README_EN.md

1408. String Matching in an Array

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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

Python3

class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:
        ans = []
        for i, w1 in enumerate(words):
            for j, w2 in enumerate(words):
                if i != j and w1 in w2:
                    ans.append(w1)
                    break
        return ans

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

function stringMatching(words: string[]): string[] {
    const res: string[] = [];
    for (const target of words) {
        for (const word of words) {
            if (word !== target && word.includes(target)) {
                res.push(target);
                break;
            }
        }
    }
    return res;
}

Rust

impl Solution {
    pub fn string_matching(words: Vec<String>) -> Vec<String> {
        let mut res = Vec::new();
        for target in words.iter() {
            for word in words.iter() {
                if word != target && word.contains(target) {
                    res.push(target.clone());
                    break;
                }
            }
        }
        res
    }
}

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