Introduction

In this blog post, we will explore how to solve the LeetCode problem "1096" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Under the grammar given below, strings can represent a set of lowercase words. Let R(expr) denote the set of words the expression represents. The grammar can best be understood through simple examples: Single letters represent a singleton set containing that word. R("a") = {"a"} R("w") = {"w"} When we take a comma-delimited list of two or more expressions, we take the union of possibilities. R("{a,b,c}") = {"a","b","c"} R("{{a,b},{b,c}}") = {"a","b","c"} (notice the final set only contains each word at most once) When we concatenate two expressions, we take the set of possible concatenations between two words where the first word comes from the first expression and the second word comes from the second expression. R("{a,b}{c,d}") = {"ac","ad","bc","bd"} R("a{b,c}{d,e}f{g,h}") = {"abdfg", "abdfh", "abefg", "abefh", "acdfg", "acdfh", "acefg", "acefh"} Formally, the three rules for our grammar: For every lowercase letter x, we have R(x) = {x}. For expressions e1, e2, ... , ek with k >= 2, we have R({e1, e2, ...}) = R(e1) ∪ R(e2) ∪ ... For expressions e1 and e2, we have R(e1 + e2) = {a + b for (a, b) in R(e1) × R(e2)}, where + denotes concatenation, and × denotes the cartesian product. Given an expression representing a set of words under the given grammar, return the sorted list of words that the expression represents. Example 1: Input: expression = "{a,b}{c,{d,e}}" Output: ["ac","ad","ae","bc","bd","be"] Example 2: Input: expression = "{{a,z},a{b,c},{ab,z}}" Output: ["a","ab","ac","z"] Explanation: Each distinct word is written only once in the final answer. Constraints: 1 <= expression.length <= 60 expression[i] consists of '{', '}', ','or lowercase English letters. The given expression represents a set of words based on the grammar given in the description.

Explanation

Here's a breakdown of the approach, followed by the code:

• Recursive Descent Parsing: The core idea is to implement a recursive descent parser to evaluate the expression according to the given grammar. The parser will have functions to handle concatenation and union (comma-separated lists).
• Set Operations: Use Python's set data structure to efficiently manage the words and perform union operations, automatically handling duplicates.

• Concatenation: Implement concatenation by iterating through the sets of words generated by the sub-expressions.

• Runtime Complexity: O(N 3^L) where N is the length of the expression string and L is the maximum length of a word in the generated set. Note: The worst case 3^L occurs from cases like "{a,b,c}{a,b,c}...{a,b,c}". While expression.length is capped at 60, the maximum number of generated words and overall runtime grows exponentially. *Storage Complexity: O(3^L), for storing the generated set of words.

Code

    class Solution:
    def braceExpansionII(self, expression: str) -> list[str]:
        def parse(s, i):
            """Parses the expression starting from index i and returns the set of words and the updated index."""
            result = set()
            while i < len(s) and s[i] != '}':
                words, i = parse_concat(s, i)
                result.update(words)
                if i < len(s) and s[i] == ',':
                    i += 1
            return result, i + 1 if i < len(s) else i

        def parse_concat(s, i):
            """Parses a concatenation of expressions."""
            result = {""}
            while i < len(s) and s[i] != '}' and s[i] != ',':
                if s[i] == '{':
                    words, i = parse(s, i + 1)
                else:
                    words = {s[i]}
                    i += 1

                new_result = set()
                for word1 in result:
                    for word2 in words:
                        new_result.add(word1 + word2)
                result = new_result
            return result, i

        result_set, _ = parse(expression, 0)
        return sorted(list(result_set))