Minimum Time to Activate String
Time complexity: $O(n\log M)$
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#include <iostream>
#include <vector>
#include <cstring>
class Solution {
public:
int minTime(std::string s, std::vector<int>& order, int k) {
if (!k) {
return 0;
}
int n = s.length();
long long max_k = ((long long)n * (n + 1)) / 2;
if (max_k < k) {
return -1;
}
else if (max_k == k) {
return n - 1;
}
str_.resize(n, 0);
int left = 1, right = n;
for (int mid = (left + right) / 2; right > left;) {
if (check(order, k, mid)) {
right = mid;
}
else {
left = mid + 1;
}
mid = (left + right) / 2;
}
return left - 1;
}
private:
bool check(std::vector<int>& order, int k, int t) {
int n = str_.size();
for (int i = 0; i < t; ++i) {
str_[order[i]] = t;
}
int cnt = 0, last = -1;
for (int i = 0; i < n; ++i) {
if (str_[i] == t) {
last = i;
}
cnt += last + 1;
if (cnt >= k) {
return true;
}
}
return false;
}
std::vector<int> str_;
};
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