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Brace Expansion II

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.

Solution Explanation for LeetCode 1096: Brace Expansion II

This problem involves parsing and evaluating a string expression based on a specific grammar involving braces {}, commas ,, and concatenation. The goal is to generate a sorted list of all possible strings that can be formed by following the grammar rules.

Understanding the Grammar

The grammar can be summarized as follows:

  1. Base Case: A single lowercase letter represents a set containing only that letter. R("a") = {"a"}

  2. Union: A comma-separated list within braces {} represents the union of the sets represented by each element. R("{a,b,c}") = {"a", "b", "c"}

  3. Concatenation: Concatenation of two expressions results in the Cartesian product of their respective sets. R("{a,b}{c,d}") = {"ac", "ad", "bc", "bd"}

Approach: Depth-First Search (DFS)

The most efficient way to solve this problem is using a recursive Depth-First Search (DFS) approach. The algorithm explores all possible combinations systematically.

Algorithm:

  1. dfs(exp) function: This recursive function takes an expression exp as input.

  2. Base Case: If exp does not contain a closing brace }, it means we have reached a valid string. Add this string to the s set (using a set to automatically handle uniqueness).

  3. Recursive Step:

    • Find the last opening brace { and the matching closing brace }.
    • Extract the part between the braces (the comma-separated options).
    • For each option in the comma-separated list, recursively call dfs with the updated expression (replacing the bracketed portion with the current option).

  4. Final Step: Once the DFS completes, convert the s set into a sorted list and return it.

Time and Space Complexity Analysis:

  • Time Complexity: The time complexity is exponential in the worst case, O(2n), where n is the number of comma-separated options within the braces. This is because each option leads to a branching path in the DFS tree. However, it's often much faster in practice if there aren't deeply nested bracketed expressions with many options.

  • Space Complexity: The space complexity is also potentially exponential, O(2n), due to the recursive calls and the s set storing all generated strings in the worst case. The space used by the recursive call stack can also contribute to the space complexity.

Code Implementation (Python):

class Solution:
    def braceExpansionII(self, expression: str) -> List[str]:
        s = set()  # Use a set to automatically handle uniqueness
 
        def dfs(exp):
            j = exp.find('}')
            if j == -1:  # Base case: No more braces
                s.add(exp)
                return
 
            i = exp.rfind('{', 0, j - 1)  # Find the last opening brace
            a, c = exp[:i], exp[j + 1:]  # Split into parts before and after braces
            for b in exp[i + 1:j].split(','):  # Iterate through options
                dfs(a + b + c)  # Recursive call
 
        dfs(expression)
        return sorted(list(s))  # Convert set to sorted list

The implementations in Java, C++, Go, and TypeScript follow a similar logic, adapting the syntax to their respective languages. The core recursive DFS algorithm remains the same. Note that the use of a TreeSet in Java provides automatic sorting, eliminating the need for a separate sort step. Similarly, set in C++ and map in Go provide automatic handling of uniqueness and implicitly sort the keys when iterated over. The TypeScript code explicitly uses a Set for uniqueness and then sorts the resulting array.