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2942. Find Words Containing Character
Description
You are given a 0-indexed array of strings words and a character x.
Return an array of indices representing the words that contain the character x.
Note that the returned array may be in any order.
Example 1:
Input: words = ["leet","code"], x = "e" Output: [0,1] Explanation: "e" occurs in both words: "leet", and "code". Hence, we return indices 0 and 1.
Example 2:
Input: words = ["abc","bcd","aaaa","cbc"], x = "a" Output: [0,2] Explanation: "a" occurs in "abc", and "aaaa". Hence, we return indices 0 and 2.
Example 3:
Input: words = ["abc","bcd","aaaa","cbc"], x = "z" Output: [] Explanation: "z" does not occur in any of the words. Hence, we return an empty array.
Constraints:
1 <= words.length <= 501 <= words[i].length <= 50xis a lowercase English letter.words[i]consists only of lowercase English letters.
Solutions
Solution 1: Traversal
We directly traverse each string words[i] in the string array words. If x appears in words[i], we add i to the answer array.
After the traversal, we return the answer array.
The time complexity is $O(L)$, where $L$ is the sum of the lengths of all strings in the array words. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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class Solution { public List<Integer> findWordsContaining(String[] words, char x) { List<Integer> ans = new ArrayList<>(); for (int i = 0; i < words.length; ++i) { if (words[i].indexOf(x) != -1) { ans.add(i); } } return ans; } } -
class Solution { public: vector<int> findWordsContaining(vector<string>& words, char x) { vector<int> ans; for (int i = 0; i < words.size(); ++i) { if (words[i].find(x) != string::npos) { ans.push_back(i); } } return ans; } }; -
class Solution: def findWordsContaining(self, words: List[str], x: str) -> List[int]: return [i for i, w in enumerate(words) if x in w] -
func findWordsContaining(words []string, x byte) (ans []int) { for i, w := range words { for _, c := range w { if byte(c) == x { ans = append(ans, i) break } } } return } -
function findWordsContaining(words: string[], x: string): number[] { const ans: number[] = []; for (let i = 0; i < words.length; ++i) { if (words[i].includes(x)) { ans.push(i); } } return ans; } -
impl Solution { pub fn find_words_containing(words: Vec<String>, x: char) -> Vec<i32> { words .into_iter() .enumerate() .filter_map(|(i, w)| w.contains(x).then(|| i as i32)) .collect() } }