Formatted question description: https://leetcode.ca/all/1898.html

# 1898. Maximum Number of Removable Characters

Medium

## 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 <= 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

Use binary search. Initially, low = 0 and high = removable.length. Each time, let mid be the mean of low and high and check whether p is still a subsequence of s after removing mid characters from s according to removable. The maximum possible k can be found in this way.

class Solution {
public int maximumRemovals(String s, String p, int[] removable) {
int low = 0, high = removable.length;
while (low < high) {
int mid = (high - low + 1) / 2 + low;
if (isPossible(s, p, removable, mid))
low = mid;
else
high = mid - 1;
}
return low;
}

public boolean isPossible(String s, String p, int[] removable, int k) {
int[] removes = new int[k];
for (int i = 0; i < k; i++)
removes[i] = removable[i];
Arrays.sort(removes);
StringBuffer sb = new StringBuffer(s);
for (int i = k - 1; i >= 0; i--)
sb.deleteCharAt(removes[i]);
return isSubsequence(p, sb.toString());
}

public boolean isSubsequence(String s, String t) {
if (s.length() == 0)
return true;
if (s.length() > t.length())
return false;
int sLength = s.length(), tLength = t.length();
int sIndex = 0, tIndex = 0;
while (sIndex < sLength && tIndex < tLength) {
char sChar = s.charAt(sIndex), tChar = t.charAt(tIndex);
if (sChar == tChar)
sIndex++;
tIndex++;
}
return sIndex == sLength;
}
}