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1870. Minimum Speed to Arrive on Time 👍

  • Time: $O(n\log(r - l))$, where r = 10^7 and l = 1
  • Space: $O(1)$
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class Solution {
 public:
  int minSpeedOnTime(vector<int>& dist, double hour) {
    int ans = -1;
    int l = 1;
    int r = 1e7;

    while (l <= r) {
      const int m = (l + r) / 2;
      if (time(dist, hour, m) > hour) {
        l = m + 1;
      } else {
        ans = m;
        r = m - 1;
      }
    }

    return ans;
  }

 private:
  double time(const vector<int>& dist, double hour, int speed) {
    double sum = 0;
    for (int i = 0; i < dist.size() - 1; ++i)
      sum += ceil((double)dist[i] / speed);
    return sum + (double)dist.back() / speed;
  }
};

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class Solution {
  public int minSpeedOnTime(int[] dist, double hour) {
    int ans = -1;
    int l = 1;
    int r = (int) 1e7;

    while (l <= r) {
      final int m = (l + r) / 2;
      if (time(dist, hour, m) > hour) {
        l = m + 1;
      } else {
        ans = m;
        r = m - 1;
      }
    }

    return ans;
  }

  private double time(int[] dist, double hour, int speed) {
    double sum = 0;
    for (int i = 0; i < dist.length - 1; ++i)
      sum += Math.ceil((double) dist[i] / speed);
    return sum + (double) dist[dist.length - 1] / speed;
  }
}

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class Solution:
  def minSpeedOnTime(self, dist: list[int], hour: float) -> int:
    ans = -1
    l = 1
    r = int(1e7)

    def time(speed: int) -> float:
      summ = 0
      for i in range(len(dist) - 1):
        summ += math.ceil(dist[i] / speed)
      return summ + dist[-1] / speed

    while l <= r:
      m = (l + r) // 2
      if time(m) > hour:
        l = m + 1
      else:
        ans = m
        r = m - 1

    return ans