Create 09-00125-valid-palindrome.py

Author: leetlab11

02-Two Pointers/09-00125-valid-palindrome.py

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+# A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, 
+# it reads the same forward and backward. Alphanumeric characters include letters and numbers.
+
+# Given a string s, return true if it is a palindrome, or false otherwise.
+
+# Example 1:
+# Input: s = "A man, a plan, a canal: Panama"
+# Output: true
+# Explanation: "amanaplanacanalpanama" is a palindrome.
+  
+# Example 2:
+# Input: s = "race a car"
+# Output: false
+# Explanation: "raceacar" is not a palindrome.
+  
+# Example 3:
+# Input: s = " "
+# Output: true
+# Explanation: s is an empty string "" after removing non-alphanumeric characters.
+# Since an empty string reads the same forward and backward, it is a palindrome.
+ 
+# Constraints:
+# 1 <= s.length <= 2 * 105
+# s consists only of printable ASCII characters.
+
+# -------------------------------------------------------------------------------------------------------------------------------------------------------------------
+# CREATING A NEW STRING:
+
+# create a new empty string
+# iterate over the current string- if a character is alphanumeric, lowercase it and it to the new string.
+# compare new string and new string in reverse; return answer
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        
+        newStr = ''
+
+        for c in s:
+            if c.isalnum():
+                newStr += c.lower()
+        
+        return newStr == newStr[::-1]
+
+Time Complexity = O(n)
+Space Complexity = O(n)
+# -------------------------------------------------------------------------------------------------------------------------------------------------------------------
+
+# TWO POINTERS:
+
+# define a function to check if a character is alphanumeric using ord(c)
+# use two pointers approach- l and r
+# keep a check for l < r
+# keep check again for l < r to ensure the pointers stay within bounds during the search for alphanumeric characters. 
+# this helps ensure that once the pointers meet or cross, the function can return without unnecessary checks.
+# if an alphanumeric character is not met, go ahead(l + 1, r -1  )
+# if an alphanumeric character is met, lowercase and compare l and r
+# if they are not equal, return False
+# if they are equal, loop continues
+# if loop does not return False and exits, return True
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        l, r = 0, len(s) - 1
+
+        while l < r:
+            while l < r and not self.alphaNum(s[l]):
+                l += 1
+            while r > l and not self.alphaNum(s[r]):
+                r -= 1
+            if s[l].lower() != s[r].lower():
+                return False
+            l, r = l + 1, r - 1
+        return True
+    
+    def alphaNum(self, c):
+        return (ord('A') <= ord(c) <= ord('Z') or 
+                ord('a') <= ord(c) <= ord('z') or 
+                ord('0') <= ord(c) <= ord('9'))
+      
+#       can use Python in-built function isalnum()
+#       return c.isalnum()
+
+Time Complexity = O(n)
+Space Complexity = O(1)
+
+# Companies:
+# Meta- 53
+# Google- 5
+# Amazon- 4
+# Apple- 2
+# Oracle- 2
+# Microsoft- 8
+# Bloomberg- 4
+# Uber- 3
+# TCS- 2
+# Goldman Sachs- 2
+# Zenefits- 2
+# Yandex- 45
+# Adobe- 20
+# Yahoo- 6
+# Toast- 6 
+# EPAM Systems- 5
+# Spotify- 4
+# Tiktok- 3
+# American Express- 3
+# Accenture- 3
+# Deloitte- 2