dfs | backtracking

leetcode 102124

python/1593.split-a-string-into-the-max-number-of-unique-substrings.py

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+#
+# @lc app=leetcode id=1593 lang=python3
+#
+# [1593] Split a String Into the Max Number of Unique Substrings
+#
+# https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/description/
+#
+# algorithms
+# Medium (57.11%)
+# Likes:    978
+# Dislikes: 44
+# Total Accepted:    52.5K
+# Total Submissions: 85.2K
+# Testcase Example:  '"ababccc"'
+#
+# 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.
+#
+#
+#
+#
+
+
+# @lc code=start
+class Solution:
+    def maxUniqueSplit(self, s: str) -> int:
+        self.max_split = 0
+        self.backtrack(s, 0, set())
+        return self.max_split
+
+    def backtrack(self, s, start, unique_substrings):
+        if start == len(s):
+            self.max_split = max(self.max_split, len(unique_substrings))
+            return
+
+        for end in range(start + 1, len(s) + 1):
+            substring = s[start:end]
+            if substring not in unique_substrings:
+                unique_substrings.add(substring)
+                self.backtrack(s, end, unique_substrings)
+                unique_substrings.remove(substring)
+
+
+# @lc code=end