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Formatted question description: https://leetcode.ca/all/500.html
500. Keyboard Row
Level
Easy
Description
Given a List of words, return the words that can be typed using letters of alphabet on only one row’s of American keyboard like the image below.
Example:
Input: [“Hello”, “Alaska”, “Dad”, “Peace”]
Output: [“Alaska”, “Dad”]
Note:
- You may use one character in the keyboard more than once.
- You may assume the input string will only contain letters of alphabet.
Solution
For each letter, store its row. Then check each word to see whether the letters are in the same row. If so, add the word to the result array.
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class Solution { public String[] findWords(String[] words) { Map<Character, Integer> letterRowMap = new HashMap<Character, Integer>(); letterRowMap.put('q', 1); letterRowMap.put('w', 1); letterRowMap.put('e', 1); letterRowMap.put('r', 1); letterRowMap.put('t', 1); letterRowMap.put('y', 1); letterRowMap.put('u', 1); letterRowMap.put('i', 1); letterRowMap.put('o', 1); letterRowMap.put('p', 1); letterRowMap.put('a', 2); letterRowMap.put('s', 2); letterRowMap.put('d', 2); letterRowMap.put('f', 2); letterRowMap.put('g', 2); letterRowMap.put('h', 2); letterRowMap.put('j', 2); letterRowMap.put('k', 2); letterRowMap.put('l', 2); letterRowMap.put('z', 3); letterRowMap.put('x', 3); letterRowMap.put('c', 3); letterRowMap.put('v', 3); letterRowMap.put('b', 3); letterRowMap.put('n', 3); letterRowMap.put('m', 3); List<String> wordsList = new ArrayList<String>(); for (String word : words) { String lowerCaseWord = word.toLowerCase(); boolean flag = true; char[] array = lowerCaseWord.toCharArray(); if (array.length == 0) { wordsList.add(word); continue; } char c0 = array[0]; int prevRow = letterRowMap.get(c0); for (char c : array) { int curRow = letterRowMap.get(c); if (curRow != prevRow) { flag = false; break; } prevRow = curRow; } if (flag) wordsList.add(word); } int length = wordsList.size(); String[] rowWords = new String[length]; for (int i = 0; i < length; i++) rowWords[i] = wordsList.get(i); return rowWords; } } ############ class Solution { public String[] findWords(String[] words) { String s = "12210111011122000010020202"; List<String> ans = new ArrayList<>(); for (var w : words) { String t = w.toLowerCase(); char x = s.charAt(t.charAt(0) - 'a'); boolean ok = true; for (char c : t.toCharArray()) { if (s.charAt(c - 'a') != x) { ok = false; break; } } if (ok) { ans.add(w); } } return ans.toArray(new String[0]); } }
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// OJ: https://leetcode.com/problems/keyboard-row/ // Time: O(N) // Space: O(1) const string keyboard[3] = {"qwertyuiop", "asdfghjkl", "zxcvbnm"}; int m[26] = {}; static const auto __init__ = []() { for (int i = 0; i < 3; ++i) { for (char c : keyboard[i]) m[c - 'a'] = i; } return 0; }(); class Solution { public: vector<string> findWords(vector<string>& A) { vector<string> ans; for (auto &s : A) { int i = -1; for (char c : s) { int j = m[tolower(c) - 'a']; if (i == -1) i = j; else if (j != i) { i = -1; break; } } if (i != -1) ans.push_back(s); } return ans; } };
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class Solution: def findWords(self, words: List[str]) -> List[str]: s1 = set('qwertyuiop') s2 = set('asdfghjkl') s3 = set('zxcvbnm') res = [] for word in words: t = set(word.lower()) if t <= s1 or t <= s2 or t <= s3: res.append(word) return res ############ class Solution(object): def findWords(self, words): """ :type words: List[str] :rtype: List[str] """ ans = [] d = {} row1 = "qwertyuiop" row2 = "asdfghjkl" row3 = "zxcvbnm" for r in row1: d[r] = 1.0 for r in row2: d[r] = 2.0 for r in row3: d[r] = 3.0 for word in words: same = True pre = d[word[0].lower()] for c in word: if pre != d[c.lower()]: same = False break pre = d[c.lower()] if same: ans.append(word) return ans
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func findWords(words []string) (ans []string) { s := "12210111011122000010020202" for _, w := range words { x := s[unicode.ToLower(rune(w[0]))-'a'] ok := true for _, c := range w[1:] { if s[unicode.ToLower(c)-'a'] != x { ok = false break } } if ok { ans = append(ans, w) } } return }
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function findWords(words: string[]): string[] { const s = '12210111011122000010020202'; const ans: string[] = []; for (const w of words) { const t = w.toLowerCase(); const x = s[t.charCodeAt(0) - 'a'.charCodeAt(0)]; let ok = true; for (const c of t) { if (s[c.charCodeAt(0) - 'a'.charCodeAt(0)] !== x) { ok = false; break; } } if (ok) { ans.push(w); } } return ans; }
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public class Solution { public string[] FindWords(string[] words) { string s = "12210111011122000010020202"; IList<string> ans = new List<string>(); foreach (string w in words) { char x = s[char.ToLower(w[0]) - 'a']; bool ok = true; for (int i = 1; i < w.Length; ++i) { if (s[char.ToLower(w[i]) - 'a'] != x) { ok = false; break; } } if (ok) { ans.Add(w); } } return ans.ToArray(); } }