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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 <= 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 \(O(1)\).

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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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