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500. Keyboard Row

Description

Given an array of strings words, return the words that can be typed using letters of the alphabet on only one row of American keyboard like the image below.

In the American keyboard:

  • the first row consists of the characters "qwertyuiop",
  • the second row consists of the characters "asdfghjkl", and
  • the third row consists of the characters "zxcvbnm".

 

Example 1:

Input: words = ["Hello","Alaska","Dad","Peace"]
Output: ["Alaska","Dad"]

Example 2:

Input: words = ["omk"]
Output: []

Example 3:

Input: words = ["adsdf","sfd"]
Output: ["adsdf","sfd"]

 

Constraints:

  • 1 <= words.length <= 20
  • 1 <= words[i].length <= 100
  • words[i] consists of English letters (both lowercase and uppercase). 

Solutions

Solution 1

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class Solution:
    def findWords(self, words: List[str]) -> List[str]:
        s1 = set('qwertyuiop')
        s2 = set('asdfghjkl')
        s3 = set('zxcvbnm')
        ans = []
        for w in words:
            s = set(w.lower())
            if s <= s1 or s <= s2 or s <= s3:
                ans.append(w)
        return ans

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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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class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        string s = "12210111011122000010020202";
        vector<string> ans;
        for (auto& w : words) {
            char x = s[tolower(w[0]) - 'a'];
            bool ok = true;
            for (char& c : w) {
                if (s[tolower(c) - 'a'] != x) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans.emplace_back(w);
            }
        }
        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();
    }
}

Solution 2

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class Solution:
    def findWords(self, words: List[str]) -> List[str]:
        ans = []
        s = "12210111011122000010020202"
        for w in words:
            x = s[ord(w[0].lower()) - ord('a')]
            if all(s[ord(c.lower()) - ord('a')] == x for c in w):
                ans.append(w)
        return ans

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