Solution Approach
Solution Overview
There are 2 main approaches to solve this problem:
- Character Frequency Counting - Complex approach
- Sorting Approach - Complex approach
Step 1: Clean the Input
Before checking if two strings are anagrams, we need to normalize them by removing spaces and converting to lowercase.
- Convert all characters to lowercase
- Remove all non-alphanumeric characters (spaces, punctuation, etc.)
- This ensures we're comparing only the relevant characters
Example:
Approach 1: Character Frequency Counting
This approach uses a hash table (or an array for limited character sets) to count the frequency of each character in both strings. If the strings are anagrams, they will have the same characters with the same frequencies.
Algorithm:
- Clean both input strings (convert to lowercase, remove spaces).
- If the cleaned strings have different lengths, return false immediately.
- Create a hash table or array to count character frequencies.
- Iterate through the first string, incrementing the count for each character.
- Iterate through the second string, decrementing the count for each character.
- If any count becomes negative or remains positive after processing both strings, return false.
- Otherwise, return true.
Time Complexity:
where n is the length of the strings. We need to iterate through each character once.
Space Complexity:
where k is the size of the character set. For English alphabet, this is O(26), which simplifies to O(1).
Approach 2: Sorting Approach
Another approach is to sort both strings and then compare them. If they are anagrams, the sorted strings will be identical.
Algorithm:
- Clean both input strings (convert to lowercase, remove spaces).
- If the cleaned strings have different lengths, return false immediately.
- Sort the characters in both strings.
- Compare the sorted strings. If they are identical, return true; otherwise, return false.
Time Complexity:
where n is the length of the strings. The sorting operation dominates the time complexity.
Space Complexity:
We need to create sorted copies of the strings, which requires space proportional to their length.
Pseudocode
12345678910111213141516171819202122232425262728function isAnagram(s, t): // Clean the input strings s = cleanString(s) t = cleanString(t) // Check if lengths are equal if s.length != t.length: return false // Create a frequency counter counter = new HashMap<Character, Integer>() // Count characters in first string for char in s: counter[char] = counter.getOrDefault(char, 0) + 1 // Decrement counts for second string for char in t: counter[char] = counter.getOrDefault(char, 0) - 1 if counter[char] < 0: return false // Check if all counts are zero for count in counter.values(): if count != 0: return false return trueString Problems
Learning Objective
Understand different approaches to solve the word puzzle challenge problem and analyze their efficiency.
Step 1: Clean the Input
Process:
- Convert all characters to lowercase
- Remove all non-alphanumeric characters (spaces, punctuation, etc.)
- This ensures we're comparing only the relevant characters
Example
Transformation:
Solution Approaches
Approach 1: Character Frequency Counting
This approach uses a hash table (or an array for limited character sets) to count the frequency of each character in both strings. If the strings are anagrams, they will have the same characters with the same frequencies.
Algorithm:
- Clean both input strings (convert to lowercase, remove spaces).
- If the cleaned strings have different lengths, return false immediately.
- Create a hash table or array to count character frequencies.
- Iterate through the first string, incrementing the count for each character.
- Iterate through the second string, decrementing the count for each character.
- If any count becomes negative or remains positive after processing both strings, return false.
- Otherwise, return true.
Approach 2: Sorting Approach
Another approach is to sort both strings and then compare them. If they are anagrams, the sorted strings will be identical.
Algorithm:
- Clean both input strings (convert to lowercase, remove spaces).
- If the cleaned strings have different lengths, return false immediately.
- Sort the characters in both strings.
- Compare the sorted strings. If they are identical, return true; otherwise, return false.
Complexity Analysis
Character Frequency Counting Approach
Time Complexity
where n is the length of the strings. We need to iterate through each character once.
Space Complexity
where k is the size of the character set. For English alphabet, this is O(26), which simplifies to O(1).
Sorting Approach Approach
Time Complexity
where n is the length of the strings. The sorting operation dominates the time complexity.
Space Complexity
We need to create sorted copies of the strings, which requires space proportional to their length.
Comparison:
The character frequency counting approach is more efficient with O(n) time complexity compared to the sorting approach's O(n log n). It also uses less space in most cases. However, the sorting approach is simpler to implement and may be preferred for its clarity when performance is not critical.
Pseudocode
Character Frequency Counting Approach
12345678910111213141516171819202122232425262728function isAnagram(s, t): // Clean the input strings s = cleanString(s) t = cleanString(t) // Check if lengths are equal if s.length != t.length: return false // Create a frequency counter counter = new HashMap<Character, Integer>() // Count characters in first string for char in s: counter[char] = counter.getOrDefault(char, 0) + 1 // Decrement counts for second string for char in t: counter[char] = counter.getOrDefault(char, 0) - 1 if counter[char] < 0: return false // Check if all counts are zero for count in counter.values(): if count != 0: return false return trueSorting Approach Approach
123456789101112131415function isAnagram(s, t): // Clean the input strings s = cleanString(s) t = cleanString(t) // Check if lengths are equal if s.length != t.length: return false // Sort both strings sortedS = sort(s) sortedT = sort(t) // Compare sorted strings return sortedS == sortedT