LeetCode 1078: Occurrences After Bigram Solution

Master LeetCode problem 1078 (Occurrences After Bigram), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1078. Occurrences After Bigram

Easy

String

Problem Explanation

Explanation

To solve this problem, we can iterate through the words in the input text. For each word, we check if the current word is equal to the first word and the next word is equal to the second word. If this condition is met, we add the word after the second word to our result list.

Time Complexity

The time complexity of this algorithm is O(n), where n is the number of words in the input text.

Space Complexity

The space complexity of this algorithm is O(k), where k is the number of occurrences of the words "first second" in the text.

Solution Code

class Solution {
    public String[] findOcurrences(String text, String first, String second) {
        String[] words = text.split(" ");
        List<String> result = new ArrayList<>();
        
        for (int i = 0; i < words.length - 2; i++) {
            if (words[i].equals(first) && words[i + 1].equals(second)) {
                result.add(words[i + 2]);
            }
        }
        
        return result.toArray(new String[result.size()]);
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 1078 (Occurrences After Bigram)?

The time complexity for LeetCode 1078 (Occurrences After Bigram) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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