Description
https://leetcode.com/problems/word-break/description/
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
Explanation
We need to determine if s can be segmented into a space-separated sequence of one or more dictionary words.
We can introduce a state variable isWordBreak[i] to indicate whether the first iith characters of the input string is able to break into words that all in the dictionary.
Set isWordBreak[0] to true, indicating that empty string ”” is also in the wordDict.
- isWordBreak[j] is used to indicate whether the first j characters of the input string is able to break into words that all in the dictionary
- Only when isWordBreak[j] = true（first j characters of the input string is word breakable) and s.substring(j, i) is also a word in the wordDict, first i characters of s can be word breakable.
Video Tutorial
Java Solution
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
boolean[] isWordBreak = new boolean[s.length() + 1];
isWordBreak[0] = true;
for (int i = 0; i < s.length() + 1; i++) {
for (int j = 0; j < i; j++) {
if (!isWordBreak[j]) {
continue;
}
if (wordDict.contains(s.substring(j, i))) {
isWordBreak[i] = true;
break;
}
}
}
return isWordBreak[s.length()];
}
}
“isWordBreak[4]==TRUE && s.substring(4,10) in WordDict”
In this statement, don’t we need to distinguish whether s.substring(5,10) is in WordDict or not?
So in the code,
if (wordDict.contains(s.substring(j+1, i))) {
isWordBreak[i] = true;
break;
}
what will be the time complexity for the algorithm?