Time: 41 ms (21.59%), Space: 16.6 MB (28.21%) - LeetHub

Author: Deepak5113

0564-find-the-closest-palindrome/0564-find-the-closest-palindrome.py

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+class Solution:
+    def nearestPalindromic(self, n: str) -> str:
+        length = len(n)
+        num = int(n)
+        
+        # Edge cases for very small numbers
+        if num == 0:
+            return "1"
+        if num == 1:
+            return "0"
+        
+        # Handle corner cases like "1000", "999"
+        candidates = set()
+        candidates.add(str(10**(length - 1) - 1))  # largest number with one less digit (e.g., 999 for 1000)
+        candidates.add(str(10**length + 1))        # smallest number with one more digit (e.g., 10001 for 9999)
+
+        # Generate the palindrome by mirroring
+        prefix = int(n[:(length + 1) // 2])
+        for i in [-1, 0, 1]:
+            new_prefix = str(prefix + i)
+            if length % 2 == 0:
+                candidate = new_prefix + new_prefix[::-1]
+            else:
+                candidate = new_prefix + new_prefix[:-1][::-1]
+            candidates.add(candidate)
+        
+        # Remove the original number from the candidates
+        candidates.discard(n)
+        
+        # Find the closest palindrome
+        closest = min(candidates, key=lambda x: (abs(int(x) - num), int(x)))
+        
+        return closest