How do you solve Palindrome Rearrangement Queries on LeetCode?

Palindrome Rearrangement Queries (LeetCode #2983) 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: A string can be rearranged into a palindrome if at most one character has an odd frequency.

What is the brute force solution for Palindrome Rearrangement Queries?

The brute force approach for Palindrome Rearrangement Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible rearrangements of the specified substrings to see if they can form a palindrome. This is straightforward but inefficient, as it requires generating permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Palindrome Rearrangement Queries?

Palindrome Rearrangement Queries uses the following concepts: Hash Table, String, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Palindrome Rearrangement Queries?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can reduce the problem size or complexity, such as using frequency counts instead of permutations. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Palindrome Rearrangement Queries?

Failing to account for character frequencies correctly. Not recognizing that rearranging substrings does not change the overall character counts.

#2983

Palindrome Rearrangement Queries

Hard

Hash TableStringPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of generating rearrangements, we can count character frequencies in the specified ranges. For a string to be rearranged into a palindrome, each character must appear an even number of times, except for at most one character.

⚙️

Algorithm

3 steps

1Step 1: For each query, count the frequency of characters in both substrings.
2Step 2: Combine the frequency counts from both substrings.
3Step 3: Check if at most one character has an odd frequency in the combined counts.

solution.py12 lines

1from collections import Counter
2
3def can_form_palindrome(s, queries):
4    n = len(s)
5    results = []
6    for a, b, c, d in queries:
7        left_count = Counter(s[a:b + 1])
8        right_count = Counter(s[c:d + 1])
9        combined_count = left_count + right_count
10        odd_count = sum(1 for count in combined_count.values() if count % 2 != 0)
11        results.append(odd_count <= 1)
12    return results

ℹ

Complexity note: This complexity is due to counting character frequencies, which requires a linear scan of the substrings, resulting in O(n) time for each query.

1A string can be rearranged into a palindrome if at most one character has an odd frequency.
2Character frequency counting is a powerful technique for solving string rearrangement problems.

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