Leetcode-3639

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_;
};

Leetcode-3639