LeetCode 1898: Maximum Number of Removable Characters Solution

Master LeetCode problem 1898 (Maximum Number of Removable Characters), a medium 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.

1898. Maximum Number of Removable Characters

Medium

Array Two Pointers String Binary Search

Problem Explanation

Explanation:

To solve this problem, we can use a binary search approach to find the maximum k such that p is still a subsequence of s after removing k characters. We can create a helper function that checks if p is a subsequence of s after removing k characters at the given indices. Then, we perform binary search on the range [0, removable.length] to find the maximum valid k.

Algorithmic Idea:

Define a helper function isSubsequence that checks if p is a subsequence of s after removing k characters at given indices.
Perform binary search on the range [0, removable.length] to find the maximum valid k.
In each iteration of binary search, check if p is still a subsequence after removing mid characters at given indices.
Update the search range based on the result of the helper function.

Time Complexity:

The time complexity of this approach is O(n * log(n)), where n is the length of removable.

Space Complexity:

The space complexity is O(n) for storing the indices of removable.

Solution Code

class Solution {
    public int maximumRemovals(String s, String p, int[] removable) {
        int low = 0, high = removable.length;
        
        while(low <= high) {
            int mid = low + (high - low) / 2;
            if(isSubsequence(s, p, removable, mid)) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        
        return high;
    }
    
    private boolean isSubsequence(String s, String p, int[] removable, int k) {
        StringBuilder sb = new StringBuilder(s);
        for (int i = 0; i < k; i++) {
            sb.setCharAt(removable[i], '*');
        }
        int j = 0;
        for (int i = 0; i < sb.length() && j < p.length(); i++) {
            if (sb.charAt(i) == p.charAt(j)) {
                j++;
            }
        }
        return j == p.length();
    }
}

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