3088. Make String Anti-palindrome 🔒

HardGreedyStringCounting SortSorting

Problem Description

We define a string s of even length n as an anti-palindrome if for each index 0 <= i < n, the condition s[i] != s[n - i - 1] holds true. Given a string s, the task is to transform it into an anti-palindrome by performing any number of operations, including zero. An operation consists of selecting two characters from s and swapping them. If more than one resultant string meets the conditions, return the lexicographically smallest one. If it cannot be transformed into an anti-palindrome, return "-1".

Intuition

To solve the problem, we begin by understanding the properties of anti-palindromes. An anti-palindrome requires that each character at the beginning half of the string is different from its corresponding character in the second half. This condition naturally suggests that reordering the characters might help.

The approach is as follows:

Sorting for Lexicographic Order: First, by sorting the string s, we ensure that any configuration we attempt to make is built from the smallest possible lexicographic arrangement. Sorting helps establish a base state to work from.
Balancing the Middle Characters: After sorting, we check the characters at the middle position. If the middle two characters are the same, it indicates that simply sorting won't be enough to achieve the anti-palindrome property.
Swapping Strategy: The strategy is to swap characters between the two halves of the string if necessary. If there are repeated characters in the middle of the sorted string that would prevent anti-palindrome properties, we attempt to swap them with characters further down the sorted sequence where possible.
Exhaust Possibilities: If, after attempting available swaps, it is impossible to achieve the anti-palindrome condition, the function will return "-1". Otherwise, the correctly swapped characters will form the lexicographically smallest anti-palindrome.

This leveraging of sorting and selective swapping ensures we can efficiently reach the desired arrangement or conclude that it is impossible.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution to this problem involves a methodical approach leveraging sorting and conditional swapping.

Initial Sorting:
- We begin by sorting the string s. Sorting the characters helps us position them in the lexicographically smallest order, which is crucial for generating the smallest possible anti-palindrome if it's feasible.
- After sorting, the string s is stored in a list cs.
Middle Character Check:
- Calculate n, the length of the list, and find m as n // 2, which represents the midpoint of the string.
- Check if the two middle characters, cs[m] and cs[m-1], are identical. If they are unequal, the sorted string can already be an anti-palindrome.
Swapping Strategy:
- If the middle characters are the same, initiate a search for the first character in the second half of the list that is different from the middle character. This is tracked using the pointer i.
- Another pointer j starts at m for possible swaps.
- Swap cs[i] and cs[j] if they're found to be suitable candidates such that swapping results in a non-palindrome property.
Termination Conditions:
- If it becomes impossible to find a suitable character to swap, indicated by reaching the end of the list with pointer i, it means it's not possible to transform s into an anti-palindrome. Hence, return "-1".
- If swaps succeed, the modified list is joined back into a string and returned as the result. This string represents the lexicographically smallest anti-palindrome, if achieved.

This structured approach to checking the middle characters and selectively swapping ensures that if an anti-palindrome can be formed, it is done efficiently and correctly.

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Example Walkthrough

Consider the string s = "aabbcc", which has an even length of 6. The task is to transform this string into an anti-palindrome.

Initial Sorting:

Sort the string s. After sorting, the string becomes cs = ['a', 'a', 'b', 'b', 'c', 'c'].

Middle Character Check: 2. Calculate the midpoint m = 6 // 2 = 3. 3. Check the middle characters: cs[2] = 'b' and cs[3] = 'b'. Since they are identical, a simple sort doesn't yield an anti-palindrome.

Swapping Strategy: 4. Look for a character in the second half of the string (indices 3 to 5) that can be swapped with the middle 'b'. Start from index 4, which holds cs[4] = 'c'. 5. Swap cs[3] ('b') with cs[4] ('c'), resulting in cs = ['a', 'a', 'b', 'c', 'b', 'c'].

Verification: 6. After the swap, verify if the string is an anti-palindrome:

Compare cs[0] ('a') with cs[5] ('c'): they are different.
Compare cs[1] ('a') with cs[4] ('b'): they are different.
Compare cs[2] ('b') with cs[3] ('c'): they are different.

This configuration satisfies the condition for an anti-palindrome. The final result is "aabcbc", which is the lexicographically smallest anti-palindrome obtainable from the original string s.

