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Maximum Palindromes After Operations

Difficulty: Medium

Problem Description

You are given a 0-indexed string array words having length n and containing 0-indexed strings. You are allowed to perform the following operation any number of times (including zero): Choose integers i, j, x, and y such that 0 <= i, j < n, 0 <= x < words[i].length, 0 <= y < words[j].length, and swap the characters words[i][x] and words[j][y]. Return an integer denoting the maximum number of palindromes words can contain, after performing some operations.

Key Insights

  • A palindrome reads the same forwards and backwards.
  • Swapping characters allows for an adjustment of character frequencies across words.
  • The ability to swap characters can help in forming pairs of characters needed for palindromes.
  • Single characters can always contribute to palindromes, allowing for a minimum count.

Space and Time Complexity

Time Complexity: O(n * m) where n is the number of words and m is the average length of the words.
Space Complexity: O(1) as we only use a fixed number of variables for counting.

Solution

To solve the problem, we can follow these steps:

  1. Count the frequency of each character in all the words.
  2. Calculate how many palindromes can be formed using these frequencies.
  3. Use a greedy approach to maximize the number of palindromes by pairing characters and taking care of single characters.

The algorithm primarily leverages a counting approach to determine character frequencies and then calculates the maximum possible palindromes based on the character pairs.

Code Solutions

def maxPalindromes(words):
    from collections import Counter

    # Count the frequency of each character in all words
    char_count = Counter()
    for word in words:
        char_count.update(word)

    palindromes = 0
    odd_count = 0

    # Count pairs of characters
    for count in char_count.values():
        palindromes += count // 2
        if count % 2 == 1:
            odd_count += 1

    # Each pair can form half a palindrome, and we can add one more if there's an odd character
    palindromes += odd_count // 2
    if odd_count > 0:
        palindromes += 1

    return palindromes
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