Maximum Palindromes After Operations

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Greedy AlgorithmsArraysStrings

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.

Note: i and j may be equal during an operation.

Example 1:

Input: words = ["abbb","ba","aa"]
Output: 3
Explanation: 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
Explanation: 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
Explanation: 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.

Try coding it in LeetCode

Maximum Palindromes After Operations

Medium

MathWorks

2 views

Topics:

Greedy AlgorithmsArraysStrings

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.

Note: i and j may be equal during an operation.

Example 1:

Input: words = ["abbb","ba","aa"]
Output: 3
Explanation: 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
Explanation: 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
Explanation: 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the string `s` and the `lengths` array, as well as the values within the `lengths` array?
Can the `lengths` array contain zero or negative values? If so, how should I handle them?
If it's impossible to create even a single palindrome with the given characters and lengths, what should I return?
Are there any restrictions on the characters within the input string `s` beyond being lowercase English letters (e.g., can it be empty)?
If there are multiple combinations of palindromes that maximize the count, is any valid combination acceptable?

Brute Force Solution

Approach

The brute force approach to maximizing palindromes involves considering every possible way to rearrange the characters in the input string. We generate all possible combinations and then check if we can form the most palindromes from each combination.

Here's how the algorithm would work step-by-step:

First, figure out all possible arrangements of the letters you have.
For each arrangement, try to make as many palindromes as possible.
To do this, start by making one palindrome. See how many letters are left.
Then, with the remaining letters, try to make another palindrome. See how many letters are left this time.
Keep repeating the previous step until you can't make any more palindromes.
Count how many palindromes you were able to make for that particular arrangement of letters.
Repeat this whole process (from the beginning) for every single possible arrangement of letters.
Finally, after trying every arrangement, choose the arrangement that allowed you to make the most palindromes.

Code Implementation

from itertools import permutations

def maximum_palindromes_brute_force(input_string):
    max_palindromes = 0
    
    # Iterate through all possible permutations of the input string
    for permutation in permutations(input_string):
        arranged_string = "".join(permutation)
        current_palindromes = 0
        remaining_string = arranged_string

        # Attempt to make as many palindromes as possible
        while len(remaining_string) > 0:
            
            # Start by trying to create a palindrome of length 1
            if len(remaining_string) == 1:
                current_palindromes += 1
                remaining_string = ""
                continue

            found_palindrome = False

            # Attempt to create palindrome with length > 1
            for i in range(1, len(remaining_string)):
                if remaining_string[0] == remaining_string[i]:

                    # If palindrome is possible, remove characters
                    current_palindromes += 1
                    remaining_string = remaining_string[1:i] + remaining_string[i+1:]
                    found_palindrome = True
                    break

            # If no palindrome could be created, stop trying
            if not found_palindrome:
                break

        # Update maximum palindromes if necessary
        max_palindromes = max(max_palindromes, current_palindromes)

    return max_palindromes

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach involves generating all possible arrangements of the input string of size n. Generating all permutations of a string of length n takes O(n!) time. For each permutation, the algorithm attempts to form palindromes, which involves iterating through the string (O(n) in the worst case for each palindrome formed, although this is dwarfed by the permutation cost). The cost of finding the maximum number of palindromes after generating each permutation doesn't significantly increase the overall complexity. Therefore, the dominant factor is generating all permutations. Thus, the overall time complexity is O(n!).

Space Complexity

O(N!) – The described brute force approach involves generating all possible arrangements (permutations) of the input string. Generating all permutations of a string of length N requires storing those permutations. In the worst-case, this involves creating and storing up to N! permutations, each having length N. While the individual palindrome checks might use relatively little space, the dominant factor is the storage of the permutations themselves, leading to a space complexity proportional to N!.

Optimal Solution

Approach

The key is to realize that we don't need to rearrange the actual strings. Instead, we just need to count how many times each letter appears across all the strings and from this information calculate the maximal number of palindromes we can create. The crucial insight is focusing on using as many pairs of letters as possible for palindromes.

Here's how the algorithm would work step-by-step:

First, count how many times each letter appears in all the given strings combined.
Next, calculate the total number of letter pairs we can make. For each letter, divide its count by two to get the number of pairs for that letter, and add these up across all letters.
Then, for each string we want to make into a palindrome, we'll use as many letter pairs as we can. Each pair can be used to form two of the same letter on either side of the center of the palindrome.
The number of palindromes we can make is limited by the total number of letter pairs and the lengths of the given strings. We want to minimize any unused letter pairs.
For each string length, if we have at least that many letter pairs available, we can form a complete palindrome of that length. If we don't have enough pairs, we use all the available pairs to form as much of the palindrome as possible.
Keep track of how many palindromes we were able to fully form (or partially form). This count is the final answer.

Code Implementation

def maximum_palindromes_after_operations(strings):
    total_letter_counts = {}
    for string in strings:
        for letter in string:
            total_letter_counts[letter] = total_letter_counts.get(letter, 0) + 1

    total_letter_pairs = 0
    for letter_count in total_letter_counts.values():
        total_letter_pairs += letter_count // 2

    strings.sort(key=len)
    palindrome_count = 0

    #Greedily use pairs to form palindromes based on string length.
    for string in strings:
        string_length = len(string)
        required_pairs = string_length // 2

        if total_letter_pairs >= required_pairs:
            palindrome_count += 1
            total_letter_pairs -= required_pairs
        else:
            #If not enough pairs, can't form the full palindrome.
            break

    return palindrome_count

Big(O) Analysis

Time Complexity

O(N + Q*M) – The initial step involves counting the frequency of each letter, which takes O(N) time, where N is the total number of characters across all input strings. Then, for each of the Q queries (string lengths), we perform a calculation related to the length M of the string for that query. This calculation involves determining if enough character pairs exist to form the palindrome. Therefore, processing all queries contributes O(Q*M) to the overall complexity. The total complexity is thus O(N + Q*M).

Space Complexity

O(1) – The solution uses a fixed-size array (or hash map) to store the counts of each letter, which has a size independent of the input string lengths or the number of strings. Since the number of distinct letters is limited (e.g., 26 for lowercase English letters), the space used for storing character counts is constant. No other data structures dependent on input size are utilized. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty string s	If the string s is empty, the number of palindromes will depend solely on the possibility of forming zero-length palindromes if lengths contains zeros, or no palindromes if lengths doesn't contain zeros.
Empty lengths array	If the lengths array is empty, no palindromes can be formed, so return 0.
String s is null or undefined	Handle null or undefined input by returning 0 or throwing an exception based on requirements.
lengths array contains zero values	Zero length strings should be considered valid palindromes, and the solution should prioritize using characters for non-zero length palindromes first.
lengths array contains lengths larger than the length of s	A length greater than the string's length cannot be fulfilled, so filter out lengths greater than the string's length or don't process them.
The number of odd character counts in s is greater than the number of lengths available	The number of palindromes that can be created is limited by the number of odd character counts in s, since each palindrome can have at most one character with odd count.
Maximum possible length of s and lengths	Ensure that the frequency map and calculations involving lengths can handle a large number of characters and lengths efficiently, without causing integer overflow or memory issues.
All characters in s are the same	This simplifies the problem, and we can form the maximum number of palindromes subject to fulfilling all provided lengths in lengths if the total string length is greater than the sum of lengths in length.