How do you solve Maximum Palindromes After Operations on LeetCode?

Maximum Palindromes After Operations (LeetCode #3035) 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: Character frequency is key to forming palindromes.

What is the brute force solution for Maximum Palindromes After Operations?

The brute force approach for Maximum Palindromes After Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible combination of swaps to see if we can form palindromes. This is straightforward but inefficient as it can lead to a lot of unnecessary computations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Palindromes After Operations?

Maximum Palindromes After Operations uses the following concepts: Array, Hash Table, String, Greedy, Sorting, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Palindromes After Operations?

Always clarify if you can swap characters freely or if there are restrictions. Think about character counts and how they relate to palindrome formation. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Maximum Palindromes After Operations?

Not considering character frequencies, leading to incorrect counts. Overcomplicating the swap logic instead of focusing on character pairs.

#3035

Maximum Palindromes After Operations

Medium

ArrayHash TableStringGreedySortingCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of checking every swap, we can count the frequency of each character across all words. This allows us to determine how many palindromes can be formed based on character pairs, making the solution efficient.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character in all words.
2Step 2: Calculate the number of pairs of characters that can form palindromes.
3Step 3: If there are leftover characters, we can use one of them to form a center of a palindrome.

solution.py12 lines

1def maxPalindromes(words):
2    from collections import Counter
3    freq = Counter()
4    for word in words:
5        freq.update(word)
6    count = 0
7    odd_found = False
8    for ch in freq:
9        count += freq[ch] // 2
10        if freq[ch] % 2 == 1:
11            odd_found = True
12    return count + (1 if odd_found else 0)

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the words and their characters once to count frequencies, making it linear with respect to the input size.

1Character frequency is key to forming palindromes.
2Swapping characters can be thought of as redistributing letters to maximize palindrome formation.

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