139. Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

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

dp, time: O(n ^ 2 * n) -> O(n ^ 3), s.substring() is O(n), space: O(n)

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        if(s == null || s.length() == 0) {
            return true;
        }
        Set<String> set = new HashSet<>(wordDict);
    // canSplit[i] represents if s.substring(0, i) can be formed by wordDict
        boolean[] canSplit = new boolean[s.length() + 1];
        canSplit[0] = true;
        
        for(int i = 1; i <= s.length(); i++) {
            if(set.contains(s.substring(0, i))) {
                canSplit[i] = true;
                continue;
            }
            for(int j = 0; j < i; j++) {
                if(canSplit[j] && set.contains(s.substring(j, i))) {
                    canSplit[i] = true;
                    break;
                }
            }
        }
        
        return canSplit[s.length()];
    }
}