README_EN.md

320. Generalized Abbreviation

中文文档

Description

A word's generalized abbreviation can be constructed by taking any number of non-overlapping and non-adjacent substrings and replacing them with their respective lengths.

• For example, "abcde" can be abbreviated into:

  <ul>
  	<li><code>&quot;a3e&quot;</code> (<code>&quot;bcd&quot;</code> turned into <code>&quot;3&quot;</code>)</li>
  	<li><code>&quot;1bcd1&quot;</code> (<code>&quot;a&quot;</code> and <code>&quot;e&quot;</code> both turned into <code>&quot;1&quot;</code>)</li>
  	<li><code>&quot;5&quot;</code> (<code>&quot;abcde&quot;</code> turned into <code>&quot;5&quot;</code>)</li>
  	<li><code>&quot;abcde&quot;</code> (no substrings replaced)</li>
  </ul>
  </li>
  <li>However, these abbreviations are <strong>invalid</strong>:
  <ul>
  	<li><code>&quot;23&quot;</code> (<code>&quot;ab&quot;</code> turned into <code>&quot;2&quot;</code> and <code>&quot;cde&quot;</code> turned into <code>&quot;3&quot;</code>) is invalid as the substrings chosen are adjacent.</li>
  	<li><code>&quot;22de&quot;</code> (<code>&quot;ab&quot;</code> turned into <code>&quot;2&quot;</code> and <code>&quot;bc&quot;</code> turned into <code>&quot;2&quot;</code>) is invalid as the substring chosen overlap.</li>
  </ul>
  </li>
  

Given a string word, return a list of all the possible generalized abbreviations of word. Return the answer in any order.

 

Example 1:

Input: word = "word"
Output: ["4","3d","2r1","2rd","1o2","1o1d","1or1","1ord","w3","w2d","w1r1","w1rd","wo2","wo1d","wor1","word"]

Example 2:

Input: word = "a"
Output: ["1","a"]

 

Constraints:

• 1 <= word.length <= 15
• word consists of only lowercase English letters.

Solutions

Python3

class Solution:
    def generateAbbreviations(self, word: str) -> List[str]:
        def dfs(s, t):
            if not s:
                ans.append(''.join(t))
                return
            for i in range(1, len(s) + 1):
                t.append(str(i))
                if i < len(s):
                    t.append(s[i])
                    dfs(s[i + 1:], t)
                    t.pop()
                else:
                    dfs(s[i:], t)
                t.pop()

            t.append(s[0])
            dfs(s[1:], t)
            t.pop()

        ans = []
        dfs(word, [])
        return ans

Java

class Solution {
    private List<String> ans;

    public List<String> generateAbbreviations(String word) {
        ans = new ArrayList<>();
        List<String> t = new ArrayList<>();
        dfs(word, t);
        return ans;
    }

    private void dfs(String s, List<String> t) {
        if ("".equals(s)) {
            ans.add(String.join("", t));
            return;
        }
        for (int i = 1; i < s.length() + 1; ++i) {
            t.add(i + "");
            if (i < s.length()) {
                t.add(String.valueOf(s.charAt(i)));
                dfs(s.substring(i + 1), t);
                t.remove(t.size() - 1);
            } else {
                dfs(s.substring(i), t);
            }
            t.remove(t.size() - 1);
        }
        t.add(String.valueOf(s.charAt(0)));
        dfs(s.substring(1), t);
        t.remove(t.size() - 1);
    }
}

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