Medium
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.removable
are distinct.class Solution {
fun maximumRemovals(s: String, p: String, removable: IntArray): Int {
if (s.isEmpty()) {
return 0
}
// binary search for the k which need to be removed
val convertedS = s.toCharArray()
var left = 0
var right = removable.size - 1
while (left <= right) {
val middle = (left + right) / 2
// remove letters from 0 to mid by changing it into some other non letters
for (i in 0..middle) {
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 (i in 0..middle) {
convertedS[removable[i]] = s[removable[i]]
}
right = middle - 1
}
}
return left
}
// simple check for subsequence
private fun isSubsequence(convertedS: CharArray, p: String): Boolean {
var p1 = 0
var p2 = 0
while (p1 < convertedS.size && p2 < p.length) {
if (convertedS[p1] != '?' && convertedS[p1] == p[p2]) {
p2 += 1
}
p1 += 1
}
return p2 == p.length
}
}