# 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)

## 4 Thoughts to “LeetCode 139. Word Break”

1. Joy Chen says:

“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;
}

2. Piyush says:

what will be the time complexity for the algorithm?