[Array] Keyboard Row - Easy

The Array Keyboard Row Easy problem is about finding words that can be typed using letters from just one row of a QWERTY keyboard. To solve this, we check each word to see if we can make it with letters from one row only. This task helps us learn about working with strings and arrays. It also shows us how to compare sets of characters in a smart way.

In this article, we will look at the Array Keyboard Row Easy problem. We will explain the overview and the problem statement in simple terms. We will give solutions in Java, Python, and C++. We will also talk about how to make each solution better. We will check how we can test and confirm these solutions. We will compare them and think about how well each one performs. By the end, we will answer some common questions to help you understand better.

  • Array Keyboard Row Easy Problem Overview
  • Understanding the Problem Statement
  • Java Solution for Array Keyboard Row Easy
  • Python Implementation of Array Keyboard Row Easy
  • C++ Approach to Array Keyboard Row Easy
  • Optimizing the Array Keyboard Row Solution
  • Testing and Validating Array Keyboard Row Implementation
  • Comparative Analysis of Solutions in Java, Python, and C++
  • Performance Considerations for Array Keyboard Row Solutions
  • Frequently Asked Questions

Understanding the Problem Statement

The “Keyboard Row” problem means we need to find which words from a list can be typed using only one row of a QWERTY keyboard. The keyboard has three rows like this:

  • Row 1: q, w, e, r, t, y, u, i, o, p
  • Row 2: a, s, d, f, g, h, j, k, l
  • Row 3: z, x, c, v, b, n, m

Problem Specification

  • Input: An array of strings.
  • Output: An array with words that can be typed using letters from one keyboard row.
  • Case Insensitivity: We need to treat both uppercase and lowercase letters the same.

Example

For the input array ["Hello", "Alaska", "Dad", "Peace"], the output should be ["Alaska", "Dad"]. Both these words can be typed with letters from one row.

The hard part is to find out which row each letter of the words belongs to. Then we can filter the input array based on that.

Java Solution for Array Keyboard Row Easy

To solve the “Array Keyboard Row Easy” problem in Java, we want to find words that can be typed using letters from one row of the keyboard. A standard QWERTY keyboard has three rows. They are:

  • Row 1: "qwertyuiop"
  • Row 2: "asdfghjkl"
  • Row 3: "zxcvbnm"

We will check each word against these rows. Here is a simple Java code for this:

import java.util.HashSet;
import java.util.Set;

public class KeyboardRow {
    public String[] findWords(String[] words) {
        Set<Character> row1 = new HashSet<>();
        Set<Character> row2 = new HashSet<>();
        Set<Character> row3 = new HashSet<>();

        for (char c : "qwertyuiop".toCharArray()) row1.add(c);
        for (char c : "asdfghjkl".toCharArray()) row2.add(c);
        for (char c : "zxcvbnm".toCharArray()) row3.add(c);

        StringBuilder result = new StringBuilder();
        
        for (String word : words) {
            char[] chars = word.toLowerCase().toCharArray();
            Set<Character> currentRow = null;

            if (row1.contains(chars[0])) currentRow = row1;
            else if (row2.contains(chars[0])) currentRow = row2;
            else if (row3.contains(chars[0])) currentRow = row3;

            boolean canBeTyped = true;
            for (char c : chars) {
                if (!currentRow.contains(c)) {
                    canBeTyped = false;
                    break;
                }
            }

            if (canBeTyped) {
                result.append(word).append(" ");
            }
        }

        return result.toString().trim().split(" ");
    }

    public static void main(String[] args) {
        KeyboardRow kr = new KeyboardRow();
        String[] words = {"Hello", "Alaska", "Dad", "Peace"};
        String[] result = kr.findWords(words);
        for (String word : result) {
            System.out.print(word + " ");
        }
    }
}

Explanation:

  • Data Structures: We create three sets for each keyboard row.
  • Logic: For each word, we check the first letter to see which row it belongs to. Then we check if all letters in the word are from that row.
  • Output: The method gives back an array of words that can be typed using one row.

This solution works well. The time it takes is O(N * M). Here N is the number of words and M is the average length of the words.

