README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2900-2999/2942.Find%20Words%20Containing%20Character/README_EN.md	1182	Biweekly Contest 118 Q1	Array String

2942. Find Words Containing Character

中文文档

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 <= 50
• 1 <= words[i].length <= 50
• x is 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 $O(1)$.

Python3

class Solution:
    def findWordsContaining(self, words: List[str], x: str) -> List[int]:
        return [i for i, w in enumerate(words) if x in w]

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}