Brace Expansion II

Hard

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StringsRecursion

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, ... , e_k 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What characters can appear within the curly braces besides comma and letters? Are there any constraints on the characters used, such as special characters or numbers?
Can the input string be empty or null? If so, what should I return in those cases?
Does the order of options within a group matter (e.g., `{a,b}` vs `{b,a}`)? Should the output be sorted alphabetically and deduplicated?
Can the expression have nested curly braces, and if so, how deeply nested can they be (is there a maximum nesting depth)?
If an expression results in the same string being generated multiple times through different expansions, should it only appear once in the final output?

Brute Force Solution

Approach

The core idea is to explore every possible combination of strings that can be formed by the expression. We break down the expression piece by piece, and for each part, we generate all possibilities, combining them until we have the complete set of results.

Here's how the algorithm would work step-by-step:

Start by identifying the simplest parts of the expression, like individual letters or strings.
For each part enclosed in curly braces, find all the different strings that can be generated within those braces.
If there are multiple comma-separated options inside the braces, treat each option as a separate possibility.
When you encounter sequential parts of the expression (meaning they are right next to each other), take all possibilities from the first part and combine them with all possibilities from the second part, creating new strings.
If the expression has nested braces, solve the innermost braces first and then work your way outwards.
As you generate new strings, keep track of all the unique strings you've found so you don't have duplicates.
Once you've processed the entire expression, sort the resulting unique strings in alphabetical order.

Code Implementation

def brace_expansion_ii(expression):
    results = set()

    def expand(sub_expression):
        options = set()

        if not sub_expression:
            return {""}

        index = 0
        while index < len(sub_expression):
            if sub_expression[index] == '{':
                brace_count = 1
                start_index = index + 1
                index += 1
                while index < len(sub_expression) and brace_count > 0:
                    if sub_expression[index] == '{':
                        brace_count += 1
                    elif sub_expression[index] == '}':
                        brace_count -= 1
                    index += 1
                end_index = index - 1

                # Recursively expand the braced expression.

                inner_options = set()
                current_option = ""
                for i in range(start_index, end_index):
                    if sub_expression[i] == ',':
                        inner_options.add(current_option)
                        current_option = ""
                    else:
                        current_option += sub_expression[i]
                inner_options.add(current_option)

                expanded_inner = set()
                for option in inner_options:
                    expanded_inner.update(expand(option))

                if index < len(sub_expression):

                    # Concatenate current brace expansion with subsequent expansions.
                    next_options = expand(sub_expression[index:])
                    new_options = set()
                    for inner_option in expanded_inner:
                        for next_option in next_options:
                            new_options.add(inner_option + next_option)
                    return new_options

                else:
                    return expanded_inner

            else:
                start_index = index
                while index < len(sub_expression) and sub_expression[index] != '{' and sub_expression[index] != '}':
                    index += 1
                
                # Extract a sequence of characters and handle it
                prefix = sub_expression[start_index:index]

                if index < len(sub_expression):
                    next_options = expand(sub_expression[index:])
                    new_options = set()
                    for next_option in next_options:
                        new_options.add(prefix + next_option)
                    return new_options

                else:
                    return {prefix}

        return options

    results = expand(expression)

    # Sort and return final results
    sorted_results = sorted(list(results))
    return sorted_results

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible combinations of strings within the input expression. In the worst-case scenario, each set of braces can double the number of potential output strings due to comma-separated options. Therefore, the number of strings generated can grow exponentially with the length of the input, n. Operations such as generating unique strings and sorting them can also contribute, but the exponential string generation dominates the time complexity. Thus, the time complexity is approximately O(2^n), where n reflects the complexity of nested options.

Space Complexity

O(2^N) – The algorithm explores all possible combinations of strings within the expression. In the worst-case scenario, each brace can double the number of possible string combinations. This leads to an exponential growth of intermediate results stored in sets and lists. Specifically, the number of unique strings stored could reach 2^N in the worst case. Therefore, the auxiliary space used to store the intermediate string combinations and final results is proportional to O(2^N), where N represents the length or nesting depth of the expression.

Optimal Solution

Approach

The challenge is to expand a string containing nested brace groups into a sorted set of all possible strings. The most efficient approach uses a stack-based strategy to manage the nested structures and combine string sets effectively, avoiding redundant calculations.

Here's how the algorithm would work step-by-step:

Imagine each part of the string represents a set of possibilities. When you see plain text, it's a single possibility: itself.
When you encounter a brace group (like '{a,b}'), treat it as a set of possibilities: 'a' and 'b'.
Use a stack to keep track of the string sets you're building as you go through the original string from left to right.
If you see plain text or a brace group, combine it with the string set at the top of the stack.
If you see a comma inside a brace group, it means you have an alternative. Start a new string set on the stack for this alternative.
When you reach the closing brace, merge all the string sets within the brace group into a single set.
After processing the entire string, you'll have a single set of all possible strings.
Finally, sort the strings in that set alphabetically to produce the final result.