Python Implementation of Array Keyboard Row Easy

To solve the “Array Keyboard Row Easy” problem in Python, we want to find words that we can type using letters from just one row of the keyboard. The rows of the keyboard look like this:

  • Row 1: qwertyuiop
  • Row 2: asdfghjkl
  • Row 3: zxcvbnm

We can make a map for each letter to show which row it belongs to. Then we check each word from the list to see if all its letters are in the same row.

Here is a simple Python code for this:

def findWords(words):
    row_map = {
        'q': 1, 'w': 1, 'e': 1, 'r': 1, 't': 1, 'y': 1, 'u': 1, 'i': 1, 'o': 1, 'p': 1,
        'a': 2, 's': 2, 'd': 2, 'f': 2, 'g': 2, 'h': 2, 'j': 2, 'k': 2, 'l': 2,
        'z': 3, 'x': 3, 'c': 3, 'v': 3, 'b': 3, 'n': 3, 'm': 3
    }

    result = []
    
    for word in words:
        # Change the word to lowercase
        lower_word = word.lower()
        # Find the row number of the first letter
        row_number = row_map[lower_word[0]]
        
        # Check if all letters are in the same row
        if all(row_map[char] == row_number for char in lower_word):
            result.append(word)
    
    return result

# Example usage
words = ["Hello", "Alaska", "Dad", "Peace"]
print(findWords(words))  # Output: ['Alaska', 'Dad']

In this code:

  • We make a dictionary called row_map. This shows which letter belongs to which row.
  • We go through each word, change it to lowercase, and check if all its letters are in the same row using a simple check.
  • If they are, we add the word to our result list.

This code checks the words against the keyboard rows and gives us the words we can type using just one row.

C++ Approach to Array Keyboard Row Easy

In this C++ approach, we will solve the Array Keyboard Row Easy problem. We need to find which words from a list can be made with letters from only one row of a standard QWERTY keyboard. We can show the keyboard rows as strings:

const std::string row1 = "qwertyuiop";
const std::string row2 = "asdfghjkl";
const std::string row3 = "zxcvbnm";

To solve this problem, we will use a set to hold the letters of each row for easy checking. The steps we take are:

  1. Change each word to lowercase.
  2. Check if all letters in the word come from any one of the keyboard rows.

Here is a simple C++ code that shows this way:

#include <iostream>
#include <vector>
#include <unordered_set>
#include <algorithm>

bool canBeTypedWithOneRow(const std::string &word, const std::unordered_set<char> &rowSet) {
    for (char c : word) {
        if (rowSet.find(tolower(c)) == rowSet.end()) {
            return false;
        }
    }
    return true;
}

std::vector<std::string> findWords(const std::vector<std::string>& words) {
    std::unordered_set<char> row1(row1.begin(), row1.end());
    std::unordered_set<char> row2(row2.begin(), row2.end());
    std::unordered_set<char> row3(row3.begin(), row3.end());
    
    std::vector<std::string> result;
    for (const std::string &word : words) {
        if (canBeTypedWithOneRow(word, row1) || 
            canBeTypedWithOneRow(word, row2) || 
            canBeTypedWithOneRow(word, row3)) {
            result.push_back(word);
        }
    }
    return result;
}

int main() {
    std::vector<std::string> words = {"Hello", "Alaska", "Dad", "Peace"};
    std::vector<std::string> result = findWords(words);
    
    for (const std::string &word : result) {
        std::cout << word << " ";
    }
    return 0;
}

Explanation:

  • The function canBeTypedWithOneRow checks if all letters in a word come from one row.
  • The findWords function goes through the list of words. It checks each word against all three rows.
  • We collect and print the result.

This C++ code helps find which words can be typed using just one row of the keyboard. For more about array problems, you can read articles like Array Two Sum - Easy or Array Contains Duplicate - Easy.

Optimizing the Array Keyboard Row Solution

We can make the solution for the “Array Keyboard Row” problem better by using sets. Sets help us check quickly if a character is in a certain row. We will represent keyboard rows as sets. This way, we can see fast if a character belongs to a row.

