Make String Anti-palindrome

Hard

Intuit

0 views

Topics:

StringsGreedy Algorithms

You are given a string s. In one operation, you can choose two distinct indices i and j, where i != j, and swap the characters s[i] and s[j].

A string is called an anti-palindrome if for every index i, s[i] != s[n - i - 1], where 0 <= i < n / 2, and n is the length of the string.

Return the minimum number of operations to make s an anti-palindrome. If it is impossible, return -1.

Example 1:

Input: s = "abba"
Output: 1
Explanation: We can swap s[0] and s[2] to get "bbaa" which is an anti-palindrome since s[0] != s[3] and s[1] != s[2].
It can be proven that one operation is the minimum number of operations to make s an anti-palindrome.

Example 2:

Input: s = "abcabc"
Output: 3
Explanation: We can swap s[0] and s[5], s[1] and s[4], and s[2] and s[3] to get "cbacba" which is an anti-palindrome since s[0] != s[5], s[1] != s[4], and s[2] != s[3].
It can be proven that three operations is the minimum number of operations to make s an anti-palindrome.

Example 3:

Input: s = "aabaa"
Output: -1
Explanation: It can be proven that it is impossible to make s an anti-palindrome.

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters.

Make String Anti-palindrome

Hard

Intuit

0 views

Topics:

StringsGreedy Algorithms

You are given a string s. In one operation, you can choose two distinct indices i and j, where i != j, and swap the characters s[i] and s[j].

A string is called an anti-palindrome if for every index i, s[i] != s[n - i - 1], where 0 <= i < n / 2, and n is the length of the string.

Return the minimum number of operations to make s an anti-palindrome. If it is impossible, return -1.

Example 1:

Input: s = "abba"
Output: 1
Explanation: We can swap s[0] and s[2] to get "bbaa" which is an anti-palindrome since s[0] != s[3] and s[1] != s[2].
It can be proven that one operation is the minimum number of operations to make s an anti-palindrome.

Example 2:

Input: s = "abcabc"
Output: 3
Explanation: We can swap s[0] and s[5], s[1] and s[4], and s[2] and s[3] to get "cbacba" which is an anti-palindrome since s[0] != s[5], s[1] != s[4], and s[2] != s[3].
It can be proven that three operations is the minimum number of operations to make s an anti-palindrome.

Example 3:

Input: s = "aabaa"
Output: -1
Explanation: It can be proven that it is impossible to make s an anti-palindrome.

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters.

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 is the maximum length of the string s? Should I be concerned about integer overflow when calculating the number of operations?
Can the input string s be empty or null? If so, what should I return?
If it's impossible to make the string an anti-palindrome, is -1 the only acceptable return value, or are there other error codes I should consider?
Are there any specific requirements for the algorithm's space complexity?
If multiple sequences of swaps achieve the minimum number of operations to make the string an anti-palindrome, is any one of them acceptable, or are there specific criteria for choosing among them?

Brute Force Solution

Approach

The brute force approach to making a string anti-palindrome involves exploring all possible ways to modify the input string. It methodically tests each alteration to see if it meets the anti-palindrome requirement. If an anti-palindrome is found, the algorithm terminates.

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

Consider all possible modifications to the string, changing each letter, one at a time, to all other possible letters.
After each potential change, check to see if the resulting string is an anti-palindrome. An anti-palindrome is a string where for every position in the string, the letter at that position is different from the letter at the corresponding position from the end of the string.
If a change results in the string becoming an anti-palindrome, you have found a solution. You may want to explore all possible anti-palindromes to potentially find the one that requires the fewest changes.
If no single change works, try changing two letters at a time, again checking after each pair of changes whether the string has become an anti-palindrome.
Continue increasing the number of changes made until you either find an anti-palindrome or exhaust all possible combinations of changes. Return the first anti-palindrome found, or the anti-palindrome that required the fewest changes if desired, or return the original string if no solution is found.

