LeetCode Solutions

500. Keyboard Row

Time: $O(n)$

Space: $O(n)$

			

class Solution {
 public:
  vector<string> findWords(vector<string>& words) {
    vector<string> ans;
    const vector<int> rows{2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3,
                           3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3};

    for (const string& word : words) {
      string lowerWord = word;
      transform(begin(lowerWord), end(lowerWord), begin(lowerWord), ::tolower);
      const int row = rows[lowerWord[0] - 'a'];
      const bool isValid = all_of(begin(lowerWord), end(lowerWord),
                                  [&](int c) { return rows[c - 'a'] == row; });
      if (isValid)
        ans.push_back(word);
    }

    return ans;
  }
};
			

class Solution {
  public String[] findWords(String[] words) {
    List<String> ans = new ArrayList<>();
    final int[] rows = {2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3,
                        3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 3};

    for (final String word : words) {
      final String lowerWord = word.toLowerCase();
      final int row = rows[lowerWord.charAt(0) - 'a'];
      final boolean isValid = lowerWord.chars().allMatch(c -> rows[c - 'a'] == row);
      if (isValid)
        ans.add(word);
    }

    return ans.toArray(new String[0]);
  }
}
			

class Solution:
  def findWords(self, words: List[str]) -> List[str]:
    ans = []
    rows = [set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm')]

    for word in words:
      lowerWord = set(word.lower())
      if any(lowerWord <= row for row in rows):
        ans.append(word)

    return ans