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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0140_word_break_ii;

// #Hard #Top_Interview_Questions #String #Hash_Table #Dynamic_Programming #Backtracking #Trie
// #Memoization #2022_06_24_Time_1_ms_(99.05%)_Space_42.2_MB_(65.12%)

import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Set;

/**
 * 140 - Word Break II\.
 *
 * Hard
 *
 * Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
 *
 * **Note** that the same word in the dictionary may be reused multiple times in the segmentation.
 *
 * **Example 1:**
 *
 * **Input:** s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
 *
 * **Output:** ["cats and dog","cat sand dog"] 
 *
 * **Example 2:**
 *
 * **Input:** s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
 *
 * **Output:** ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
 *
 * **Explanation:** Note that you are allowed to reuse a dictionary word. 
 *
 * **Example 3:**
 *
 * **Input:** s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
 *
 * **Output:** [] 
 *
 * **Constraints:**
 *
 * *   `1 <= s.length <= 20`
 * *   `1 <= wordDict.length <= 1000`
 * *   `1 <= wordDict[i].length <= 10`
 * *   `s` and `wordDict[i]` consist of only lowercase English letters.
 * *   All the strings of `wordDict` are **unique**.
**/
public class Solution {
    public List wordBreak(String s, List wordDict) {
        List result = new LinkedList<>();
        Set wordSet = new HashSet<>(wordDict);
        dfs(s, wordSet, 0, new StringBuilder(), result);
        return result;
    }

    private void dfs(
            String s, Set wordSet, int index, StringBuilder sb, List result) {
        if (index == s.length()) {
            if (sb.charAt(sb.length() - 1) == ' ') {
                sb.setLength(sb.length() - 1);
            }
            result.add(sb.toString());
            return;
        }
        int len = sb.length();
        for (int i = index + 1; i <= s.length(); ++i) {
            String subs = s.substring(index, i);
            if (wordSet.contains(subs)) {
                sb.append(subs).append(" ");
                dfs(s, wordSet, i, sb, result);
            }
            sb.setLength(len);
        }
    }
}