1078. Occurrences After Bigram
Description
Given two strings 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.
Return an array of all the words third for each occurrence of "first second third".
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"]
Constraints:
1 <= text.length <= 1000textconsists of lowercase English letters and spaces.- All the words in
textare separated by a single space. 1 <= first.length, second.length <= 10firstandsecondconsist of lowercase English letters.textwill not have any leading or trailing spaces.
Solutions
Solution 1
Python3
class Solution:
def findOcurrences(self, text: str, first: str, second: str) -> List[str]:
words = text.split()
ans = []
for i in range(len(words) - 2):
a, b, c = words[i : i + 3]
if a == first and b == second:
ans.append(c)
return ans
Java
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]);
}
}
C++
class Solution {
public:
vector<string> findOcurrences(string text, string first, string second) {
istringstream is(text);
vector<string> words;
string word;
while (is >> word) {
words.emplace_back(word);
}
vector<string> ans;
int n = words.size();
for (int i = 0; i < n - 2; ++i) {
if (words[i] == first && words[i + 1] == second) {
ans.emplace_back(words[i + 2]);
}
}
return ans;
}
};
Go
func findOcurrences(text string, first string, second string) (ans []string) {
words := strings.Split(text, " ")
n := len(words)
for i := 0; i < n-2; i++ {
if words[i] == first && words[i+1] == second {
ans = append(ans, words[i+2])
}
}
return
}
TypeScript
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;
}