How do you solve Occurrences After Bigram on LeetCode?

Occurrences After Bigram (LeetCode #1078) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the structure of the problem helps in identifying patterns.

What is the brute force solution for Occurrences After Bigram?

The brute force approach for Occurrences After Bigram is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will split the text into words and check every possible triplet of words to see if they match the pattern 'first second third'. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Occurrences After Bigram?

Occurrences After Bigram uses the following concepts: String. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Occurrences After Bigram?

Always clarify the problem statement and constraints before jumping into coding. Think aloud during the interview to show your thought process. Consider edge cases and test your solution against them.

What are common mistakes when solving Occurrences After Bigram?

Not handling edge cases where there are not enough words to form a triplet. Overlooking the need to check bounds when accessing the list.

#1078

Occurrences After Bigram

Easy

StringSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

By iterating through the words only once and checking adjacent pairs, we can efficiently find the occurrences without needing nested loops. This reduces our time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Split the input text into a list of words.
2Step 2: Initialize an empty result list.
3Step 3: Iterate through the list of words from the first to the second last index, checking if the current word and the next word match 'first' and 'second'. If they do, add the word after the second to the result list.

solution.py7 lines

1def findOcurrences(text, first, second):
2    words = text.split()
3    result = []
4    for i in range(len(words) - 2):
5        if words[i] == first and words[i + 1] == second:
6            result.append(words[i + 2])
7    return result

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the list of words. The space complexity is O(n) due to the storage of the result list, which can grow depending on the number of matches found.

1Understanding the structure of the problem helps in identifying patterns.
2Recognizing that we can reduce complexity by avoiding nested loops is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.