- Difficulty: Hard
- Tags: LeetCode, Hard, Two Pointers, String, Dynamic Programming, leetcode-2911, O(n^3), O(n^2 * logn), Number Theory, DP
Problem
Given a string s and an integer k, partition s into k substrings such that the letter changes needed to make each substring a semi-palindrome are minimized.
Return the minimum number of letter changes required.
A semi-palindrome is a special type of string that can be divided into palindromes based on a repeating pattern. To check if a string is a semi-palindrome:
- Choose a positive divisor
dof the string's length.dcan range from1up to, but not including, the string's length. For a string of length1, it does not have a valid divisor as per this definition, since the only divisor is its length, which is not allowed. - For a given divisor
d, divide the string into groups where each group contains characters from the string that follow a repeating pattern of lengthd. Specifically, the first group consists of characters at positions1,1 + d,1 + 2d, and so on; the second group includes characters at positions2,2 + d,2 + 2d, etc. - The string is considered a semi-palindrome if each of these groups forms a palindrome.
Consider the string "abcabc":
- The length of
"abcabc"is6. Valid divisors are1,2, and3. - For
d = 1: The entire string"abcabc"forms one group. Not a palindrome. - For
d = 2:- Group 1 (positions
1, 3, 5):"acb" - Group 2 (positions
2, 4, 6):"bac" - Neither group forms a palindrome.
- Group 1 (positions
- For
d = 3:- Group 1 (positions
1, 4):"aa" - Group 2 (positions
2, 5):"bb" - Group 3 (positions
3, 6):"cc" - All groups form palindromes. Therefore,
"abcabc"is a semi-palindrome.
- Group 1 (positions
Example 1:
Input: s = "abcac", k = 2
Output: 1
Explanation: Divide s into "ab" and "cac". "cac" is already semi-palindrome. Change "ab" to "aa", it becomes semi-palindrome with d = 1.
Example 2:
Input: s = "abcdef", k = 2
Output: 2
Explanation: Divide s into substrings "abc" and "def". Each needs one change to become semi-palindrome.
Example 3:
Input: s = "aabbaa", k = 3
Output: 0
Explanation: Divide s into substrings "aa", "bb" and "aa". All are already semi-palindromes.
Constraints:
2 <= s.length <= 2001 <= k <= s.length / 2scontains only lowercase English letters.