## # 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: 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
``````

### # Solution

Approach 1: DP. Time: `O(N^3)` or `O(N*L*L)` where L is the avg length of the words -- if using a trie.

### # Code (Python)

Approach 1:

``````class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
if not s:
return True
if not wordDict:
return False
dictionary = set(wordDict)
table = [False for _ in range(len(s) + 1)] # table[i] -- s[0:i] can be broken down
table = True
for i in range(1, len(s) + 1):
for j in range(i - 1, -1, -1): # optimization -- work from the back because word lengths are usually short
if table[j] and s[j:i] in dictionary: # essentially O(n^3) time because s[j:i] is O(n)
table[i] = True
break
return table[-1]

# alternatively can use a trie to put wordDict
# entire program is now O(N*L*L) where L is the avg length of the words
#for i in range(1, len(s) + 1): # O(N)
#    for j in range(i - 1, -1, -1): # O(L)
#        if table[j] and trie.search(s, j, i-1) # O(L)
#            table[i] = True
#            break
``````

### # Code (C++)

Approach 1:

``````class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
int sSize = s.size();
bool match[sSize + 1] = {false};
match[sSize] = true;
for (int i = sSize - 1; i >= 0; --i)
{
for (int j = i; j < sSize; ++j)
{
string subS = s.substr(i, j - i + 1);
if (std::find(wordDict.begin(), wordDict.end(), subS) != wordDict.end())
{
match[i] = match[j+1];
if (match[i])
break;
}
}
}
return match;
}
};
``````