Code Implementation

def make_string_anti_palindrome_brute_force(input_string):    best_solution = None
    minimum_changes = float('inf')
    string_length = len(input_string)

    for char_index in range(string_length):
        original_character = input_string[char_index]

        for ascii_value in range(ord('a'), ord('z') + 1):
            new_character = chr(ascii_value)
            modified_string = list(input_string)

            # Change the character at current index
            modified_string[char_index] = new_character
            modified_string = "".join(modified_string)

            is_anti_palindrome = True
            for index in range(string_length // 2):
                if modified_string[index] == modified_string[string_length - 1 - index]:

                    # String is not anti-palindrome
                    is_anti_palindrome = False
                    break

            if is_anti_palindrome:

                # Check if current change is better than the existing solution
                number_of_changes = 0
                for index in range(string_length):
                    if input_string[index] != modified_string[index]:
                        number_of_changes += 1

                if number_of_changes < minimum_changes:
                    minimum_changes = number_of_changes
                    best_solution = modified_string

    return best_solution

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(26^n) – The brute force approach iterates through all possible modifications of the string. For each position in the string of length n, we can potentially change it to any of the other 25 characters (assuming a 26-character alphabet). This gives us approximately 26^n possible combinations to check. Each of these combinations requires verifying if it is an anti-palindrome, which takes O(n) time, but the dominant factor is the number of combinations. Therefore, the overall time complexity is O(26^n).

Space Complexity

O(N!) – The brute force approach explores all possible modifications of the input string. In the worst case, the algorithm might need to consider all possible combinations of changes to the string, implicitly using a recursion tree. The maximum depth of this tree could be related to the length of the string N, with each level branching out to explore all possible character substitutions. This leads to a space complexity that grows factorially with the size of the input string, hence O(N!).

Optimal Solution

Approach

The goal is to change a string so that it's no longer a palindrome (reads the same forwards and backward). The optimal strategy is to carefully count how many of each character we have and then rearrange them to avoid creating a palindrome. We prioritize making sure the start and end of the string are different.

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

First, count how many times each letter appears in the original string.
If every letter appears an even number of times, there's no way to make it anti-palindrome (different from palindrome), so return the empty string or null to indicate failure.
Now, build the new string. Start by placing the most frequent letter at the beginning of the new string.
Then, place the second most frequent letter at the end of the new string. This ensures the first and last characters are different.
Continue placing letters, alternating between adding to the middle, making sure at no point are characters equidistant from the center the same until all instances of letters are placed.
Lastly, verify the rearranged string is truly an anti-palindrome. If by chance the rearrangement failed, then return the empty string or null to indicate failure.

Code Implementation

def make_string_antipalindrome(input_string):
    string_length = len(input_string)
    number_of_changes = 0
    modified_string = list(input_string)

    for index in range(string_length // 2):
        left_character = modified_string[index]
        right_character = modified_string[string_length - 1 - index]

        # Characters at both ends are equal
        if left_character == right_character:

            # Need to change one character to break the palindrome
            for char_code in range(ord('a'), ord('z') + 1):
                replacement_character = chr(char_code)
                if replacement_character != right_character:
                    modified_string[index] = replacement_character
                    break

            number_of_changes += 1

    return number_of_changes

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log n) – Counting character frequencies involves iterating through the string once, taking O(n) time. Sorting the frequencies (implicitly done when finding most/second most frequent) typically takes O(n log n) time using efficient sorting algorithms. Building the new string by placing characters primarily involves insertion operations, which are interleaved. The final verification iterates through the rearranged string, taking O(n) time. Therefore, the dominant operation is sorting the character frequencies, resulting in O(n log n) overall time complexity.

Space Complexity

O(1) – The algorithm primarily uses a character count data structure which, given the fixed alphabet size, occupies constant space. While rearranging the string, the algorithm primarily manipulates the input string itself (in-place). Additional variables used for intermediate calculations or indexing during rearrangement contribute constant space. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty string or null input	Return 0 if the string is empty or null, as it's trivially an anti-palindrome (or invalid input should be rejected).
String with length 1	Return 0 as it is trivially an anti-palindrome.
String with length 2 and identical characters	Return -1, as no number of swaps can make the string anti-palindrome.
String where all characters are identical	Return -1, as no anti-palindrome can be created.
String with even length and all characters appearing exactly twice	Return 0 because if the characters are already in the correct order, then no operations are needed.
String with even length and character counts where one character appears more than n/2 times	Return -1, because an anti-palindrome cannot be formed.
Very long string requiring many swaps, potentially leading to integer overflow in cost calculation.	Use a data type that supports a large range of numbers (e.g., long) to avoid potential overflow.
String with multiple valid solutions (different swap sequences)	The algorithm should return the minimum number of swaps among all valid solutions.