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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/
// Author: github.com/lzl124631x
// 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;
    }
};