Keyboard Row - Practice Coding Problems

Keyboard Row - Problem

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.

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"
The third row consists of the characters "zxcvbnm"

Input & Output

Example 1 — Mixed Rows

$ Input: words = ["Hello", "Alaska", "Dad", "Peace"]

› Output: ["Alaska", "Dad"]

💡 Note: "Alaska" uses only row 1 letters (a,s,k), "Dad" uses only row 2 letters (d,a). "Hello" and "Peace" use multiple rows.

Example 2 — Single Character

$ Input: words = ["omk"]

› Output: []

💡 Note: "omk" contains 'o' (row 1), 'm' (row 3), 'k' (row 2) - uses all three keyboard rows, so not included.

Example 3 — Case Insensitive

$ Input: words = ["adsdf", "sfd"]

› Output: ["adsdf", "sfd"]

💡 Note: Both words use only row 2 letters (a,d,s,f). Case doesn't matter as stated in problem.

Constraints

1 ≤ words.length ≤ 20
1 ≤ words[i].length ≤ 100
words[i] consists of English letters.

Visualization

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Understanding the Visualization

Input

Array of words to check

Process

Check if each word uses only one keyboard row

Output

Return words that pass the single-row test

Key Takeaway

🎯 Key Insight: Pre-map letters to rows for instant validation rather than searching strings repeatedly

Asked in

G Google 15 a Amazon 12 M Microsoft 8 🍎 Apple 6

The key insight is to pre-map keyboard letters to their rows for fast lookup. Best approach uses a hash map to store each letter's row number, then validates each word by checking if all characters belong to the same row. Time: O(n × m), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force String Matching	O(n × m × k)	O(1)	For each word, check every character against all three keyboard rows
Set-Based Row Validation	O(n × m)	O(1)	Use sets for each keyboard row to enable faster membership testing
Hash Map Lookup	O(n × m)	O(1)	Pre-map each letter to its keyboard row for O(1) lookup

Brute Force String Matching — Algorithm Steps

For each word, assume it belongs to row 1
Check if any character is not in row 1, then try row 2
If not in row 2, try row 3
If all characters belong to same row, add word to result

Visualization

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Step-by-Step Walkthrough

Input Word

Take first word and convert to lowercase

Find Row

Search which row contains first character

Validate

Check if all characters belong to same row

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    char* rows[] = {"qwertyuiop", "asdfghjkl", "zxcvbnm"};
    char result[1000][100];
    int resultCount = 0;
    
    // Simple JSON parsing for array of strings
    char* token = strtok(line + 1, ","); // Skip first bracket
    while (token != NULL) {
        // Remove quotes and whitespace
        while (*token == ' ' || *token == '"') token++;
        int len = strlen(token);
        while (len > 0 && (token[len-1] == ' ' || token[len-1] == '"' || token[len-1] == ']')) {
            token[len-1] = 0;
            len--;
        }
        
        if (strlen(token) > 0) {
            // Convert to lowercase for checking
            char wordLower[100];
            strcpy(wordLower, token);
            for (int i = 0; wordLower[i]; i++) {
                wordLower[i] = tolower(wordLower[i]);
            }
            
            // Find which row the first character belongs to
            int targetRow = -1;
            for (int i = 0; i < 3; i++) {
                if (strchr(rows[i], wordLower[0]) != NULL) {
                    targetRow = i;
                    break;
                }
            }
            
            // Check if all characters belong to the same row
            int valid = 1;
            for (int i = 0; wordLower[i]; i++) {
                if (strchr(rows[targetRow], wordLower[i]) == NULL) {
                    valid = 0;
                    break;
                }
            }
            
            if (valid) {
                strcpy(result[resultCount], token);
                resultCount++;
            }
        }
        
        token = strtok(NULL, ",");
    }
    
    // Print result
    printf("[");
    for (int i = 0; i < resultCount; i++) {
        if (i > 0) printf(",");
        printf("\"%s\"", result[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m × k)

n words × m average word length × k characters per row (30 total)

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force String Matching — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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