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1408. String Matching in an Array

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

Solution 1: Brute Force Enumeration

We directly enumerate all strings $words[i]$, and check whether it is a substring of other strings. If it is, we add it to the answer.

The time complexity is $O(n^3)$, and the space complexity is $O(n)$. Where $n$ is the length of the string array.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

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

• 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
  }
  

• function stringMatching(words: string[]): string[] {
      const ans: string[] = [];
      const n = words.length;
      for (let i = 0; i < n; ++i) {
          for (let j = 0; j < n; ++j) {
              if (words[j].includes(words[i]) && i !== j) {
                  ans.push(words[i]);
                  break;
              }
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn string_matching(words: Vec<String>) -> Vec<String> {
          let mut ans = Vec::new();
          let n = words.len();
          for i in 0..n {
              for j in 0..n {
                  if i != j && words[j].contains(&words[i]) {
                      ans.push(words[i].clone());
                      break;
                  }
              }
          }
          ans
      }
  }
  
  

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