Solution Implementation

1class Solution:
2    def makeAntiPalindrome(self, s: str) -> str:
3        # Sort the characters in the string
4        sorted_chars = sorted(s)
5        n = len(sorted_chars)
6        half_n = n // 2
7
8        # Check if the middle elements in the sorted list are the same
9        if sorted_chars[half_n] == sorted_chars[half_n - 1]:
10            # Initialize indices for swapping
11            i = half_n
12            # Increase i until characters differ or end is reached
13            while i < n and sorted_chars[i] == sorted_chars[i - 1]:
14                i += 1
15
16            j = half_n
17            # Try to swap characters across the second half
18            # Until the list can no longer be rearranged
19            while j < n and sorted_chars[j] == sorted_chars[n - j - 1]:
20                if i >= n:
21                    return "-1"  # Return "-1" if no valid transformation is possible
22                sorted_chars[i], sorted_chars[j] = sorted_chars[j], sorted_chars[i]
23                i += 1
24                j += 1
25
26        return "".join(sorted_chars)
27

1import java.util.Arrays;
2
3class Solution {
4    public String makeAntiPalindrome(String s) {
5        // Convert the string into a character array and sort it.
6        char[] characters = s.toCharArray();
7        Arrays.sort(characters);
8      
9        int n = characters.length;
10        int midpoint = n / 2;
11      
12        // Check if the middle characters in sorted array are the same.
13        if (characters[midpoint] == characters[midpoint - 1]) {
14            int i = midpoint;
15          
16            // Move the pointer i to find a non-repeating character.
17            while (i < n && characters[i] == characters[i - 1]) {
18                ++i;
19            }
20          
21            // Swap characters from the middle to the end with those from the opposite end.
22            for (int j = midpoint; j < n && characters[j] == characters[n - j - 1]; ++i, ++j) {
23                if (i >= n) {
24                    return "-1"; // Return "-1" if not possible to rearrange to make it anti-palindrome.
25                }
26              
27                // Swap characters to disrupt any chance of palindrome formation.
28                char temp = characters[i];
29                characters[i] = characters[j];
30                characters[j] = temp;
31            }
32        }
33      
34        // Return the modified string.
35        return new String(characters);
36    }
37}
38

1class Solution {
2public:
3    string makeAntiPalindrome(string s) {
4        // Sort the string to arrange characters in increasing order
5        sort(s.begin(), s.end());
6
7        int n = s.length(); // Get the length of the string
8        int half_length = n / 2; // Calculate the middle index
9      
10        // Check if two middle characters are the same
11        if (s[half_length] == s[half_length - 1]) {
12            int i = half_length;
13          
14            // Find the next character that is different from the middle
15            while (i < n && s[i] == s[i - 1]) {
16                ++i;
17            }
18          
19            // Attempt to create an anti-palindrome by swapping symmetrically
20            for (int j = half_length; j < n && s[j] == s[n - j - 1]; ++i, ++j) {
21                // If i reaches or exceeds the string length, return "-1"
22                if (i >= n) {
23                    return "-1";
24                }
25                // Swap characters to try to prevent palindrome formation
26                swap(s[i], s[j]);
27            }
28        }
29      
30        // Return the resulting string
31        return s;
32    }
33};
34

1function makeAntiPalindrome(s: string): string {
2    // Convert the input string into an array of characters and sort them lexicographically
3    const cs: string[] = s.split('').sort();
4    const n: number = cs.length;
5    const m: number = n >> 1; // Find the middle index for the sorted array
6
7    // Check the condition where more than half of the characters are identical
8    if (cs[m] === cs[m - 1]) {
9        let i: number = m;
10        // Find the first character that differs from the middle character
11        while (i < n && cs[i] === cs[i - 1]) {
12            i++;
13        }
14
15        // Try to adjust the array for anti-palindrome by swapping
16        for (let j: number = m; j < n && cs[j] === cs[n - j - 1]; i++, j++) {
17            if (i >= n) {
18                // If no suitable configuration is found, return "-1"
19                return '-1';
20            }
21            // Swap characters to achieve potential anti-palindrome
22            [cs[j], cs[i]] = [cs[i], cs[j]];
23        }
24    }
25
26    // Join the array back into a string and return it
27    return cs.join('');
28}
29

Time and Space Complexity

The time complexity of the makeAntiPalindrome function is O(n \times \log n). This is primarily due to the sorted(s) function, which sorts the string s and operates in O(n \times \log n). The subsequent operations, including while loops and swapping elements, are bounded by O(n).

The space complexity is O(n), attributed to the storage requirement for the sorted list cs, which needs additional space proportional to the length of the input string s.

Learn more about how to find time and space complexity quickly.

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Question 1 out of 10

How does merge sort divide the problem into subproblems?

Divide the array into two equal halves by index

Divide the array into two based on whether an element is smaller than an arbitrary value

Divide the array into a stray element and the rest of the array

Merge sort does not use divide and conquer

I don't know