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1898. Maximum Number of Removable Characters

Description

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 <= 105
  • 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.

Solutions

We notice that if removing the characters at the first $k$ indices in $\textit{removable}$ still makes $p$ a subsequence of $s$, then removing the characters at $k \lt k' \leq \textit{removable.length}$ indices will also satisfy the condition. This monotonicity allows us to use binary search to find the maximum $k$.

We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = \textit{removable.length}$. Then we perform binary search. In each search, we take the middle value $mid = \left\lfloor \frac{l + r + 1}{2} \right\rfloor$ and check if removing the characters at the first $mid$ indices in $\textit{removable}$ still makes $p$ a subsequence of $s$. If it does, we update the left boundary $l = mid$; otherwise, we update the right boundary $r = mid - 1$.

After the binary search ends, we return the left boundary $l$.

The time complexity is $O(k \times \log k)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$, and $k$ is the length of $\textit{removable}$.

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class Solution:
    def maximumRemovals(self, s: str, p: str, removable: List[int]) -> int:
        def check(k: int) -> bool:
            rem = [False] * len(s)
            for i in removable[:k]:
                rem[i] = True
            i = j = 0
            while i < len(s) and j < len(p):
                if not rem[i] and p[j] == s[i]:
                    j += 1
                i += 1
            return j == len(p)

        l, r = 0, len(removable)
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

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class Solution {
    private char[] s;
    private char[] p;
    private int[] removable;

    public int maximumRemovals(String s, String p, int[] removable) {
        int l = 0, r = removable.length;
        this.s = s.toCharArray();
        this.p = p.toCharArray();
        this.removable = removable;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int k) {
        boolean[] rem = new boolean[s.length];
        for (int i = 0; i < k; ++i) {
            rem[removable[i]] = true;
        }
        int i = 0, j = 0;
        while (i < s.length && j < p.length) {
            if (!rem[i] && p[j] == s[i]) {
                ++j;
            }
            ++i;
        }
        return j == p.length;
    }
}

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class Solution {
public:
    int maximumRemovals(string s, string p, vector<int>& removable) {
        int m = s.size(), n = p.size();
        int l = 0, r = removable.size();
        bool rem[m];

        auto check = [&](int k) {
            memset(rem, false, sizeof(rem));
            for (int i = 0; i < k; i++) {
                rem[removable[i]] = true;
            }
            int i =