Approach

  1. Define Keyboard Rows: We make each keyboard row a set of characters.
  2. Iterate Through Words: For each word, we check if all characters are from the same row.
  3. Use Set for Fast Lookup: Sets let us check character presence in O(1) average time.

Implementation

Here is a simple implementation in Python:

def findWords(words):
    # Define the keyboard rows
    row1 = set("qwertyuiop")
    row2 = set("asdfghjkl")
    row3 = set("zxcvbnm")
    
    result = []
    
    for word in words:
        # Convert word to lowercase to handle case insensitivity
        lower_word = word.lower()
        
        # Determine the row of the first character
        if lower_word[0] in row1:
            row = row1
        elif lower_word[0] in row2:
            row = row2
        elif lower_word[0] in row3:
            row = row3
        else:
            continue
        
        # Check if all characters belong to the same row
        if all(char in row for char in lower_word):
            result.append(word)
    
    return result

# Example usage
words = ["Hello", "Alaska", "Dad", "Peace"]
print(findWords(words))  # Output: ['Alaska', 'Dad']

Key Points

  • Time Complexity: This way takes O(n * m) time. Here, n is the number of words and m is the average length of the words. This is good given the set work.
  • Space Complexity: We only need O(1) space for the character sets. They have a fixed number of elements (26 letters).
  • Case Insensitivity: We change words to lowercase to handle both upper and lower case letters.

This way makes fewer checks for each word. We also make sure the performance stays good as the input size increases. By using sets for checking characters, we make a better solution for the “Array Keyboard Row” problem.

Testing and Validating Array Keyboard Row Implementation

To make sure the Array Keyboard Row solution works correctly, we need to do careful testing and validation. We will create a list of test cases that cover different situations, including some special cases.

Test Cases

  1. Basic cases:
    • Input: ["Hello", "Alaska", "Dad", "Peace"]
      • Expected Output: ["Alaska", "Dad"]
    • Input: ["omk"]
      • Expected Output: []
    • Input: ["adsdf", "sfd"]
      • Expected Output: ["adsdf", "sfd"]
  2. Single Character Input:
    • Input: ["A", "b", "C"]
      • Expected Output: ["A", "C"]
  3. Mixed cases:
    • Input: ["qwerty", "QWERTY", "123"]
      • Expected Output: ["qwerty", "QWERTY"]
  4. Empty Input:
    • Input: []
      • Expected Output: []
  5. Input with Non-Alphabetic Characters:
    • Input: ["abc123", "def$", "gh!i"]
      • Expected Output: []

Validation Process

  • We should use assert statements in the code to help automate the testing.
  • It is good to use a testing framework like JUnit for Java, unittest for Python, or Google Test for C++. This way, we can run the test cases more easily.

Example Validation Code (Python)

def findWords(words):
    keyboard_rows = ["qwertyuiop", "asdfghjkl", "zxcvbnm"]
    rows_set = [set(row) for row in keyboard_rows]
    
    result = []
    for word in words:
        char_set = set(word.lower())
        if any(char_set <= row for row in rows_set):
            result.append(word)
    return result

# Testing Function
def test_findWords():
    assert findWords(["Hello", "Alaska", "Dad", "Peace"]) == ["Alaska", "Dad"]
    assert findWords(["omk"]) == []
    assert findWords(["adsdf", "sfd"]) == ["adsdf", "sfd"]
    assert findWords([]) == []
    assert findWords(["abc123", "def$", "gh!i"]) == []

test_findWords()

Performance Testing

  • We need to check the time complexity. It should be O(n * m). Here, n is the number of words and m is the average length of the words.
  • It is useful to measure the execution time with a big set of inputs. This way, we can see if the performance is good enough.

Edge Cases

  • We should test with different input sizes, such as:
    • A single word.
    • Large lists of words like 10,000 entries.
    • Words that use more than one keyboard row.

By testing the Array Keyboard Row solution carefully, we can check its reliability and efficiency in different situations.

Comparative Analysis of Solutions in Java Python and C++

When we solve the Array Keyboard Row problem, we can use Java, Python, or C++. Each language has its own way to write code and its own performance. We will compare the key parts of each solution.

