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)
    

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