Add No_125: Valid Palindrome

Author: brianchiang-tw

No_125_Valid Palindrome/No_125_Valid Palindrome_description.md

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+'''
+
+Description:
+
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+Note: For the purpose of this problem, we define empty string as valid palindrome.
+
+Example 1:
+
+Input: "A man, a plan, a canal: Panama"
+Output: true
+Example 2:
+
+Input: "race a car"
+Output: false
+
+'''

No_125_Valid Palindrome/valid_palindrome_by_regex.py

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+'''
+
+Description:
+
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+Note: For the purpose of this problem, we define empty string as valid palindrome.
+
+Example 1:
+
+Input: "A man, a plan, a canal: Panama"
+Output: true
+Example 2:
+
+Input: "race a car"
+Output: false
+
+'''
+
+
+import re
+
+class Solution:
+    
+    def refine(self, s:str) ->str:
+    
+        regex = r'[^A-Za-z0-9]'
+        pattern = re.compile(regex)
+        
+        return pattern.sub('', s)
+    
+    def isPalindrome(self, s: str) -> bool:
+        
+        alphanumeric_s = self.refine(s).lower()
+        
+        return alphanumeric_s == alphanumeric_s[::-1]
+
+
+
+# n : the length of inputs string s
+
+## Time Complexity: O( n )
+#
+# The major overhead in time is the refine function and [::-1] string copy, which is of O( n ).
+
+
+## Space Complexity: O( n )
+#
+# The compare is out-of-place, need extra space of O( n )
+# The major overhead in space is storage for out-of-place comparison, which is of O( n )
+
+
+
+def test_bench():
+
+    test_data = [
+                    "A man, a plan, a canal: Panama",
+                    "race a car"
+                ]
+
+    # expected output:
+    '''
+    True
+    False
+    '''
+
+    for test_s in test_data:
+
+        print( Solution().isPalindrome(test_s) )
+    
+    return 
+
+
+
+if __name__ == '__main__':
+
+    test_bench()

No_125_Valid Palindrome/valid_palindrome_by_two_pointers.py

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+'''
+
+Description:
+
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+Note: For the purpose of this problem, we define empty string as valid palindrome.
+
+Example 1:
+
+Input: "A man, a plan, a canal: Panama"
+Output: true
+Example 2:
+
+Input: "race a car"
+Output: false
+
+'''
+
+class Solution:
+    
+
+    def isPalindrome(self, s: str) -> bool:
+        
+        left, right = 0, len(s)-1
+        
+        while left < right:
+            
+            # skip non-alphanumeric characters on left hand side
+            while not s[left].isalnum() and left < right:
+                left += 1
+            
+            # skip non-alphanumeric characters on right hand side
+            while not s[right].isalnum() and left <right:
+                right -=1
+            
+            # compare
+            if s[left].lower() != s[right].lower():
+                return False
+            
+            left += 1
+            right -= 1
+            
+        
+        return True
+
+
+
+# n : the length of inputs string s
+
+## Time Complexity: O( n )
+#
+# The major overhead in time is the while loop iterating on (left < right), which is of O( n ).
+
+
+## Space Complexity: O( 1 )
+#
+# The compare is in-place, no need of extra space to copy string
+# The major overhead in space is two-pointers (i.e., left and right), which is of O( 1 ).
+
+
+
+def test_bench():
+
+    test_data = [
+                    "A man, a plan, a canal: Panama",
+                    "race a car"
+                ]
+
+    # expected output:
+    '''
+    True
+    False
+    '''
+
+    for test_s in test_data:
+
+        print( Solution().isPalindrome(test_s) )
+    
+    return 
+
+
+
+if __name__ == '__main__':
+
+    test_bench()