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_;
};
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