Solution Explanation for Valid Anagram

This problem asks whether two strings, s and t, are anagrams of each other. An anagram is a word or phrase formed by rearranging the letters of another word or phrase. This means that s and t must have the same length, and each character in s must have a corresponding character in t with the same frequency.

We explore two main approaches: counting character frequencies and sorting.

Approach 1: Counting Character Frequencies

This approach uses a hash table (or an array in this case since we only have lowercase English letters) to store the frequency of each character in string s. Then, we iterate through string t, decrementing the count for each character encountered. If at any point a count becomes negative, it means t has more occurrences of a character than s, and they are not anagrams. If after iterating through t all counts are zero, s and t are anagrams.

Algorithm:

1. Check Lengths: If the lengths of s and t are different, they cannot be anagrams; return false.
2. Initialize Counter: Create an array cnt of size 26 (for lowercase English letters) and initialize all elements to 0.
3. Count Characters in s: Iterate through s, incrementing the count for each character (cnt[s[i] - 'a']++).
4. Decrement Counts with t: Iterate through t, decrementing the count for each character (cnt[t[i] - 'a']--).
5. Check for Differences: After processing t, iterate through cnt. If any element is not 0, it means there's an unequal character frequency; return false.
6. Return True: If all counts are 0, return true.

Time Complexity: O(n), where n is the length of the strings. We iterate through each string once. Space Complexity: O(1). We use a fixed-size array of size 26.

Code Examples:

(See the Python, Java, C++, Go, TypeScript, Rust, Javascript, C#, and C code examples provided in the original response. These all implement this approach efficiently).

Approach 2: Sorting

This approach leverages the fact that anagrams have the same characters when sorted. We sort both strings and then compare them for equality.

Algorithm:

1. Check Lengths: Same as in Approach 1.
2. Sort Strings: Sort both strings s and t alphabetically.
3. Compare Strings: Compare the sorted strings. If they are equal, return true; otherwise, return false.

Time Complexity: O(n log n), where n is the length of the strings, due to the sorting step. Space Complexity: O(n) in some implementations due to the creation of new sorted strings (although in place sorting algorithms can lower this). However, in languages like C++ that use std::sort, space complexity can be considered O(log n) due to recursive calls.

Code Examples (Illustrative):

(The Rust example from the original response utilizes this sorting approach. Note that this is less efficient than the counting approach for this specific problem).

Choosing the Best Approach:

For this specific problem, where the character set is limited to lowercase English letters, the counting character frequencies approach (Approach 1) is significantly more efficient due to its linear time complexity. The sorting approach becomes more competitive when dealing with larger character sets or when sorting is already a necessary operation within a broader context. In many cases, the improved time complexity of O(n) vs O(n log n) is preferred for improved efficiency.