Problem

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

Examples

Example 1:

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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:

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

Example 3:

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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.

Solution

Method 1 - Naive approach

For each word in the array, we can check if it is a substring of any other word. Here is what we can do:

  • Iterate through the list of words.
  • For each word, compare it against every other word to check if it is a substring.
  • If it is a substring of any other word, add it to the result list.

Video explanation

Here is the video explaining this method in detail. Please check it out:

Code

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class Solution {
    public List<String> stringMatching(String[] words) {
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words.length; j++) {
                if (i != j && words[j].contains(words[i])) {
                    ans.add(words[i]);
                    break;
                }
            }
        }
        return ans;
    }
}
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class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:
        ans: List[str] = []
        for i in range(len(words)):
            for j in range(len(words)):
                if i != j and words[i] in words[j]:
                    ans.append(words[i])
                    break
        return ans

Complexity

  • ⏰ Time complexity: O(n^2 * l^2) where n is the number of words and l is the average length of a word in words. Because we run 2 loops to match the words and then contains method takes O(l1*l2) and assuming average length of words is l, it becomes O(l^2).
  • 🧺 Space complexity: O(n) for storing the result list.