0139. Word Break

https://leetcode.com/problems/word-break

Description

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.

ac1: DP

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> dict = new HashSet<String>(wordDict);
        int n = s.length();
        boolean[] dp = new boolean[n+1];
        dp[0] = true;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if(dp[j] && dict.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }                
            }
        }
        return dp[n];
    }
}

ac2: DFS + Memorization

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        // edge cases
        if (s == null || wordDict == null || s.length() == 0 || wordDict.size() == 0) 
            return false;

        // initialize
        Set<String> dict = new HashSet<String>(wordDict);
        Boolean[] cache = new Boolean[s.length()];
        return dfs(s, 0, dict, cache);
    }

    private boolean dfs(String s, int start, Set<String> dict, Boolean[] cache) {
        // exit
        if (start == s.length()) return true;

        // check cache, avoid repeat
        // careful, must use Boolean, instead of boolean, because need to check if cache[start] has been filled before.
        if (cache[start] != null) return cache[start];  

        // process
        for (int e = start + 1; e <= s.length(); e++) {
            String w = s.substring(start, e);
            if (dict.contains(w) && dfs(s, e, dict, cache)) {
                cache[start] = true;
                return true;
            }
        }

        // string invalid
        cache[start] = false;

        return false;
    }
}