## Tuesday, February 7, 2017

### LeetCode 500 - Keyboard Row

https://leetcode.com/problems/keyboard-row/
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 1:
```Input: ["Hello", "Alaska", "Dad", "Peace"]
```
Note:

1. You may use one character in the keyboard more than once.
2. You may assume the input string will only contain letters of alphabet.
http://www.cnblogs.com/grandyang/p/6421749.html

```    vector<string> findWords(vector<string>& words) {
vector<string> res;
unordered_set<char> row1{'q','w','e','r','t','y','u','i','o','p'};
unordered_set<char> row2{'a','s','d','f','g','h','j','k','l'};
unordered_set<char> row3{'z','x','c','v','b','n','m'};
for (string word : words) {
int one = 0, two = 0, three = 0;
for (char c : word) {
if (c < 'a') c += 32;
if (row1.count(c)) one = 1;
if (row2.count(c)) two = 1;
if (row3.count(c)) three = 1;
if (one + two + three > 1) break;
}
if (one + two + three == 1) res.push_back(word);
}
return res;
}```

``````    public String[] findWords(String[] words) {
String[] strs = {"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"};
Map<Character, Integer> map = new HashMap<>();
for(int i = 0; i<strs.length; i++){
for(char c: strs[i].toCharArray()){
map.put(c, i);//put <char, rowIndex> pair into the map
}
}
for(String w: words){
if(w.equals("")) continue;
int index = map.get(w.toUpperCase().charAt(0));
for(char c: w.toUpperCase().toCharArray()){
if(map.get(c)!=index){
index = -1; //don't need a boolean flag.
break;
}
}
if(index!=-1) res.add(w);//if index != -1, this is a valid string
}
return res.toArray(new String[0]);
}``````
def findWords(self, words): """ :type words: List[str] :rtype: List[str] """ rs = map(set, ['qwertyuiop','asdfghjkl','zxcvbnm']) ans = [] for word in words: wset = set(word.lower()) if any(wset <= rset for rset in rs): ans.append(word) return ans
``````public String[] findWords(String[] words) {
return Stream.of(words).filter(s -> s.toLowerCase().matches("[qwertyuiop]*|[asdfghjkl]*|[zxcvbnm]*")).toArray(String[]::new);
}``````