From 43b54571cfee95c7be8e973ff5f81d6b2dd40793 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Sat, 23 Apr 2022 18:23:57 +0530
Subject: [PATCH] valid-palindrome py3

---
 0125_valid-palindrome/README.md           | 30 +++++++++++++++++++++++
 0125_valid-palindrome/python3/solution.py | 23 +++++++++++++++++
 2 files changed, 53 insertions(+)
 create mode 100644 0125_valid-palindrome/README.md
 create mode 100644 0125_valid-palindrome/python3/solution.py

diff --git a/0125_valid-palindrome/README.md b/0125_valid-palindrome/README.md
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+++ b/0125_valid-palindrome/README.md
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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.
\ No newline at end of file
diff --git a/0125_valid-palindrome/python3/solution.py b/0125_valid-palindrome/python3/solution.py
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+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        i = 0
+        j = len(s) - 1
+        
+        while i < j:
+            left = s[i].lower()
+            right = s[j].lower()
+            
+            if not left.isalnum():
+                i += 1
+                continue
+            elif not right.isalnum():
+                j -= 1
+                continue
+            
+            if left != right:
+                return False
+            
+            i += 1
+            j -= 1
+        
+        return True