Leetcode: 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: trueExplanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]Output: trueExplanation: 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

class Solution:

    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        length = len(s)
        res = [0] * (length+1)
        res[0] = 1
        for j in range(1,length+1):
            for i in range(j):
                if s[i:j] in wordDict and res[i]:
                    print(s[i:j])
                    res[j] = 1
                    
        return res[length]

Dynamic Programing:

All the after-result depend on the previous calculation.

For example: 

For res[j] to be True, res[i] and string s[i:j] in wordDict hace to be true.

Basic case is res[0] = 1 and goes through the string and the substring.

Time complexity is O(n^2) and space complexity is N