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 firstk 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 maximumk 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
Solution 1: Binary Search
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 $\textit{removable}$.