Code Implementation

def brace_expansion_ii(expression):
    stack_of_sets = [{""}]
    
    for char in expression:
        if char == '{':
            stack_of_sets.append({""})

        elif char == ',':
            # Start a new set for the alternative.
            stack_of_sets.append({""})

        elif char == '}':
            # Merge all sets within the group.
            combined_set = set()
            while len(stack_of_sets) > 0 and stack_of_sets[-1] != 'marker':
                combined_set.update(stack_of_sets.pop())
            stack_of_sets.append(combined_set)

        else:
            # Combine with the current set at the top.
            current_set = stack_of_sets[-1]
            new_set = set()
            for string in current_set:
                new_string = string + char
                new_set.add(new_string)
            stack_of_sets[-1] = new_set
    
    # Merge the stack.
    final_set = set()
    while stack_of_sets:
        final_set.update(stack_of_sets.pop())
        
    return sorted(list(final_set))

Big(O) Analysis

Time Complexity

O(N*M*Llog(L)) – Let N be the length of the input string `expression`. Let M be the maximum number of string sets within any brace group, representing the number of alternative branches at the deepest level of nesting. Let L be the maximum length of any individual string generated during the expansion. The algorithm iterates through the input string (O(N)). Within each iteration, set operations (union, concatenation) occur. These set operations, specifically the union of string sets, can have a time complexity proportional to the size of the sets being merged, and since each string in the set can have a maximum length L, operations can be bounded by O(M*L) where M accounts for string set size. At the end, the set of expanded strings needs to be sorted, contributing a factor of O(Llog(L)) to the total complexity. Therefore, the overall time complexity is approximately O(N*M*Llog(L)).

Space Complexity

O(N) – The algorithm utilizes a stack to store sets of strings, where N represents the length of the input string. In the worst-case scenario (e.g., a string with deeply nested brace groups), the stack's height could grow proportionally to N, holding intermediate sets of strings. Additionally, each set within the stack can contain strings whose combined length might also be proportional to N, leading to O(N) space for each set. Therefore, the overall auxiliary space complexity is O(N) due to the stack and the sets it contains.

Edge Cases

Case	How to Handle
Empty input string	Return an empty set, as there are no expressions to expand.
Input string containing only an empty curly brace pair '{}'	Return a set containing only the empty string ''.
Deeply nested expressions (e.g., {{{a}}} ) exceeding recursion limits	The solution should be iterative using a stack to avoid stack overflow errors.
Extremely long concatenated strings leading to memory exhaustion	Implement optimizations to avoid creating unnecessarily large intermediate strings; consider using generators or iterators for efficient memory usage.
Input with extremely large number of union options (e.g., {a,b,c,...,z} repeated multiple times)	Ensure the solution's memory usage doesn't explode; using set operations and efficient concatenation methods are key.
Input containing invalid characters other than lowercase letters, commas, and curly braces.	Raise an exception, return an error set, or filter invalid characters based on problem requirements; specify what behavior is preferred.
Unbalanced curly braces (e.g., '{a,b' or 'a,b}')	Raise an error indicating malformed input or handle the cases gracefully assuming an empty string in such scenarios.
Duplicate strings produced during expansion	Use a set data structure to store the results, automatically eliminating duplicates.

Empty input string

How to Handle:

Return an empty set, as there are no expressions to expand.

Input string containing only an empty curly brace pair '{}'

How to Handle:

Return a set containing only the empty string ''.

Deeply nested expressions (e.g., {{{a}}} ) exceeding recursion limits

How to Handle:

The solution should be iterative using a stack to avoid stack overflow errors.

Extremely long concatenated strings leading to memory exhaustion

How to Handle:

Implement optimizations to avoid creating unnecessarily large intermediate strings; consider using generators or iterators for efficient memory usage.

Input with extremely large number of union options (e.g., {a,b,c,...,z} repeated multiple times)

How to Handle:

Ensure the solution's memory usage doesn't explode; using set operations and efficient concatenation methods are key.

Input containing invalid characters other than lowercase letters, commas, and curly braces.

How to Handle:

Raise an exception, return an error set, or filter invalid characters based on problem requirements; specify what behavior is preferred.

Unbalanced curly braces (e.g., '{a,b' or 'a,b}')

How to Handle:

Raise an error indicating malformed input or handle the cases gracefully assuming an empty string in such scenarios.

Duplicate strings produced during expansion

How to Handle:

Use a set data structure to store the results, automatically eliminating duplicates.