- Palindromic Substrings DaleStudy#267

Author: ayosecu

palindromic-substrings/ayosecu.py

@@ -0,0 +1,31 @@
+class Solution:
+    """
+        - Time Complexity: O(n^2), n = len(s)
+        - Space Complexity: O(1)
+    """
+    def countSubstrings(self, s: str) -> int:
+        count = 0
+
+        def checkSide(l, r):
+            cnt = 0
+            while l >= 0 and r < len(s) and s[l] == s[r]:
+                cnt += 1
+                l -= 1
+                r += 1
+            return cnt
+        
+        for i in range(len(s)):
+            count += checkSide(i, i)        # Odd case
+            count += checkSide(i, i + 1)    # Even case
+        
+        return count
+
+tc = [
+        ("abc", 3),
+        ("aaa", 6)
+]
+
+sol = Solution()
+for i, (s, e) in enumerate(tc, 1):
+    r = sol.countSubstrings(s)
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")