1898. Maximum Number of Removable Characters

Problem

You are given two strings s and p where p is a **subsequence **of s. You are also given a **distinct 0-indexed **integer array removable containing a subset of indices of s (s is also 0-indexed).

You want to choose an integer k (0 <= k <= removable.length) such that, after removing k characters from s using the first k indices in removable, p is still a subsequence of s. More formally, you will mark the character at s[removable[i]] for each 0 <= i < k, then remove all marked characters and check if p is still a subsequence.

Return **the *maximum* k you can choose such that p is still a subsequence of s after the removals**.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

  Example 1:

Input: s = "abcacb", p = "ab", removable = [3,1,0]
Output: 2
Explanation: After removing the characters at indices 3 and 1, "abcacb" becomes "accb".
"ab" is a subsequence of "accb".
If we remove the characters at indices 3, 1, and 0, "abcacb" becomes "ccb", and "ab" is no longer a subsequence.
Hence, the maximum k is 2.

Example 2:

Input: s = "abcbddddd", p = "abcd", removable = [3,2,1,4,5,6]
Output: 1
Explanation: After removing the character at index 3, "abcbddddd" becomes "abcddddd".
"abcd" is a subsequence of "abcddddd".

Example 3:

Input: s = "abcab", p = "abc", removable = [0,1,2,3,4]
Output: 0
Explanation: If you remove the first index in the array removable, "abc" is no longer a subsequence.

  Constraints:

• 1 <= p.length <= s.length <= 10^5

• 0 <= removable.length < s.length

• 0 <= removable[i] < s.length

• p is a subsequence of s.

• s and p both consist of lowercase English letters.

• The elements in removable are distinct.

Solution (Java)

class Solution {
    public int maximumRemovals(String s, String p, int[] removable) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        // binary search for the k which need to be removed
        char[] convertedS = s.toCharArray();
        int left = 0;
        int right = removable.length - 1;
        while (left <= right) {
            int middle = (left + right) / 2;
            // remove letters from 0 to mid by changing it into some other non letters
            for (int i = 0; i <= middle; i++) {
                convertedS[removable[i]] = '?';
            }
            // if it is still subsequence change left boundary
            // else replace all removed ones and change right boundary
            if (isSubsequence(convertedS, p)) {
                left = middle + 1;
            } else {
                for (int i = 0; i <= middle; i++) {
                    convertedS[removable[i]] = s.charAt(removable[i]);
                }
                right = middle - 1;
            }
        }
        return left;
    }

    // simple check for subsequence
    private boolean isSubsequence(char[] convertedS, String p) {
        int p1 = 0;
        int p2 = 0;
        while (p1 < convertedS.length && p2 < p.length()) {
            if (convertedS[p1] != '?' && convertedS[p1] == p.charAt(p2)) {
                p2 += 1;
            }
            p1 += 1;
        }
        return p2 == p.length();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).