Question

1408. String Matching in an Array

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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.

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

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;
  }
};

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
  }
}