Solution 1 - Keyboard Row Set Check

A word is valid if all its letters belong to the same keyboard row.

TimeComplexity: O(total\ characters)

SpaceComplexity: O(1)

""" 0500.1 - Keyboard Row - Solution 1 - Row Set Check """

from typing import List

class Solution:
    def findWords(self, words: List[str]) -> List[str]:
        rows = [set("qwertyuiop"), set("asdfghjkl"), set("zxcvbnm")]
        out = []
        for w in words:
            s = set(w.lower())
            if any(s <= r for r in rows):
                out.append(w)
        return out

if __name__ == "__main__":
    print(Solution().findWords(["Hello","Alaska","Dad","Peace"]))  # ["Alaska","Dad"]

function findWords(words){
 const rows=["qwertyuiop","asdfghjkl","zxcvbnm"].map(r=>new Set(r.split("")));
 return words.filter(w=>{
  const s=new Set(w.toLowerCase().split(""));
  return rows.some(r => [...s].every(ch => r.has(ch)));
 });
}

function findWords(words:string[]):string[]{
 const rows=[new Set("qwertyuiop"), new Set("asdfghjkl"), new Set("zxcvbnm")];
 return words.filter(w=>{
  const s=new Set(w.toLowerCase());
  return rows.some(r => [...s].every(ch => r.has(ch)));
 });
}

package main

import "strings"

func findWords(words []string) []string {
	rows := []string{"qwertyuiop", "asdfghjkl", "zxcvbnm"}
	out := []string{}

	for _, w := range words {
		lw := strings.ToLower(w)
		for _, row := range rows {
			ok := true
			for i := 0; i < len(lw); i++ {
				if !strings.ContainsRune(row, rune(lw[i])) {
					ok = false
					break
				}
			}
			if ok {
				out = append(out, w)
				break
			}
		}
	}
	return out
}

func main() {}

fn find_words(words: Vec<&str>) -> Vec<&str> {
	let rows = ["qwertyuiop", "asdfghjkl", "zxcvbnm"];
	words.into_iter().filter(|w| {
		let lw = w.to_lowercase();
		rows.iter().any(|r| lw.chars().all(|c| r.contains(c)))
	}).collect()
}

fn main() {}

0500 - Keyboard Row

Solution 1 - Keyboard Row Set Check