LeetCode Solutions

524. Longest Word in Dictionary through Deleting

Time: $O(|\texttt{d}||\texttt{s}|)$

Space: $O(|\texttt{s}|)$

			

class Solution {
 public:
  string findLongestWord(string s, vector<string>& d) {
    string ans;

    for (const string& word : d)
      if (isSubsequence(word, s))
        if (word.length() > ans.length() ||
            word.length() == ans.length() && word.compare(ans) < 0)
          ans = word;

    return ans;
  }

 private:
  // Returns true if a is a subsequence of b
  bool isSubsequence(const string& a, const string& b) {
    int i = 0;
    for (const char c : b)
      if (i < a.length() && c == a[i])
        ++i;
    return i == a.length();
  };
};
			

class Solution {
  public String findLongestWord(String s, List<String> d) {
    String ans = "";

    for (final String word : d)
      if (isSubsequence(word, s))
        if (word.length() > ans.length() ||
            word.length() == ans.length() && word.compareTo(ans) < 0)
          ans = word;

    return ans;
  }

  // Returns true if a is a subsequence of b
  private boolean isSubsequence(final String a, final String b) {
    int i = 0;
    for (final char c : b.toCharArray())
      if (i < a.length() && c == a.charAt(i))
        ++i;
    return i == a.length();
  }
}
			

class Solution:
  def findLongestWord(self, s: str, d: List[str]) -> str:
    ans = ''

    for word in d:
      i = 0
      for c in s:
        if i < len(word) and c == word[i]:
          i += 1
      if i == len(word):
        if len(word) > len(ans) or len(word) == len(ans) and word < ans:
          ans = word

    return ans