feat: add two-pointer approach for isPalindrome function with detailed comments on complexity

Author: hongseoupyun

valid-palindrome/hongseoupyun.py

@@ -1,13 +1,44 @@
+
+import re
+
+class Solution:
 # Palindrome is a word, phrase, number, or other sequences of characters that reads the same forward and backward (ignoring non-alphanumeric characters - letters and number).
 # the solution should first filter out non-alphanumeric characters and convert the string to lowercase.
 # Then, it should check if the cleaned string is equal to its reverse.
 
 # Time complexity = O(n), as it checks, replaces, converts to lowercase, and reverses all characters in the string
 # Space complexity = O(n), as it creates a new string with a maximum length equal to the original string
-import re
-
-class Solution:
     def isPalindrome(self, s: str) -> bool:
         filtered_string = re.sub(r'[^a-zA-Z0-9]','',s).lower()
         reversed_string = filtered_string[::-1]
         return reversed_string == filtered_string
+
+
+# Using Two-pointer; It searches from given string directly so the Space complexity is O(n) - better method
+# By using two index, searchs and compares from start and end of the string like this -><- 
+# Do this while start index is less than end index
+# Skip non-alphanumeric characters
+# If both pointer are in characters, then convert to lowercase and compare
+# If not equal, return false
+# If all characters are equal, return true
+
+# Time complexity = O(n), as it checks and converts to lowercase all characters in the string
+# Space complexity = O(1), as it uses only a constant amount of extra space
+    def isPalindromeTwoPointer(self, s: str) -> bool:
+        left = 0
+        right = len(s) - 1
+
+        while left < right:
+            if not s[left].isalnum():
+                left+=1
+                continue
+            if not s[right].isalnum():
+                right-=1
+                continue
+            if s[left].lower() != s[right].lower():
+                return False
+            
+            left+=1
+            right-=1
+
+        return True

valid-palindrome/hongseoupyun.ts

@@ -15,3 +15,42 @@ function isPalindrome(s: string): boolean {
         return false
     }
 };
+
+//Palindrom example: "A man, a plan, a canal: Panama"
+
+// Using Two-pointer; It searches from given string directly so the Space complexity is O(n) - better method
+// By using two index, searchs and compares from start and end of the string like this-><- 
+// Do this while start index is less than end index
+// Skip non-alphanumeric characters
+// If both pointer are in characters, then convert to lowercase and compare
+// If not equal, return false
+// If all characters are equal, return true
+
+// Time complexity = O(n), as it checks and converts to lowercase all characters in the string
+// Space complexity = O(1), as it uses only a constant amount of extra space
+function isPalindrome2(s: string): boolean {
+
+    let left: number = 0;
+    let right: number = s.length - 1;
+
+    let isAlphaNum = (str: string): boolean => {
+        return /^[a-zA-Z0-9]$/.test(str)
+    }
+
+    while (left < right) {
+        if (!isAlphaNum(s[left])) {
+            left++
+            continue
+        }
+        if (!isAlphaNum(s[right])) {
+            right--
+            continue
+        }
+        if (s[left].toLowerCase() !== s[right].toLowerCase()) {
+            return false
+        }
+        left++
+        right--
+    }
+     return true
+};