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Formatted question description: https://leetcode.ca/all/1078.html

1078. Occurrences After Bigram (Easy)

Given words first and second, consider occurrences in some text of the form "first second third", where second comes immediately after first, and third comes immediately after second.

For each such occurrence, add "third" to the answer, and return the answer.

 

Example 1:

Input: text = "alice is a good girl she is a good student", first = "a", second = "good"
Output: ["girl","student"]

Example 2:

Input: text = "we will we will rock you", first = "we", second = "will"
Output: ["we","rock"]

 

Note:

  1. 1 <= text.length <= 1000
  2. text consists of space separated words, where each word consists of lowercase English letters.
  3. 1 <= first.length, second.length <= 10
  4. first and second consist of lowercase English letters.

Solution 1.

  • class Solution {
    public String[] findOcurrences(String text, String first, String second) {
        List<String> occurrencesList = new ArrayList<String>();
        String[] textArray = text.split(" ");
        int length = textArray.length;
        for (int i = 2; i < length; i++) {
            String word1 = textArray[i - 2], word2 = textArray[i - 1];
            if (word1.equals(first) && word2.equals(second))
                occurrencesList.add(textArray[i]);
        }
        int size = occurrencesList.size();
        String[] occurrencesArray = new String[size];
        for (int i = 0; i < size; i++)
            occurrencesArray[i] = occurrencesList.get(i);
        return occurrencesArray;
    }
}

############

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

  • // OJ: https://leetcode.com/problems/occurrences-after-bigram/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<string> findOcurrences(string text, string first, string second) {
        vector<string> ans;
        istringstream ss(text);
        string prev2, prev, word;
        while (ss >> word) {
            if (prev2 == first && prev == second) ans.push_back(word);
            prev2 = prev;
            prev = word;
        }
        return ans;
    }
};

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

############

# 1078. Occurrences After Bigram
# https://leetcode.com/problems/occurrences-after-bigram/

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


  • func findOcurrences(text string, first string, second string) []string {
	words := strings.Split(text, " ")
	var ans []string
	for i := 0; i < len(words)-2; i++ {
		if words[i] == first && words[i+1] == second {
			ans = append(ans, words[i+2])
		}
	}
	return ans
}

  • function findOcurrences(text: string, first: string, second: string): string[] {
    const words = text.split(' ');
    const n = words.length;
    const ans: string[] = [];
    for (let i = 0; i < n - 2; i++) {
        if (words[i] === first && words[i + 1] === second) {
            ans.push(words[i + 2]);
        }
    }
    return ans;
}

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