Formatted question description: https://leetcode.ca/all/1408.html

1408. String Matching in an Array (Easy)

Given an array of string words. Return all strings in words which is substring of another word in any order.

String words[i] is substring of words[j], if can be obtained removing some characters to left and/or right side of words[j].

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: []

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 30
words[i] contains only lowercase English letters.
It's guaranteed that words[i] will be unique.

Related Topics:
String

Solution 1. Brute force

For each words[i], check if it’s a substring of words[j] (0 <= j < N && i != j). Once find a match, we can push words[i] to result and continue to word[i + 1].

// OJ: https://leetcode.com/problems/string-matching-in-an-array/
// Time: O(N^2 * D)
// Space: O(1)
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) continue;
                if (words[j].find(words[i]) != string::npos) {
                    ans.push_back(words[i]);
                    break;
                }
            }
        }
        return ans;
    }
};

Java

class Solution {
    public List<String> stringMatching(String[] words) {
        Arrays.sort(words, new Comparator<String>() {
            public int compare(String word1, String word2) {
                return word1.length() - word2.length();
            }
        });
        List<String> list = new ArrayList<String>();
        int length = words.length;
        for (int i = 0; i < length; i++) {
            String word1 = words[i];
            boolean flag = false;
            for (int j = i + 1; j < length; j++) {
                String word2 = words[j];
                if (word2.indexOf(word1) >= 0) {
                    flag = true;
                    break;
                }
            }
            if (flag)
                list.add(word1);
        }
        return list;
    }
}

// OJ: https://leetcode.com/problems/string-matching-in-an-array/
// Time: O(N^2 * D)
// Space: O(1)
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) continue;
                if (words[j].find(words[i]) != string::npos) {
                    ans.push_back(words[i]);
                    break;
                }
            }
        }
        return ans;
    }
};

class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:
        res = set()
        for word in words:
            for cur in words:
                if cur in word and cur != word:
                    res.add(cur)
        return list(res)

1408 - String Matching in an Array

1408. String Matching in an Array (Easy)

Solution 1. Brute force

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