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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0139_word_break;

// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #String #Hash_Table
// #Dynamic_Programming #Trie #Memoization #Algorithm_II_Day_15_Dynamic_Programming
// #Dynamic_Programming_I_Day_9 #Udemy_Dynamic_Programming #Big_O_Time_O(M+max*N)_Space_O(M+N+max)
// #2022_06_24_Time_2_ms_(97.08%)_Space_42.1_MB_(90.92%)

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

/**
 * 139 - Word Break\.
 *
 * Medium
 *
 * Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
 *
 * **Note** that the same word in the dictionary may be reused multiple times in the segmentation.
 *
 * **Example 1:**
 *
 * **Input:** s = "leetcode", wordDict = ["leet","code"]
 *
 * **Output:** true
 *
 * **Explanation:** Return true because "leetcode" can be segmented as "leet code". 
 *
 * **Example 2:**
 *
 * **Input:** s = "applepenapple", wordDict = ["apple","pen"]
 *
 * **Output:** true
 *
 * **Explanation:** Return true because "applepenapple" can be segmented as "apple pen apple". Note that you are allowed to reuse a dictionary word. 
 *
 * **Example 3:**
 *
 * **Input:** s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
 *
 * **Output:** false 
 *
 * **Constraints:**
 *
 * *   `1 <= s.length <= 300`
 * *   `1 <= wordDict.length <= 1000`
 * *   `1 <= wordDict[i].length <= 20`
 * *   `s` and `wordDict[i]` consist of only lowercase English letters.
 * *   All the strings of `wordDict` are **unique**.
**/
public class Solution {
    public boolean wordBreak(String s, List wordDict) {
        Set set = new HashSet<>();
        int max = 0;
        boolean[] flag = new boolean[s.length() + 1];
        for (String st : wordDict) {
            set.add(st);
            if (max < st.length()) {
                max = st.length();
            }
        }
        for (int i = 1; i <= max; i++) {
            if (dfs(s, 0, i, max, set, flag)) {
                return true;
            }
        }
        return false;
    }

    private boolean dfs(String s, int start, int end, int max, Set set, boolean[] flag) {
        if (!flag[end] && set.contains(s.substring(start, end))) {
            flag[end] = true;
            if (end == s.length()) {
                return true;
            }
            for (int i = 1; i <= max; i++) {
                if (end + i <= s.length() && dfs(s, end, end + i, max, set, flag)) {
                    return true;
                }
            }
        }
        return false;
    }
}