LeetCode 139. 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.

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()];
    }
}

Python Solution

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        
        f = [False for i in range(len(s) + 1)]
        f[0] = True
        
        for i in range(len(s)):
            for j in range(i, len(s)):
                if f[i] and s[i:j + 1] in wordDict:
                    f[j + 1] = True

        return f[len(s)]

• Time Complexity: O(N^2)
• Space Complexity: O(N)