LeetCode 2911: Minimum Changes to Make K Semi-palindromes Solution

Master LeetCode problem 2911 (Minimum Changes to Make K Semi-palindromes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2911. Minimum Changes to Make K Semi-palindromes

Hard

Two Pointers String Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will iterate through all possible divisors to find the minimum changes needed for each substring to become a semi-palindrome. We will maintain a 2D array to store the minimum changes for each substring length and number of partitions. At the end, we will return the minimum changes needed for k partitions.

Algorithmic Idea:
1. Initialize a 2D array dp of size n x k where n is the length of the input string s.
2. Iterate through all divisors from 1 to n.
3. For each divisor, calculate the minimum changes needed to make each substring a semi-palindrome.
4. Update the dp array by considering all possible partitions at each position.
5. Return the minimum changes for k partitions.
Time Complexity: O(n^2 * k)
Space Complexity: O(n * k)

Solution Code

class Solution {
    public int minChangesToMakeSemiPalindrome(String s, int k) {
        int n = s.length();
        int[][] dp = new int[n + 1][k + 1];
        
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++) {
                dp[i][j] = n;
            }
        }
        
        for (int d = 1; d < n; d++) {
            for (int i = d; i <= n; i++) {
                int numChanges = 0;
                for (int j = i; j >= d; j -= d) {
                    if (s.charAt(j - 1) != s.charAt(i - 1)) {
                        numChanges++;
                    }
                    dp[i][1] = Math.min(dp[i][1], dp[j - 1][0] + numChanges);
                }
                for (int j = 2; j <= k; j++) {
                    for (int p = i; p >= d; p -= d) {
                        dp[i][j] = Math.min(dp[i][j], dp[p - 1][j - 1] + dp[i][1]);
                    }
                }
            }
        }
        
        return dp[n][k];
    }
}

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