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.
Note that the strings are case-insensitive, both lowercased and uppercased of the same letter are treated as if they are at the same row.
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"]
Explanation:
Both "a" and "A" are in the 2nd row of the American keyboard due to case insensitivity.
Example 2:
Input: words = ["omk"]
Output: []
Example 3:
Input: words = ["adsdf","sfd"]
Output: ["adsdf","sfd"]
Constraints:
1 <= words.length <= 201 <= words[i].length <= 100words[i] consists of English letters (both lowercase and uppercase).
Solutions
Solution 1
PythonJavaC++GoTypeScriptC#
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(code-box)
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]);
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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();
}
}(code-box)
Solution 2
Python
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(code-box)
