564

Author: OfficialCodeVoyage

564 Find the Closest Palindrome/solution.py

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+# Given a string n representing an integer, return the closest integer (not including itself), which is a palindrome. If there is a tie, return the smaller one.
+#
+# The closest is defined as the absolute difference minimized between two integers.
+#
+#
+#
+# Example 1:
+#Input: n = "123"
+#Output: "121"
+
+# need to find 2 polindromes first
+# return max(save(pol1, pol2))
+
+n = '120' # output 121
+
+def forward_cheking(x) -> int:
+
+    while True :
+        number_to_check = int(x) + 1
+        number = [i for i in str(number_to_check)]
+        po1, po2 = 0, len(number) - 1
+
+        while po1 < po2:
+
+            if number[po1] == number[po2]:
+                po1 +=1
+                po2 -=1
+            else:
+                number_to_check +=1
+                number = [i for i in str(number_to_check)]
+                po1, po2 = 0, len(number) - 1
+
+
+        return number_to_check
+
+
+def backward_cheking(x) -> int:
+
+    while True :
+        number_to_check = int(x) - 1
+        number = [i for i in str(number_to_check)]
+        po1, po2 = 0, len(number) - 1
+
+        while po1 < po2:
+
+            if number[po1] == number[po2]:
+                po1 +=1
+                po2 -=1
+            else:
+                number_to_check -=1
+                number = [i for i in str(number_to_check)]
+                po1, po2 = 0, len(number) - 1
+
+
+        return number_to_check
+
+
+def nearestPalindromic(n: str) -> str:
+    if int(n) < 0:
+        return '0'
+    elif 0 < int(n) < 10:
+        return str(int(n) - 1)
+    elif int(n) == 10 or int(n) == 11:
+        return '9'
+
+    forward_palindrome = forward_cheking(n)
+    backward_palindrome = backward_cheking(n)
+
+    if abs(forward_palindrome - int(n)) == abs(int(n) - backward_palindrome):
+        return str(min(forward_palindrome, backward_palindrome))
+    elif abs(forward_palindrome - int(n)) < abs(int(n) - backward_palindrome):
+        return str(forward_palindrome)
+    else:
+        return str(backward_palindrome)
+
+nearestPalindromic(n)