Add valid-palindrome solution in TypeScript

Author: Jeehay28

valid-palindrome/Jeehay28.ts

@@ -0,0 +1,55 @@
+// Approach 2
+// 🗓️ 2025-04-14
+// ⏳ Time Complexity: O(n)
+// 💾 Space Complexity: O(1)
+
+function isPalindrome(s: string): boolean {
+  let low = 0, high = s.length - 1;
+  const reg = /[0-9a-zA-Z]/;
+
+  while (low < high) {
+    while (low < high && !reg.test(s[low])) {
+      low += 1;
+    }
+
+    while (low < high && !reg.test(s[high])) {
+      high -= 1;
+    }
+
+    if (s[low].toLowerCase() !== s[high].toLowerCase()) {
+      return false;
+    }
+    low += 1;
+    high -= 1;
+  }
+
+  return true;
+}
+
+
+// Approach 1
+// 🗓️ 2025-04-14
+// ⏳ Time Complexity: O(n)
+// 💾 Space Complexity: O(n)
+
+// function isPalindrome(s: string): boolean {
+//   // cover all uppercase letters into lowercasse letters
+//   // remove all non-alphanumeric characters (letters and numbers)
+//   // it reads the same forward and backward
+
+//   const cleaned = s.toLowerCase().match(/[0-9a-z]/g);
+
+//   // Converting to lowercase (toLowerCase()): O(n), where n is the length of the string
+//   // Matching with regex (match(/[0-9a-z]/g)): O(n), where n is the length of the string, as the regex engine needs to check each character in the string
+//   // join("") operation: O(m), where m is the length of the cleaned array.
+//   // reverse() operation: O(m), where m is the length of the cleaned array.
+//   // Comparison (s_forward === s_backward): O(n) because it checks each character one by one
+
+//   if (cleaned !== null) {
+//     const s_forward = cleaned.join("");
+//     const s_backward = cleaned.reverse().join("");
+//     return s_forward === s_backward;
+//   } else {
+//     return true;
+//   }
+// }