Question

1870. Minimum Speed to Arrive on Time

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Description

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.

Solutions

Solution 1

class Solution:
  def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
    def check(speed):
      res = 0
      for i, d in enumerate(dist):
        res += (d / speed) if i == len(dist) - 1 else math.ceil(d / speed)
      return res <= hour

    r = 10**7 + 1
    ans = bisect_left(range(1, r), True, key=check) + 1
    return -1 if ans == r else ans

class Solution {
  public int minSpeedOnTime(int[] dist, double hour) {
    int left = 1, right = (int) 1e7;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (check(dist, mid, hour)) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return check(dist, left, hour) ? left : -1;
  }

  private boolean check(int[] dist, int speed, double hour) {
    double res = 0;
    for (int i = 0; i < dist.length; ++i) {
      double cost = dist[i] * 1.0 / speed;
      res += (i == dist.length - 1 ? cost : Math.ceil(cost));
    }
    return res <= hour;
  }
}

class Solution {
public:
  int minSpeedOnTime(vector<int>& dist, double hour) {
    int left = 1, right = 1e7;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (check(dist, mid, hour)) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return check(dist, left, hour) ? left : -1;
  }

  bool check(vector<int>& dist, int speed, double hour) {
    double res = 0;
    for (int i = 0; i < dist.size(); ++i) {
      double cost = dist[i] * 1.0 / speed;
      res += (i == dist.size() - 1 ? cost : ceil(cost));
    }
    return res <= hour;
  }
};

func minSpeedOnTime(dist []int, hour float64) int {
  n := len(dist)
  const mx int = 1e7
  x := sort.Search(mx, func(s int) bool {
    s++
    var cost float64
    for _, v := range dist[:n-1] {
      cost += math.Ceil(float64(v) / float64(s))
    }
    cost += float64(dist[n-1]) / float64(s)
    return cost <= hour
  })
  if x == mx {
    return -1
  }
  return x + 1
}

impl Solution {
  pub fn min_speed_on_time(dist: Vec<i32>, hour: f64) -> i32 {
    let n = dist.len();

    let check = |speed| {
      let mut cur = 0.0;
      for (i, &d) in dist.iter().enumerate() {
        if i == n - 1 {
          cur += (d as f64) / (speed as f64);
        } else {
          cur += ((d as f64) / (speed as f64)).ceil();
        }
      }
      cur <= hour
    };

    let mut left = 1;
    let mut right = 1e7 as i32;
    while left < right {
      let mid = left + (right - left) / 2;
      if !check(mid) {
        left = mid + 1;
      } else {
        right = mid;
      }
    }

    if check(left) {
      return left;
    }
    -1
  }
}

/**
 * @param {number[]} dist
 * @param {number} hour
 * @return {number}
 */
var minSpeedOnTime = function (dist, hour) {
  if (dist.length > Math.ceil(hour)) return -1;
  let left = 1,
    right = 10 ** 7;
  while (left < right) {
    let mid = (left + right) >> 1;
    if (arriveOnTime(dist, mid, hour)) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }
  return left;
};

function arriveOnTime(dist, speed, hour) {
  let res = 0.0;
  let n = dist.length;
  for (let i = 0; i < n; i++) {
    let cost = parseFloat(dist[i]) / speed;
    if (i != n - 1) {
      cost = Math.ceil(cost);
    }
    res += cost;
  }
  return res <= hour;
}