LeetCode in Kotlin

1898. Maximum Number of Removable Characters

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:

Solution

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
    }
}