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

1898. Maximum Number of Removable Characters

Level

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;
    }
}

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