Leetcode-1898

Maximum Number of Removable Characters

Time complexity: $O(n\log K)$

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#include <iostream>
#include <cstring>
#include <vector>

class Solution {
public:
    int maximumRemovals(std::string s, std::string p, std::vector<int>& removable) {
        int n = s.length(), m = removable.size();
        mask_.resize(n, 0x3fff'ffff);
        for (int i = 0; i < m; ++i) {
            mask_[removable[i]] = i;
        }
        
        int left = 0, right = std::min(m, n - (int)p.length());
        for (int mid = (left + right) / 2; left <= right;) {
            if (check(s, p, mid)) {
                left = mid + 1;
            }
            else {
                right = mid - 1;
            }
            mid = (left + right) / 2;
        }
        
        return right;
    }
    
private:
    bool check(std::string& s, std::string& p, int k) {
        int n = s.length();
        int m = p.length();
        int a = 0;
        
        for (int i = 0; i < n; ++i) {
            if (mask_[i] >= k && s[i] == p[a]) {
                ++a;
                if (a == m) {
                    return true;
                }
            }
        }
        return false;
    }
    
    std::vector<int> mask_;
};

Leetcode-1898