Java Solution

import java.util.HashSet;
import java.util.Set;

public class KeyboardRow {
    public String[] findWords(String[] words) {
        Set<Character> row1 = new HashSet<>(); // QWERTYUIOP
        Set<Character> row2 = new HashSet<>(); // ASDFGHJKL
        Set<Character> row3 = new HashSet<>(); // ZXCVBNM

        // Fill the sets
        for (char c : "qwertyuiop".toCharArray()) row1.add(c);
        for (char c : "asdfghjkl".toCharArray()) row2.add(c);
        for (char c : "zxcvbnm".toCharArray()) row3.add(c);

        List<String> result = new ArrayList<>();
        for (String word : words) {
            char[] chars = word.toLowerCase().toCharArray();
            Set<Character> rowSet = null;

            // Determine which row to use
            if (row1.contains(chars[0])) rowSet = row1;
            else if (row2.contains(chars[0])) rowSet = row2;
            else if (row3.contains(chars[0])) rowSet = row3;

            // Check if all characters are in the same row
            boolean valid = true;
            for (char ch : chars) {
                if (!rowSet.contains(ch)) {
                    valid = false;
                    break;
                }
            }

            if (valid) result.add(word);
        }
        return result.toArray(new String[0]);
    }
}

Python Implementation

def findWords(words):
    row1 = set("qwertyuiop")
    row2 = set("asdfghjkl")
    row3 = set("zxcvbnm")
    
    result = []
    for word in words:
        chars = set(word.lower())
        if chars <= row1 or chars <= row2 or chars <= row3:
            result.append(word)
    return result

C++ Approach

#include <vector>
#include <unordered_set>
#include <string>

using namespace std;

vector<string> findWords(vector<string>& words) {
    unordered_set<char> row1{'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'};
    unordered_set<char> row2{'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'};
    unordered_set<char> row3{'z', 'x', 'c', 'v', 'b', 'n', 'm'};

    vector<string> result;
    for (const string& word : words) {
        char first = tolower(word[0]);
        unordered_set<char>* rowSet = nullptr;

        if (row1.count(first)) rowSet = &row1;
        else if (row2.count(first)) rowSet = &row2;
        else if (row3.count(first)) rowSet = &row3;

        bool valid = true;
        for (char ch : word) {
            if (!rowSet->count(tolower(ch))) {
                valid = false;
                break;
            }
        }
        if (valid) result.push_back(word);
    }
    return result;
}

Performance Considerations

  • Time Complexity: All three solutions run in O(N * M). Here N is number of words and M is the average length of the words.
  • Space Complexity: Each solution uses extra space for character sets. This is O(1) because size of keyboard rows is constant.

Final Thoughts

Java, Python, and C++ solutions give us similar performance and function for the Array Keyboard Row problem. The choice of language often depends on what the developer knows best and the needs of the project. If we want to read more about array problems, we can check articles like Array Two Sum and Array Contains Duplicate.

Performance Considerations for Array Keyboard Row Solutions

When we work on solutions for the Array Keyboard Row problem, we must think about performance. This is important to keep things running fast as the input size gets bigger. Here are some key points to think about:

  • Time Complexity:
    • The basic method is to go through the words and check if each letter is in one of the three keyboard rows. If we have n words and m is the longest word, the time complexity is O(n * m). We can make this faster if we prepare the character row mapping in advance.
  • Space Complexity:
    • Using a set to keep characters of each row takes O(1) space. This is because the maximum number of unique characters is fixed at 26 (the English letters). So, the space complexity is O(1).
  • Data Structures:
    • Using a set helps us check membership in O(1) average time, which can make our solution much faster. For example, we can create three sets for each keyboard row.
  • Input Size:
    • As the input size gets larger, how well the algorithm works becomes even more important. We should make sure our implementation deals with special cases, like empty strings or very short words, in a good way.
  • Batch Processing:
    • If we need to process many words, we can use batch processing methods. This helps cut down the extra work from checking the same characters again.

