onenoughtone
Code Implementation
Word Puzzle Challenge Implementation
Below is the implementation of the word puzzle challenge:
solution.js
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849function isAnagram(s, t) { // Clean the input strings s = s.toLowerCase().replace(/[^a-z0-9]/g, ''); t = t.toLowerCase().replace(/[^a-z0-9]/g, ''); // Check if lengths are equal if (s.length !== t.length) { return false; } // Approach 1: Character Frequency Counting const counter = {}; // Count characters in first string for (let char of s) { counter[char] = (counter[char] || 0) + 1; } // Decrement counts for second string for (let char of t) { if (!counter[char]) { return false; } counter[char]--; } // Check if all counts are zero return Object.values(counter).every(count => count === 0);} // Alternative approach using sortingfunction isAnagramAlt(s, t) { // Clean the input strings s = s.toLowerCase().replace(/[^a-z0-9]/g, ''); t = t.toLowerCase().replace(/[^a-z0-9]/g, ''); // Check if lengths are equal if (s.length !== t.length) { return false; } // Sort both strings and compare return s.split('').sort().join('') === t.split('').sort().join('');} // Test casesconsole.log(isAnagram("listen", "silent")); // trueconsole.log(isAnagram("Astronomer", "Moon starer")); // trueconsole.log(isAnagram("hello", "world")); // falseStep-by-Step Explanation
Let's break down the implementation:
- Clean the Input: Normalize both strings by converting to lowercase and removing non-alphanumeric characters.
- Check Lengths: If the cleaned strings have different lengths, they cannot be anagrams, so return false.
- Count Characters: Create a hash table to count the frequency of each character in the first string.
- Verify Counts: Iterate through the second string, decrementing the counts. If any character is missing or has a count of zero, return false.
- Final Check: Ensure all character counts are zero, which confirms the strings are anagrams.
String Problems
Learning Objective
Implement the word puzzle challenge solution in different programming languages.
Word Puzzle Challenge Implementation
Below is the implementation of the word puzzle challenge in different programming languages. Select a language tab to view the corresponding code.
solution.js
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849function isAnagram(s, t) { // Clean the input strings s = s.toLowerCase().replace(/[^a-z0-9]/g, ''); t = t.toLowerCase().replace(/[^a-z0-9]/g, ''); // Check if lengths are equal if (s.length !== t.length) { return false; } // Approach 1: Character Frequency Counting const counter = {}; // Count characters in first string for (let char of s) { counter[char] = (counter[char] || 0) + 1; } // Decrement counts for second string for (let char of t) { if (!counter[char]) { return false; } counter[char]--; } // Check if all counts are zero return Object.values(counter).every(count => count === 0);} // Alternative approach using sortingfunction isAnagramAlt(s, t) { // Clean the input strings s = s.toLowerCase().replace(/[^a-z0-9]/g, ''); t = t.toLowerCase().replace(/[^a-z0-9]/g, ''); // Check if lengths are equal if (s.length !== t.length) { return false; } // Sort both strings and compare return s.split('').sort().join('') === t.split('').sort().join('');} // Test casesconsole.log(isAnagram("listen", "silent")); // trueconsole.log(isAnagram("Astronomer", "Moon starer")); // trueconsole.log(isAnagram("hello", "world")); // falseStep-by-Step Explanation
1Clean the Input
Normalize both strings by converting to lowercase and removing non-alphanumeric characters.
2Check Lengths
If the cleaned strings have different lengths, they cannot be anagrams, so return false.
3Count Characters
Create a hash table to count the frequency of each character in the first string.
4Verify Counts
Iterate through the second string, decrementing the counts. If any character is missing or has a count of zero, return false.
5Final Check
Ensure all character counts are zero, which confirms the strings are anagrams.
Language-Specific Notes
JavaScript
- Uses regular expressions with
replace()for string cleaning - Uses an object as a hash table for character counting
- Leverages
Object.values()andevery()for final verification - Alternative approach uses
split(''),sort(), andjoin('')
Python
- Uses list comprehension with
isalnum()for string cleaning - Uses a dictionary for character counting
- Leverages
all()with a generator expression for final verification - Alternative approach uses
sorted()for direct comparison
Java
- Uses
replaceAll()with regex for string cleaning - Uses
HashMapfor character counting - Requires explicit iteration over values for final verification
- Alternative approach uses
Arrays.sort()andArrays.equals()
C++
- Uses a custom
cleanString()function withisalnum()andtolower() - Uses
unordered_mapfor character counting - Requires explicit iteration over map entries for final verification
- Alternative approach uses
sort()and direct string comparison
Key Takeaways
- There are two main approaches to check if strings are anagrams: character frequency counting and sorting.
- The frequency counting approach is more efficient with O(n) time complexity.
- Proper input cleaning is essential for handling case sensitivity and non-alphanumeric characters.
- Hash tables (or dictionaries/maps) are ideal for counting character frequencies.
- This problem demonstrates the importance of string normalization in text processing applications.
Try It Yourself
Modify the code to implement an alternative approach and test it with the same examples.
Challenge:
Implement a function that solves the word puzzle challenge problem using a different approach than shown above.
Common Edge Cases
Empty Strings
Two empty strings are anagrams of each other.
isAnagram("", "") → true
Case Sensitivity
The function should be case-insensitive.
isAnagram("Tea", "Eat") → true
Special Characters
Special characters and spaces should be ignored.
isAnagram("A b c!", "cba") → true