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 maximumk 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

Solution: Binary Search

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string} s
 * @param {string} p
 * @param {number[]} removable
 * @return {number}
 */
const maximumRemovals = function (s, p, removable) {
  if (removable.length === 0) return 0;
  const isSubsequence = (target, sub) => {
    let subIndex = 0;

    for (const element of target) {
      if (element === sub[subIndex]) subIndex += 1;
    }
    return subIndex === sub.length;
  };
  let left = 0;
  let right = removable.length - 1;

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    const string = [...s];

    for (let index = 0; index <= mid; index++) {
      string[removable[index]] = '';
    }

    isSubsequence(string.join(''), p) ? (left = mid + 1) : (right = mid - 1);
  }
  return left;
};