LeetCode 3003: Maximize the Number of Partitions After Operations Solution

Master LeetCode problem 3003 (Maximize the Number of Partitions After Operations), 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.

3003. Maximize the Number of Partitions After Operations

Hard

String Dynamic Programming Bit Manipulation Bitmask

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible characters to change in the input string s. For each character change, we calculate the maximum number of partitions that can be achieved by applying the partitioning operation as described in the problem. We keep track of the maximum partitions obtained over all possible character changes and return this maximum value as the final answer.

Here is a high-level overview of the algorithm:

Iterate over all lowercase English letters.
For each character change, simulate the partitioning operation and calculate the maximum number of partitions.
Keep track of the maximum partitions obtained and return this value.

The time complexity of this algorithm is O(26 * N), where N is the length of the input string s. The space complexity is O(N) for storing the input string.

Solution Code

class Solution {
    public int maxPartitions(String s, int k) {
        int maxPartitions = 0;
        
        for (char c = 'a'; c <= 'z'; c++) {
            StringBuilder sb = new StringBuilder(s);
            for (int i = 0; i < sb.length(); i++) {
                if (sb.charAt(i) != c) {
                    sb.setCharAt(i, c);
                    int partitions = calculatePartitions(sb.toString(), k);
                    maxPartitions = Math.max(maxPartitions, partitions);
                }
            }
        }
        
        return maxPartitions;
    }
    
    private int calculatePartitions(String s, int k) {
        int partitions = 0;
        
        while (!s.isEmpty()) {
            int distinctChars = 0;
            int i = 0;
            Set<Character> seen = new HashSet<>();
            while (i < s.length() && distinctChars <= k) {
                if (!seen.contains(s.charAt(i))) {
                    seen.add(s.charAt(i));
                    distinctChars++;
                }
                if (distinctChars <= k) {
                    i++;
                }
            }
            partitions++;
            s = s.substring(i);
        }
        
        return partitions;
    }
}

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