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Leetcode #1870: Minimum Speed to Arrive on Time

In this guide, we solve Leetcode #1870 Minimum Speed to Arrive on Time in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Array, Binary Search

Intuition

The problem structure suggests a monotonic decision, which makes binary search a natural fit.

By halving the search space each step, we reach the answer efficiently.

Approach

Search either directly on a sorted array or on the answer space using a check function.

Each check is fast, and the logarithmic search keeps the overall runtime low.

Steps:

  • Define the search bounds.
  • Check the mid point condition.
  • Narrow the bounds until convergence.

Example

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.

Python Solution

class Solution: def minSpeedOnTime(self, dist: List[int], hour: float) -> int: def check(v: int) -> bool: s = 0 for i, d in enumerate(dist): t = d / v s += t if i == len(dist) - 1 else ceil(t) return s <= hour if len(dist) > ceil(hour): return -1 r = 10**7 + 1 ans = bisect_left(range(1, r), True, key=check) + 1 return -1 if ans == r else ans

Complexity

The time complexity is O(n×log⁡M)O(n \times \log M)O(n×logM), where nnn and MMM are the number of train trips and the upper bound of the speed, respectively. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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