Keyboard Row – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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).

C++ Keyboard Row LeetCode Solution

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<int> dict(26);
        vector<string> rows = {"QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM"};
        for (int i = 0; i < rows.size(); i++) {
            for (auto c : rows[i]) dict[c-'A'] = 1 << i;
        }
        vector<string> res;
        for (auto w : words) {
            int r = 7;
            for (char c : w) {
                r &= dict[toupper(c)-'A'];
                if (r == 0) break;
            }
            if (r) res.push_back(w);
        }
        return res;
    }
};

Java Keyboard Row LeetCode Solution

public class Solution {
    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
            }
        }
        List<String> res = new LinkedList<>();
        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]);
    }
}
 

Python 3 Keyboard Row LeetCode Solution

class Solution(object):
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        a=set('qwertyuiop')
        b=set('asdfghjkl')
        c=set('zxcvbnm')
        ans=[]
        for word in words:
            t=set(word.lower())
            if a&t==t:
                ans.append(word)
            if b&t==t:
                ans.append(word)
            if c&t==t:
                ans.append(word)
        return ans

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264

Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195

Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140

Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96

Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66

Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40

Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31

Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21

Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16

Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9

Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7

Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4

Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3

Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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