Keyboard Row

LeetCode 500 | Difficulty: Easy

Easy

Problem 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 <= 20`

- `1 <= words[i].length <= 100`

- `words[i]` consists of English letters (both lowercase and uppercase).

Topics: Array, Hash Table, String

---

Approach

Hash Map

Use a hash map for O(1) average lookups. Store seen values, frequencies, or indices. The key question: what should I store as key, and what as value?

When to use

Need fast lookups, counting frequencies, finding complements/pairs.

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

---

Solutions

Solution 1: C# (Best: 228 ms)

Metric	Value
Runtime	228 ms
Memory	42.9 MB
Date	2022-02-02

Solution

public class Solution {
    public string[] FindWords(string[] words) {
       string option1 = "qwertyuiop";
    string option2 = "asdfghjkl";
    string option3 = "zxcvbnm";
    List<string> result = new List<string>();
    foreach (var word in words)
    {
        var word1 = word.ToLower();
        if (!word1.Where(x => !option1.Contains(x)).Any() || 
            !word1.Where(x => !option2.Contains(x)).Any() || 
            !word1.Where(x => !option3.Contains(x)).Any())
        result.Add(word);
    }
    return result.ToArray();    
    }
}

📜 1 more C# submission(s)

Submission (2022-02-02) — 315 ms, 42.9 MB

public class Solution {
    public string[] FindWords(string[] words) {
        string option1 = "qwertyuiop";
    string option2 = "asdfghjkl";
    string option3 = "zxcvbnm";
    List<string> result = new List<string>();
    foreach (var word in words)
    {
        var word1 = word.ToLower();
        if (word1.Where(x => !option1.Contains(x)).ToList().Count == 0) result.Add(word);
        if (word1.Where(x => !option2.Contains(x)).ToList().Count == 0) result.Add(word);
        if (word1.Where(x => !option3.Contains(x)).ToList().Count == 0) result.Add(word);
    }
    return result.ToArray();
    }
}

---

Complexity Analysis

Approach	Time	Space
Hash Map	$O(n)$	$O(n)$

---

Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Hash map gives O(1) lookup — think about what to use as key vs value.