Example Implementations

Here is a quick look at how we can implement this in different programming languages, focusing on making it run better:

Java Implementation:

import java.util.*;

public class KeyboardRow {
    public String[] findWords(String[] words) {
        Set<Character> row1 = new HashSet<>(Arrays.asList('q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'));
        Set<Character> row2 = new HashSet<>(Arrays.asList('a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'));
        Set<Character> row3 = new HashSet<>(Arrays.asList('z', 'x', 'c', 'v', 'b', 'n', 'm'));
        List<String> result = new ArrayList<>();

        for (String word : words) {
            char[] chars = word.toLowerCase().toCharArray();
            Set<Character> rowUsed = null;

            for (char c : chars) {
                if (row1.contains(c)) {
                    if (rowUsed == null) rowUsed = row1;
                    else if (rowUsed != row1) break;
                } else if (row2.contains(c)) {
                    if (rowUsed == null) rowUsed = row2;
                    else if (rowUsed != row2) break;
                } else if (row3.contains(c)) {
                    if (rowUsed == null) rowUsed = row3;
                    else if (rowUsed != row3) break;
                }
            }
            if (rowUsed != null) result.add(word);
        }
        return result.toArray(new String[0]);
    }
}

Python Implementation:

def find_words(words):
    row1 = set("qwertyuiop")
    row2 = set("asdfghjkl")
    row3 = set("zxcvbnm")
    result = []

    for word in words:
        chars = set(word.lower())
        if chars <= row1 or chars <= row2 or chars <= row3:
            result.append(word)
    return result

C++ Implementation:

#include <vector>
#include <string>
#include <unordered_set>
#include <algorithm>

class KeyboardRow {
public:
    std::vector<std::string> findWords(std::vector<std::string>& words) {
        std::unordered_set<char> row1 = {'q','w','e','r','t','y','u','i','o','p'};
        std::unordered_set<char> row2 = {'a','s','d','f','g','h','j','k','l'};
        std::unordered_set<char> row3 = {'z','x','c','v','b','n','m'};
        std::vector<std::string> result;

        for (auto& word : words) {
            std::string lowerWord = word;
            std::transform(lowerWord.begin(), lowerWord.end(), lowerWord.begin(), ::tolower);
            std::unordered_set<char> chars(lowerWord.begin(), lowerWord.end());

            if (std::includes(row1.begin(), row1.end(), chars.begin(), chars.end()) ||
                std::includes(row2.begin(), row2.end(), chars.begin(), chars.end()) ||
                std::includes(row3.begin(), row3.end(), chars.begin(), chars.end())) {
                result.push_back(word);
            }
        }
        return result;
    }
};

By looking at these performance points, we can make better solutions for the Array Keyboard Row problem. This will help us improve speed and manage our resources better. For more info on similar array problems, we can check out Array Two Sum and Array Best Time to Buy and Sell Stock.

Frequently Asked Questions

What is the Array Keyboard Row Problem?

The Array Keyboard Row problem is about finding which words from a list can be typed using letters from just one row of a QWERTY keyboard. This problem is easy to solve. When we learn how to tackle this problem, we can improve our skills in handling arrays and processing strings.

How do I solve the Array Keyboard Row problem in Java?

To solve the Array Keyboard Row problem in Java, we can use a set to show the letters in each keyboard row. Then we check each word to see if all its letters are from one row. This way is simple and works well. It is a nice way to practice Java coding and array use.

Can I implement the Array Keyboard Row solution in Python?

Yes, we can easily implement the Array Keyboard Row solution in Python. We can use sets to store the letters of each keyboard row. Then we check if a word can be made using letters from just one row. This method is short and uses Python’s strong data tools for good performance.

What is the C++ approach for solving the Array Keyboard Row problem?

In C++, we can solve the Array Keyboard Row problem by making arrays or sets for each keyboard row. Next, we go through the words given and check if each word can be made with only the letters from one row. This way takes advantage of C++’s good data handling for effective solutions.

How can I optimize the Array Keyboard Row solution?

To make the Array Keyboard Row solution better, we should try to lower the number of checks for each word. We can use hash sets for each row. This helps us quickly see if letters are in that row. Also, we can stop early if we find a letter that does not match. This will help a lot, especially with bigger lists.

For more reading on array challenges, look at Array Two Sum and Array Contains Duplicate to practice more algorithms.