leetcode 1096. Brace Expansion II

Under a grammar given below, strings can represent a set of lowercase words.  Let's use R(expr) to denote the set of words the expression represents.

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 2 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 3 rules for our grammar:

• For every lowercase letter x, we have R(x) = {x}
• For expressions e_1, e_2, ... , e_k with k >= 2, we have R({e_1,e_2,...}) = R(e_1) ∪ R(e_2) ∪ ...
• For expressions e_1 and e_2, we have R(e_1 + e_2) = {a + b for (a, b) in R(e_1) × R(e_2)}, 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: "{a,b}{c,{d,e}}"
Output: ["ac","ad","ae","bc","bd","be"]

Example 2:

Input: "{{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. 1 <= expression.length <= 50
2. expression[i] consists of '{', '}', ','or lowercase English letters.
3. The given expression represents a set of words based on the grammar given in the description.

解題思路：

一對花括號便是完整的表達式 ，重點在於如何處理逗號之間表達式的並集和非逗號間隔表達式的連接集。

從題意中可以瞭解到只要有逗號出現，逗號之前的表達式便與之後的表達式不存在連接的關係了，可以直接將逗號之前組成的字符串的集合加到結果中，但是如果是非逗號間隔的表達式就需要知道之前的表達式組成了什麼樣的字符串，也就是需要一個前綴，所以定義一個前綴所有字符串的集合unordered_set<string> prefixes,但遇到逗號時，將prefixes加入到res中，並清空prefixes ;

class Solution {
public:
    vector<string> braceExpansionII(string expression) 
    {
        int pos = 0 ;
        unordered_set<string> strs = brace(expression , pos) ;
        vector<string> res(strs.begin() , strs.end()) ;
        sort(res.begin() , res.end()) ;
        return res ;
    }
    
    unordered_set<string> brace(string &expression , int &pos)
    {
        unordered_set<string> res , prefixes ;
        int len = expression.size() ;
        
        while(pos < len)
        {
            if(expression[pos] == '{')
            {
                pos++ ;
                unordered_set<string> words = brace(expression , pos) ;
                if(!prefixes.empty())
                {
                    str_multi(prefixes , words) ;
                }
                else
                    prefixes = words ;
            }
            else if(expression[pos] == ',')
            {
                str_add(res , prefixes) ;
                prefixes.clear() ;
                pos++ ;
            }
            else if(expression[pos] >='a' && expression[pos] <= 'z')
            {
                string s = "" ;
                while(expression[pos] >='a' && expression[pos] <= 'z')
                {
                    s += expression[pos] ;
                    pos++ ;
                }
                
                if(!prefixes.empty())
                {
                    unordered_set<string> new_prefixes ;
                    for(auto str : prefixes)
                        new_prefixes.insert(str + s) ;
                    
                    prefixes = new_prefixes ;
                }
                else
                    prefixes.insert(s) ;
            }
            else 
            {
                pos++ ;
                break ;    
            }
        }
        
        str_add(res , prefixes) ;
        
        return res ;
    }
    
    void str_multi(unordered_set<string> &prefixes , unordered_set<string> &words)
    {
        unordered_set<string> res ;
        
        if(words.empty()) return ;
        
        for(auto prefix : prefixes)
        {
            for(auto word : words)
                res.insert(prefix + word) ;
        }
        
        prefixes = res ;
    }
    
    void str_add(unordered_set<string> &res , unordered_set<string> &prefixes)
    {
        for(auto prefix : prefixes)
            res.insert(prefix) ;
    }
};