[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. You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

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

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Thought process:

1. Sub-problem: determine if a substring of s can be segmented into a sequence of one or more dictionary words.
2. Function: f[i] = if there exists (f[j] && s[j, i] is a dictionary word).
3. Initialization: f[0] = true. Add a dummy character at the start of the string.
4. Answer: f[s.length].

Solution:

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

Time complexity: 

Say m = size of wordDict and n = length of s, the overall time complexity is O(mn^2).