Is Minimum Operations to Make a Rotated Palindrome II easy or hard?

It is rated Hard on FleetCode with an acceptance rate of 78.3%. The difficulty comes from optimizing the brute force O(n^2) solution to O(n) using clever precomputation.

Minimum Operations to Make a Rotated Palindrome II Python/Java solution

In Python, you can use a list of length n for difference array and iterate over pairs in the doubled string. In Java, use an int[] array. Both implementations run in O(n) time and O(n) space.

How to solve Minimum Operations to Make a Rotated Palindrome II in O(n)?

Create a doubled string s+s. For each possible rotation start index i (0 to n-1), you need mismatches between pairs (i+l, i+n-1-l) for l from 0 to n/2-1. Use a difference array to add 1 to all start indices that include a given mismatching pair, then take the minimum cumulative value.

What is the best approach for Minimum Operations to Make a Rotated Palindrome II?

The optimal approach uses a two-pointer technique on a doubled string. By concatenating the string with itself, every rotation corresponds to a substring of length n. Precomputing mismatch counts for all rotations using difference arrays yields O(n) time and O(n) space.

Is Minimum Operations to Make a Rotated Palindrome II asked at Google/Amazon/Meta?

This problem is tagged as General and is Hard difficulty. While not explicitly tied to a specific company, similar string manipulation and palindrome problems are common in interviews at top tech companies like Google, Amazon, and Meta.

What data structure is used in Minimum Operations to Make a Rotated Palindrome II?

The optimal solution uses an array (or list) as a difference array to accumulate mismatch counts across all rotations. The doubled string is also used to avoid explicit rotations.

What is the time complexity of Minimum Operations to Make a Rotated Palindrome II?

The optimal solution runs in O(n) time, where n is the length of the input string. The brute force approach takes O(n^2) time.

Description Solution

4028. Minimum Operations to Make a Rotated Palindrome II

Hard65.8% acceptancePremium elsewhere, free here

You are given a string s consisting of lowercase English letters.

You can perform the following operations any number of times (including zero) and in any order:

Increment: Choose any index i and replace s[i] with the next lowercase English letter. The letter after 'z' is 'a'.
Left rotate: Move the first character of the string to the end.

Return the minimum number of operations required to make s a palindrome.

Example 1:

Input: s = "abc"

Output: 2

Explanation:

One optimal solution:

Left rotate the string: "abc" -> "bca".
Increment 'a' to 'b': "bca" -> "bcb".
"bcb" is a palindrome. Thus, the answer is 2.

Example 2:

Input: s = "yb"

Output: 3

Explanation:

Increment the first character three times: "yb" -> "zb" -> "ab" -> "bb".
"bb" is a palindrome. Thus, the answer is 3.

Constraints:

2 <= s.length <= 5 * 10⁴
s consists only of lowercase English letters.

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