Solved Problem1 - Group Anagrams, Problem 2 - Isomorphic strings and Problem 3 - Word pattern

[anirudhv98]

Completed Hashing-1, please review

[super30admin]

Grouping Anagrams Together (Problem1.cs)

Your solution is correct and handles the problem well. However, there is an opportunity to improve the time efficiency. The current approach sorts each string, which takes O(K log K) time per string. For large strings, this could be a bottleneck.

Consider using an alternative method to generate a key for anagrams without sorting. One common approach is to use a frequency array or a prime number hashing technique. For example, you can assign each letter a prime number and compute the product of primes for all characters in the string. Since multiplication of primes is commutative and unique (due to the fundamental theorem of arithmetic), this product will be the same for anagrams. This method would reduce the key generation time to O(K) per string.

Here's an example of how you might implement the prime hashing in C#:

private int[] primes = new int[] {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101};

public IList<IList<string>> GroupAnagrams(string[] strs) {
    Dictionary<double, List<string>> dict = new Dictionary<double, List<string>>();
    foreach (string str in strs) {
        double hash = 1;
        foreach (char c in str) {
            hash *= primes[c - 'a'];
        }
        if (!dict.ContainsKey(hash)) {
            dict[hash] = new List<string>();
        }
        dict[hash].Add(str);
    }
    return dict.Values.ToList<IList<string>>();
}

Note: Be cautious with large numbers; using double might lead to precision issues for very long strings. Alternatively, you can use a BigInteger in C# to avoid overflow, but for the constraints (string length <= 100), double should be sufficient.

Another alternative is to use a frequency string as the key. For each string, create a character count array and convert it to a string (e.g., "#2#0#0...#1" for "aab" where 2 is count of 'a', 1 is count of 'b', etc.). This also runs in O(K) time per string.

Your current solution is acceptable and passes on LeetCode, but for optimal performance, especially in scenarios with long strings, consider one of these alternatives.

VERDICT: PASS

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Isomorphic Strings (Problem2.cs)

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VERDICT: NEEDS_IMPROVEMENT

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Word Pattern (Problem3.cs)

Your solution is correct and efficient. You have correctly implemented the bijection check by using a dictionary for pattern-to-word mapping and a hash set to ensure that no two pattern characters map to the same word. This is a good approach.

Strengths:

• You correctly check for the length mismatch at the beginning, which is important.
• The code is clean and well-commented.
• You have considered both directions of mapping: pattern to word (using dictionary) and word to pattern (using hash set to ensure uniqueness).

Areas for Improvement:

• While your solution is correct, it is slightly different from the common approach of using two dictionaries (one for each direction). Using two dictionaries (one for pattern char to word and one for word to pattern char) might make the code more symmetric and easier to understand. For example, in your current code, the hash set only stores words to check for duplicates, but it doesn't store which pattern character they are mapped to. This is sufficient for uniqueness, but if you needed to retrieve the pattern character for a word (for some reason), you wouldn't have that information. However, for this problem, it is not necessary.
• You might want to consider edge cases more explicitly, such as when the pattern is empty or the string is empty, but your current code should handle these because the length check would catch them.

Overall, your solution is excellent. Keep up the good work!

VERDICT: PASS