Palindrome Partitioning/ Palindrome Partitioning II

Author: gavinfish

131 Palindrome Partitioning.py

@@ -0,0 +1,38 @@
+'''
+Given a string s, partition s such that every substring of the partition is a palindrome.
+
+Return all possible palindrome partitioning of s.
+
+For example, given s = "aab",
+Return
+
+  [
+    ["aa","b"],
+    ["a","a","b"]
+  ]
+'''
+
+class Solution(object):
+    def partition(self, s):
+        """
+        :type s: str
+        :rtype: List[List[str]]
+        """
+        if not s:
+            return [[]]
+        result = []
+        for i in range(len(s)):
+            if self.isPalindrome(s[:i + 1]):
+                for r in self.partition(s[i + 1:]):
+                    result.append([s[:i + 1]] + r)
+        return result
+
+    def isPalindrome(self, s):
+        return s == s[::-1]
+
+
+if __name__ == "__main__":
+    assert Solution().partition("aab") == [
+        ["a", "a", "b"],
+        ["aa", "b"]
+    ]

132 Palindrome Partitioning II.py

@@ -0,0 +1,30 @@
+'''
+Given a string s, partition s such that every substring of the partition is a palindrome.
+
+Return the minimum cuts needed for a palindrome partitioning of s.
+
+For example, given s = "aab",
+Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.
+'''
+
+class Solution(object):
+    def minCut(self, s):
+        """
+        :type s: str
+        :rtype: List[List[str]]
+        """
+        n = len(s)
+        dp = [0 for __ in range(n)]
+        isPal = [[False for __ in range(n)] for __ in range(n)]
+        for i in range(n):
+            m = i
+            for j in range(i + 1):
+                if s[j] == s[i] and (j + 1 > i - 1 or isPal[j + 1][i - 1]):
+                    isPal[j][i] = True
+                    m = 0 if j == 0 else min(m, dp[j - 1] + 1)
+            dp[i] = m
+        return dp[-1]
+
+
+if __name__ == "__main__":
+    assert Solution().minCut("aab") == 1