# 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

Binary search.

Template 1:

boolean check(int x) {
}

int search(int left, int right) {
while (left < right) {
int mid = (left + right) >> 1;
if (check(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}


Template 2:

boolean check(int x) {
}

int search(int left, int right) {
while (left < right) {
int mid = (left + right + 1) >> 1;
if (check(mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}

• class Solution {
public:
int maximumRemovals(string s, string p, vector<int>& removable) {
int left = 0, right = removable.size();
while (left < right) {
int mid = left + right + 1 >> 1;
if (check(s, p, removable, mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}

bool check(string s, string p, vector<int>& removable, int mid) {
int m = s.size(), n = p.size(), i = 0, j = 0;
unordered_set<int> ids;
for (int k = 0; k < mid; ++k) {
ids.insert(removable[k]);
}
while (i < m && j < n) {
if (ids.count(i) == 0 && s[i] == p[j]) {
++j;
}
++i;
}
return j == n;
}
};

• class Solution:
def maximumRemovals(self, s: str, p: str, removable: List[int]) -> int:
def check(k):
i = j = 0
ids = set(removable[:k])
while i < m and j < n:
if i not in ids and s[i] == p[j]:
j += 1
i += 1
return j == n

m, n = len(s), len(p)
left, right = 0, len(removable)
while left < right:
mid = (left + right + 1) >> 1
if check(mid):
left = mid
else:
right = mid - 1
return left


• func maximumRemovals(s string, p string, removable []int) int {
check := func(k int) bool {
ids := make(map[int]bool)
for _, r := range removable[:k] {
ids[r] = true
}
var i, j int
for i < len(s) && j < len(p) {
if !ids[i] && s[i] == p[j] {
j++
}
i++
}
return j == len(p)
}

left, right := 0, len(removable)
for left < right {
mid := (left + right + 1) >> 1
if check(mid) {
left = mid
} else {
right = mid - 1
}
}
return left
}

• function maximumRemovals(s: string, p: string, removable: number[]): number {
let left = 0,
right = removable.length;
while (left < right) {
let mid = (left + right + 1) >> 1;
if (isSub(s, p, new Set(removable.slice(0, mid)))) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}

function isSub(str: string, sub: string, idxes: Set<number>): boolean {
let m = str.length,
n = sub.length;
let i = 0,
j = 0;
while (i < m && j < n) {
if (!idxes.has(i) && str.charAt(i) == sub.charAt(j)) {
++j;
}
++i;
}
return j == n;
}


• use std::collections::HashSet;

impl Solution {
pub fn maximum_removals(s: String, p: String, removable: Vec<i32>) -> i32 {
let m = s.len();
let n = p.len();
let s = s.as_bytes();
let p = p.as_bytes();

let check = |k| {
let mut i = 0;
let mut j = 0;
let ids: HashSet<i32> = removable[..k].iter().cloned().collect();
while i < m && j < n {
if !ids.contains(&(i as i32)) && s[i] == p[j] {
j += 1;
}
i += 1;
}
j == n
};

let mut left = 0;
let mut right = removable.len();
while left + 1 < right {
let mid = left + (right - left) / 2;
if check(mid) {
left = mid;
} else {
right = mid;
}
}

if check(right) {
return right as i32;
}
left as i32
}
}