How do you solve Keyboard Row on LeetCode?

Keyboard Row (LeetCode #500) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding character mapping is crucial for efficient lookups.

What is the brute force solution for Keyboard Row?

The brute force approach for Keyboard Row is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each word individually to see if all its letters belong to one of the three keyboard rows. This is straightforward but can be slow due to repeated checks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Keyboard Row?

Keyboard Row uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Keyboard Row?

Explain your thought process clearly before coding. Consider edge cases, such as single-letter words. Discuss time and space complexity as you develop your solution.

What are common mistakes when solving Keyboard Row?

Not considering case insensitivity. Failing to check all characters belong to the same row.

#500

Keyboard Row

Easy

ArrayHash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a set to map each character to its respective row, allowing for quick lookups. This reduces the need for multiple checks and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a mapping of each character to its corresponding keyboard row.
2Step 2: For each word, convert it to lowercase and check the row of the first character.
3Step 3: Verify if all characters belong to the same row using the mapping.

solution.py11 lines

1def findWords(words):
2    row_map = {c: 1 for c in 'qwertyuiop'}
3    row_map.update({c: 2 for c in 'asdfghjkl'})
4    row_map.update({c: 3 for c in 'zxcvbnm'})
5    result = []
6    for word in words:
7        lower_word = word.lower()
8        row = row_map[lower_word[0]]
9        if all(row_map[c] == row for c in lower_word):
10            result.append(word)
11    return result

ℹ

Complexity note: The time complexity is O(n) since we only traverse each word once and perform constant-time lookups for each character. The space complexity is O(n) for the row mapping.

1Understanding character mapping is crucial for efficient lookups.
2Case insensitivity can be handled by converting to lowercase.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.