#3035

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Maximum Palindromes After Operations

The problem focuses on maximizing the number of palindromes that can be formed from a given list of words through specific letter swaps.

array hash table string greedy sorting

Problem Statement

You are given a 0-indexed string array words with length n, each containing lowercase English letters. You are allowed to perform the following operation any number of times: swap any letter from one word with a letter from another word. Your task is to return the maximum number of palindromes that can be achieved in the array after performing some or all of these operations.

A palindrome is a word that reads the same forwards and backwards. The goal is to identify how many words can be transformed into palindromes by redistributing characters across the list of words through these swaps. The problem tests your ability to efficiently scan arrays and utilize hash lookups to determine the feasibility of forming palindromes.

Examples

Example 1

Input: words = ["abbb","ba","aa"]

Output: 3

In this example, one way to get the maximum number of palindromes is: Choose i = 0, j = 1, x = 0, y = 0, so we swap words[0][0] and words[1][0]. words becomes ["bbbb","aa","aa"]. All strings in words are now palindromes. Hence, the maximum number of palindromes achievable is 3.

Example 2

Input: words = ["abc","ab"]

Output: 2

In this example, one way to get the maximum number of palindromes is: Choose i = 0, j = 1, x = 1, y = 0, so we swap words[0][1] and words[1][0]. words becomes ["aac","bb"]. Choose i = 0, j = 0, x = 1, y = 2, so we swap words[0][1] and words[0][2]. words becomes ["aca","bb"]. Both strings are now palindromes. Hence, the maximum number of palindromes achievable is 2.

Example 3

Input: words = ["cd","ef","a"]

Output: 1

In this example, there is no need to perform any operation. There is one palindrome in words "a". It can be shown that it is not possible to get more than one palindrome after any number of operations. Hence, the answer is 1.

Constraints

1 <= words.length <= 1000
1 <= words[i].length <= 100
words[i] consists only of lowercase English letters.

Solution Approach

Array Scanning and Character Counting

Scan through each word and count the occurrences of each character. The key is recognizing that a palindrome requires pairs of characters, and you can freely redistribute letters between words. By counting character frequencies globally across all words, it becomes easier to identify how many pairs and single characters are available for palindrome formation.

Greedy Swapping Strategy

Using a greedy approach, consider performing character swaps only when necessary. Attempt to match characters from different words to form palindromes, prioritizing those words with an odd number of characters. If characters from multiple words can be swapped to create more palindromes, proceed with this approach, ensuring minimal operations to maximize the result.

Hash Table Optimization

To track character frequencies efficiently, use a hash table to store the count of each letter across all words. This approach allows you to quickly access and update the frequency of each character as you simulate the redistributions, ensuring that you can calculate the maximum number of palindromes without redundant recalculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the number of words n and the average length of words m. Efficient use of hash tables to count characters leads to a time complexity of O(n * m), where each letter in each word is processed once. Space complexity is O(26) due to the fixed size of the alphabet used for hashing.

What Interviewers Usually Probe

Check if the candidate effectively uses greedy methods for maximizing palindromes.
Look for clear understanding of how to count and redistribute characters using hash tables.
Evaluate how well the candidate balances time complexity with the need for multiple character swaps.

Common Pitfalls or Variants

Common pitfalls

Failing to properly account for global character counts and only focusing on individual words.
Overcomplicating the solution by trying to optimize too early, missing simple swaps.
Not considering edge cases where no swaps are needed, such as when all words are already palindromes.

Follow-up variants

Consider variations with different input constraints, such as allowing some fixed swaps or limiting the number of operations.
Modify the problem by introducing additional restrictions on swap operations (e.g., restricting which words can swap with which).
Expand the problem to count and return the exact words that can be palindromes after the operation.

FAQ

How can I optimize the solution for maximum palindromes?

The key is efficiently counting characters across all words and applying a greedy strategy for redistributing letters to form palindromes.

What patterns should I focus on for this problem?

The main pattern involves array scanning and hash table usage for frequency counting and character redistribution.

How do I handle cases where no operations are needed?

In such cases, ensure your solution can return the result without performing any swaps, checking for pre-existing palindromes first.

Can this problem be solved using dynamic programming?

Dynamic programming is not necessary here; the problem is efficiently solved with greedy swaps and hash table lookups.

What if there are large words in the input?

The solution handles large words efficiently by processing each character once and using hash tables for fast lookups, ensuring scalability.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def maxPalindromesAfterOperations(self, words: List[str]) -> int:
        s = mask = 0
        for w in words:
            s += len(w)
            for c in w:
                mask ^= 1 << (ord(c) - ord("a"))
        s -= mask.bit_count()
        words.sort(key=len)
        ans = 0
        for w in words:
            s -= len(w) // 2 * 2
            if s < 0:
                break
            ans += 1
        return ans

Continue Practicing

#2131 longest palindrome by concatenating two letter words #3186 maximum total damage with spell casting #2856 minimum array length after pair removals