Question

1087. Brace Expansion

中文文档

Description

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.

Solutions

Solution 1

class Solution:
  def expand(self, s: str) -> List[str]:
    def convert(s):
      if not s:
        return
      if s[0] == '{':
        j = s.find('}')
        items.append(s[1:j].split(','))
        convert(s[j + 1 :])
      else:
        j = s.find('{')
        if j != -1:
          items.append(s[:j].split(','))
          convert(s[j:])
        else:
          items.append(s.split(','))

    def dfs(i, t):
      if i == len(items):
        ans.append(''.join(t))
        return
      for c in items[i]:
        t.append(c)
        dfs(i + 1, t)
        t.pop()

    items = []
    convert(s)
    ans = []
    dfs(0, [])
    ans.sort()
    return ans

class Solution {
  private List<String> ans;
  private List<String[]> items;

  public String[] expand(String s) {
    ans = new ArrayList<>();
    items = new ArrayList<>();
    convert(s);
    dfs(0, new ArrayList<>());
    Collections.sort(ans);
    return ans.toArray(new String[0]);
  }

  private void convert(String s) {
    if ("".equals(s)) {
      return;
    }
    if (s.charAt(0) == '{') {
      int j = s.indexOf("}");
      items.add(s.substring(1, j).split(","));
      convert(s.substring(j + 1));
    } else {
      int j = s.indexOf("{");
      if (j != -1) {
        items.add(s.substring(0, j).split(","));
        convert(s.substring(j));
      } else {
        items.add(s.split(","));
      }
    }
  }

  private void dfs(int i, List<String> t) {
    if (i == items.size()) {
      ans.add(String.join("", t));
      return;
    }
    for (String c : items.get(i)) {
      t.add(c);
      dfs(i + 1, t);
      t.remove(t.size() - 1);
    }
  }
}