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1408. String Matching in an Array

Description

Given an array of string words, return all strings in words that is a substring of another word. You can return the answer in any order.

A substring is a contiguous sequence of characters within a string

 

Example 1:

Input: words = ["mass","as","hero","superhero"]
Output: ["as","hero"]
Explanation: "as" is substring of "mass" and "hero" is substring of "superhero".
["hero","as"] is also a valid answer.

Example 2:

Input: words = ["leetcode","et","code"]
Output: ["et","code"]
Explanation: "et", "code" are substring of "leetcode".

Example 3:

Input: words = ["blue","green","bu"]
Output: []
Explanation: No string of words is substring of another string.

 

Constraints:

  • 1 <= words.length <= 100
  • 1 <= words[i].length <= 30
  • words[i] contains only lowercase English letters.
  • All the strings of words are unique.

Solutions

Solution 1: Brute Force Enumeration

We directly enumerate all strings $words[i]$, and check whether it is a substring of other strings. If it is, we add it to the answer.

The time complexity is $O(n^3)$, and the space complexity is $O(n)$. Where $n$ is the length of the string array.

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class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:
        ans = []
        for i, s in enumerate(words):
            if any(i != j and s in t for j, t in enumerate(words)):
                ans.append(s)
        return ans

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class Solution {
    public List<String> stringMatching(String[] words) {
        List<String> ans = new ArrayList<>();
        int n = words.length;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i != j && words[j].contains(words[i])) {
                    ans.add(words[i]);
                    break;
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> stringMatching(vector<string>& words) {
        vector<string> ans;
        int n = words.size();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i != j && words[j].find(words[i]) != string::npos) {
                    ans.push_back(words[i]);
                    break;
                }
            }
        }
        return ans;
    }
};

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func stringMatching(words []string) []string {
    ans := []string{}
    for i, w1 := range words {
        for j, w2 := range words {
            if i != j && strings.Contains(w2, w1) {
                ans = append(ans, w1)
                break
            }
        }
    }
    return ans
}

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function stringMatching(words: string[]): string[] {
    const ans: string[] = [];
    const n = words.length;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (words[j].includes(words[i]) && i !== j) {
                ans.push(words[i]);
                break;
            }
        }
    }
    return ans;
}

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impl Solution {
    pub fn string_matching(words: Vec<String>) -> Vec<String> {
        let mut ans = Vec::new();
        let n = words.len();
        for i in 0..n {
            for j in 0..n {
                if i != j && words[j].contains(&words[i]) {
                    ans.push(words[i].clone());
                    break;
                }
            }
        }
        ans
    }
}

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