LeetCode 3517: Smallest Palindromic Rearrangement I ·

Table of Contents

LeetCode Weekly Contest 445 - This article is part of a series.

Part 1: LeetCode 3516: Find Closest Person

Part 2: This Article

Meta Data
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Difficulty: medium First Attempt: 2025-04-13

Total time: 05:00.00

Intuition
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We need to determine whether the string has an odd or even length in order to check if there’s a middle character that needs special handling.
Then, we sort the first half of the string based on ord(), and construct the final result by concatenating the left half, the middle character (if any), and the reversed left half.

Approach
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class Solution:
    def smallestPalindrome(self, s: str) -> str:
        odd = len(s)%2
        helf_index = len(s)//2
        left = list(s[:helf_index])
        
        middle_str = s[helf_index] if odd else ""
        
        left_sorted = "".join(sorted(left, key=ord))
        
        return left_sorted+middle_str+left_sorted[::-1]