How do you solve Smallest Palindromic Rearrangement II on LeetCode?

Smallest Palindromic Rearrangement II (LeetCode #3518) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n!) time complexity and O(n) space complexity. Key insight: Only half of the characters are needed to form a palindrome.

What is the time complexity of Smallest Palindromic Rearrangement II?

The optimal solution for Smallest Palindromic Rearrangement II runs in O(n!) time and uses O(n) space. Generating permutations of half the string is factorial, but we avoid duplicates using a set.

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

The brute force approach for Smallest Palindromic Rearrangement II is Brute Force, with O(n!) time complexity and O(n) space complexity. Generate all permutations of the string and filter for palindromic ones. This method is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n!).

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

Smallest Palindromic Rearrangement II uses the following concepts: Hash Table, Math, String, Combinatorics, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest Palindromic Rearrangement II?

Explain your thought process clearly. Be prepared to discuss time and space complexities. Practice generating permutations and combinations.

What are common mistakes when solving Smallest Palindromic Rearrangement II?

Not accounting for duplicate characters in permutations. Forgetting to check the k-th index against the number of unique palindromes.

#3518

Smallest Palindromic Rearrangement II

Hard

Hash TableMathStringCombinatoricsCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n!)
Space	O(n)	O(n)

💡

Intuition

Time O(n!)Space O(n)

Count character frequencies, build half the palindrome, and generate permutations efficiently. This reduces complexity by leveraging symmetry in palindromes.

⚙️

Algorithm

3 steps

1Step 1: Count character frequencies and determine half counts for the palindrome.
2Step 2: Generate the half string and create permutations of it.
3Step 3: Construct full palindromes and return the k-th lexicographically smallest.

solution.py9 lines

1from collections import Counter
2from itertools import permutations
3
4def kth_palindrome(s, k):
5    count = Counter(s)
6    half = ''.join(char * (count[char] // 2) for char in sorted(count))
7    unique_half = sorted(set(permutations(half)))
8    palindromes = [''.join(h) + ''.join(h)[::-1] for h in unique_half]
9    return palindromes[k-1] if k <= len(palindromes) else ''

ℹ

Complexity note: Generating permutations of half the string is factorial, but we avoid duplicates using a set.

1Only half of the characters are needed to form a palindrome.
2Distinct arrangements can be generated using character counts.

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