How do you solve Smallest Palindromic Rearrangement I on LeetCode?

Smallest Palindromic Rearrangement I (LeetCode #3517) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Palindromes are symmetrical.

What is the time complexity of Smallest Palindromic Rearrangement I?

The optimal solution for Smallest Palindromic Rearrangement I runs in O(n log n) time and uses O(n) space. Counting characters is O(n), sorting keys is O(k log k) where k is the number of unique characters.

What is the brute force solution for Smallest Palindromic Rearrangement I?

The brute force approach for Smallest Palindromic Rearrangement I is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all permutations of the string and filter for palindromes. This is inefficient but straightforward. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Smallest Palindromic Rearrangement I?

Smallest Palindromic Rearrangement I uses the following concepts: String, Sorting, Counting Sort. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest Palindromic Rearrangement I?

Explain your thought process clearly. Consider edge cases. Optimize as you go.

What are common mistakes when solving Smallest Palindromic Rearrangement I?

Forgetting to sort characters. Not handling odd-length strings correctly.

#3517

Smallest Palindromic Rearrangement I

Medium

StringSortingCounting SortHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

Count characters, build half the palindrome, then mirror it. This is efficient and leverages the properties of palindromes.

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Algorithm

3 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: Construct half of the palindrome using the sorted characters.
3Step 3: Mirror the half to form the full palindrome.

solution.py6 lines

1from collections import Counter
2
3def smallest_palindrome(s):
4    count = Counter(s)
5    half = ''.join(sorted(c * (freq // 2) for c, freq in count.items()))
6    return half + half[::-1] if len(s) % 2 == 0 else half + min(count) + half[::-1]

ℹ

Complexity note: Counting characters is O(n), sorting keys is O(k log k) where k is the number of unique characters.

1Palindromes are symmetrical.
2Character frequency determines palindrome structure.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.