2023-07-26 Daily Challenge

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Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 26

Description

Minimum Speed to Arrive on Time

You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the ith train ride.

Each train can only depart at an integer hour, so you may need to wait in between each train ride.

  • For example, if the 1st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd train ride at the 2 hour mark.

Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.

Tests are generated such that the answer will not exceed 107 and hour will have at most two digits after the decimal point.

 

Example 1:

Input: dist = [1,3,2], hour = 6
Output: 1
Explanation: At speed 1:
- The first train ride takes 1/1 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours.
- Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours.
- You will arrive at exactly the 6 hour mark.

Example 2:

Input: dist = [1,3,2], hour = 2.7
Output: 3
Explanation: At speed 3:
- The first train ride takes 1/3 = 0.33333 hours.
- Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours.
- You will arrive at the 2.66667 hour mark.

Example 3:

Input: dist = [1,3,2], hour = 1.9
Output: -1
Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.

 

Constraints:

  • n == dist.length
  • 1 <= n <= 105
  • 1 <= dist[i] <= 105
  • 1 <= hour <= 109
  • There will be at most two digits after the decimal point in hour.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  const double EPS = 1e-10;
  bool satisfy(const vector<int>& dist, double hour, double speed) {
    double result = 0;
    for(int i = 0; i + 1 < dist.size(); ++i) {
      result += max<double>(1, ceil(dist[i] / speed));
    }
    result += dist.back() / speed;
    return result <= hour;
  }
public:
  int minSpeedOnTime(vector<int>& dist, double hour) {
    if(dist.size() > hour + 1) return -1;
    int high = 100000000;
    int low = 0;
    while(high > low) {
      double mid = (high + low) / 2;
      if(satisfy(dist, hour, mid)) {
        high = mid;
      } else {
        low = mid + 1;
      }
    }
    return low > 1e7 ? -1 : low;
  }
};

// Accepted
// 65/65 cases passed (275 ms)
// Your runtime beats 92.81 % of cpp submissions
// Your memory usage beats 100 % of cpp submissions (101.1 MB)
Algorithm