1087. Brace Expansion

1087. Brace Expansion (Medium)

You are given a string s representing a list of words. Each letter in the word has one or more options.

• If there is one option, the letter is represented as is.
• If there is more than one option, then curly braces delimit the options. For example, "{a,b,c}" represents options ["a", "b", "c"].

For example, if s = "a{b,c}", the first character is always 'a', but the second character can be 'b' or 'c'. The original list is ["ab", "ac"].

Return all words that can be formed in this manner, sorted in lexicographical order.

 

Example 1:

Input: s = "{a,b}c{d,e}f"
Output: ["acdf","acef","bcdf","bcef"]

Example 2:

Input: s = "abcd"
Output: ["abcd"]

 

Constraints:

• 1 <= s.length <= 50
• s consists of curly brackets '{}', commas ',', and lowercase English letters.
• s is guaranteed to be a valid input.
• There are no nested curly brackets.
• All characters inside a pair of consecutive opening and ending curly brackets are different.

Companies:
Google, Amazon

Related Topics:
String, Backtracking, Breadth-First Search

Similar Questions:

• Decode String (Medium)
• Letter Case Permutation (Medium)
• Brace Expansion II (Hard)

Solution 1. Backtracking

// OJ: https://leetcode.com/problems/brace-expansion/
// Author: github.com/lzl124631x
// Time: O(NK) where K is the total number of possible branches
// Space: O(N)
class Solution {
public:
    vector<string> expand(string s) {
        vector<vector<char>> v;
        for (int i = 0, N = s.size(); i < N; ++i) {
            if (s[i] == '{') {
                ++i;
                v.emplace_back();
                while (s[i] != '}') {
                    v.back().push_back(s[i++]);
                    if (s[i] == ',') ++i;
                }
                sort(begin(v.back()), end(v.back()));
            } else v.push_back({ s[i] });
        }
        vector<string> ans;
        string tmp;
        function<void(int)> dfs = [&](int i) {
            if (i == v.size()) {
                ans.push_back(tmp);
                return;
            }
            for (int j = 0; j < v[i].size(); ++j) {
                tmp += v[i][j];
                dfs(i + 1);
                tmp.pop_back();
            }
        };
        dfs(0);
        return ans;
    }
};