 1078: Occurrences After Bigram

Meta Data

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Difficulty: easy First Attempt: 2022-10-04

• Submissions: Runtime Error → Accepted ×3

Approach

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Split into words and check each i (String Template 6): if words[i] == first and words[i + 1] == second, collect words[i + 2]. Looping only to len(words) - 2 guarantees the third word exists.

My Attempts

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Attempt 1 checked only i + 1 < len(text) but read text[i + 2]: "b b" with first = second = "b" raised IndexError. Checking i + 2 passed.

Solution

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class Solution:
    def findOcurrences(self, text: str, first: str, second: str) -> List[str]:
        w = text.split()
        return [w[i + 2] for i in range(len(w) - 2)
                if w[i] == first and w[i + 1] == second]

Complexity

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

Mistakes to Avoid

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• Since i + 2 is accessed, bound-check i + 2.

Interview Explanation

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I split into words and check every window of three, collecting the third when the first two equal first and second, looping only up to the third-to-last word. Time O(n).