[LeetCode] 1898. Maximum Number of Removable Characters

[frdmu]

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.

二分法，寻找第一个不满足条件的，还考到了判断子序列。代码如下：

class Solution {
public:
    bool is_subseq(string s, string p, vector<int> removable, int mid) {
        vector<char> tmp;
        string str; 
        
        for (auto c: s) {
            tmp.push_back(c);
        }     
        for (int i = 0; i <= mid; i++) {
            tmp[removable[i]] = '0'; 
        }
        for (auto e: tmp) {
            if (e != '0') {
                str += e;
            }
        }
        
        int m = str.size();
        int n = p.size();
        int i = 0, j = 0; 
        while (i < m && j < n) {
            if (str[i] == p[j]) j++;
            i++; 
        }
        
        return j == n; 
    }
    int maximumRemovals(string s, string p, vector<int>& removable) {
        int n = removable.size(); 
        int left = 0, right = n; 
        
        while (left < right) {
            int mid = left +((right - left) >> 1);
            if (is_subseq(s, p, removable, mid)) left = mid + 1;
            else right = mid; 
        }
        
        return right;
    }
};