Given a string s, return the maximum number of unique substrings that the given string can be split into.
You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = "ababccc"
Output: 5
Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times.
Example 2:
Input: s = "aba"
Output: 2
Explanation: One way to split maximally is ['a', 'ba'].
Example 3:
Input: s = "aa"
Output: 1
Explanation: It is impossible to split the string any further.
Constraints:
1 <= s.length <= 16
s contains only lower case English letters.
class Solution:
    def maxUniqueSplit(self, s: str) -> int:
        res = 0
        dfs = [[set(),s]]
        while dfs:
            curSet, curS = dfs.pop()            
            res = max(res, len(curSet))
            if len(curSet)+len(curS) <= res or len(curS) == 0:
                   continue
                      
            for i in range(len(curS)):
                if curS[:i+1] in curSet: continue
                    
                curSet.add(curS[:i+1])
                dfs.append([curSet.copy(), curS[i+1:]])
                curSet.remove(curS[:i+1])
                   
        return res
        
            
            
                
            

class Solution:
    def maxUniqueSplit(self, s: str) -> int:
        res = 0
        dfs = [[set(),s]]
        while dfs:
            curSet, curS = dfs.pop()            
            res = max(res, len(curSet))
            if len(curSet)+len(curS) <= res or len(curS) == 0:
                   continue
                      
            for i in range(len(curS)):
                if curS[:i+1] in curSet: continue
                    
                curSet.add(curS[:i+1])
                dfs.append([curSet.copy(), curS[i+1:]])
                curSet.remove(curS[:i+1])
                   
        return res
        
            
            
                
            
