LeetCode 139. Word Break (Medium)

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

 方法1： DFS - TLE

方法2： dp

1 dp[i] represent if s[:i] could be constructed using the wordDict 

2 for j that is smaller then I, if we have dp[j] that is True and s[j:I] exists in wordDict, then we update dp[i] = True 

3 initialize dp[i] = False dp[0] = True 

4 final answer dp[n]

time complexity: O(N^2) space complexity: O(N) 

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        if not s or len(s) == 0: return True 
        
        wordSet = set(wordDict)
        
        n = len(s)
        dp = [False for _ in range(n+1)]
        
        dp[0] = True 
        
        for i in range(1, n+1):
            for j in range(i):
                if dp[j] and s[j:i] in wordSet:
                    dp[i] = True 
                    
        return dp[n]