LeetCode 3018: Maximum Number of Removal Queries That Can Be Processed I Solution

Master LeetCode problem 3018 (Maximum Number of Removal Queries That Can Be Processed I), 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.

3018. Maximum Number of Removal Queries That Can Be Processed I

Hard

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use binary search along with the concept of prefix sums. We will first calculate the prefix sums of the removal queries and sort them. Then, we can iterate over the queries in reverse order, removing elements from the prefix sums array and checking if the resulting array is still sorted. We keep track of the maximum number of removal queries that can be processed.

Algorithm:

Calculate the prefix sums of the removal queries.
Sort the prefix sums array.
Iterate over the removal queries in reverse order.
For each query, remove the corresponding element from the prefix sums array.
Check if the resulting array is still sorted.
Update the maximum number of removal queries that can be processed.

Time complexity: O(n log n) where n is the number of removal queries. Space complexity: O(n) for storing the prefix sums array.

Solution Code

class Solution {
    public int maxRemovals(String s, int[] p, int[] q) {
        int n = s.length();
        int m = p.length;
        int[] prefixSums = new int[m + 1];
        
        for (int i = 0; i < m; i++) {
            prefixSums[i + 1] = prefixSums[i] + 1;
        }
        
        int left = 0, right = m;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (canRemove(s, p, q, prefixSums, mid)) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return right;
    }
    
    private boolean canRemove(String s, int[] p, int[] q, int[] prefixSums, int k) {
        int[] removed = new int[k];
        Arrays.sort(removed);
        
        for (int i = 0; i < k; i++) {
            prefixSums[removed[i] + 1]--;
        }
        
        int j = 0;
        for (int i = 0; i < s.length(); i++) {
            if (j < k && i == q[removed[j]]) {
                j++;
            } else if (s.charAt(i) == s.charAt(p[i])) {
                prefixSums[j + 1]++;
            }
        }
        
        for (int i = 1; i <= k; i++) {
            if (prefixSums[i] < prefixSums[i - 1]) {
                return false;
            }
        }
        
        return true;
    }
}

Try It Yourself

Loading code editor...

Frequently Asked Questions

How to solve LeetCode 3018 (Maximum Number of Removal Queries That Can Be Processed I)?

This page provides optimized solutions for LeetCode problem 3018 (Maximum Number of Removal Queries That Can Be Processed I) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 3018 (Maximum Number of Removal Queries That Can Be Processed I)?

The time complexity for LeetCode 3018 (Maximum Number of Removal Queries That Can Be Processed I) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 3018 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 3018 in Java, C++, or Python.

Back to LeetCode Solutions