Welcome to Subscribe On Youtube

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

• 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;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // 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]:
          ans = []
          for i, w1 in enumerate(words):
              for j, w2 in enumerate(words):
                  if i != j and w1 in w2:
                      ans.append(w1)
                      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)
  

• 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 res: string[] = [];
      for (const target of words) {
          for (const word of words) {
              if (word !== target && word.includes(target)) {
                  res.push(target);
                  break;
              }
          }
      }
      return res;
  }
  
  

• impl Solution {
      pub fn string_matching(words: Vec<String>) -> Vec<String> {
          let mut res = Vec::new();
          for target in words.iter() {
              for word in words.iter() {
                  if word != target && word.contains(target) {
                      res.push(target.clone());
                      break;
                  }
              }
          }
          res
      }
  }
  
  

All Problems

All Solutions