Introduction

In this blog post, we will explore how to solve the LeetCode problem "1883" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer hoursBefore, the number of hours you have to travel to your meeting. To arrive at your meeting, you have to travel through n roads. The road lengths are given as an integer array dist of length n, where dist[i] describes the length of the ith road in kilometers. In addition, you are given an integer speed, which is the speed (in km/h) you will travel at. After you travel road i, you must rest and wait for the next integer hour before you can begin traveling on the next road. Note that you do not have to rest after traveling the last road because you are already at the meeting. For example, if traveling a road takes 1.4 hours, you must wait until the 2 hour mark before traveling the next road. If traveling a road takes exactly 2 hours, you do not need to wait. However, you are allowed to skip some rests to be able to arrive on time, meaning you do not need to wait for the next integer hour. Note that this means you may finish traveling future roads at different hour marks. For example, suppose traveling the first road takes 1.4 hours and traveling the second road takes 0.6 hours. Skipping the rest after the first road will mean you finish traveling the second road right at the 2 hour mark, letting you start traveling the third road immediately. Return the minimum number of skips required to arrive at the meeting on time, or -1 if it is impossible. Example 1: Input: dist = [1,3,2], speed = 4, hoursBefore = 2 Output: 1 Explanation: Without skipping any rests, you will arrive in (1/4 + 3/4) + (3/4 + 1/4) + (2/4) = 2.5 hours. You can skip the first rest to arrive in ((1/4 + 0) + (3/4 + 0)) + (2/4) = 1.5 hours. Note that the second rest is shortened because you finish traveling the second road at an integer hour due to skipping the first rest. Example 2: Input: dist = [7,3,5,5], speed = 2, hoursBefore = 10 Output: 2 Explanation: Without skipping any rests, you will arrive in (7/2 + 1/2) + (3/2 + 1/2) + (5/2 + 1/2) + (5/2) = 11.5 hours. You can skip the first and third rest to arrive in ((7/2 + 0) + (3/2 + 0)) + ((5/2 + 0) + (5/2)) = 10 hours. Example 3: Input: dist = [7,3,5,5], speed = 1, hoursBefore = 10 Output: -1 Explanation: It is impossible to arrive at the meeting on time even if you skip all the rests. Constraints: n == dist.length 1 <= n <= 1000 1 <= dist[i] <= 105 1 <= speed <= 106 1 <= hoursBefore <= 107

Explanation

• Binary Search for Skips: The core idea is to use binary search to find the minimum number of skips required. The search space is from 0 to n - 1 (inclusive) because you can skip at most n - 1 rests.

• Check Feasibility: For a given number of skips, we need to check if it's possible to arrive on time. We calculate the total time taken with that many skips and compare it with hoursBefore.

• Time Calculation: The time calculation involves iterating through the distances and adding the travel time for each road. When the number of skips is not exhausted, we can skip the rest. The last road has no rest.

• Runtime Complexity: O(n log n) where n is the number of roads. Binary search takes O(log n) and the check function for each skip takes O(n)

• Storage Complexity: O(1)

Code

    def minSkips(dist, speed, hoursBefore):
    n = len(dist)

    def check(skips):
        time = 0.0
        for i in range(n - 1):
            travel_time = dist[i] / speed
            time += travel_time
            if time == int(time):
                continue
            if skips > 0:
                skips -= 1
            else:
                time = int(time + 1)
        time += dist[n - 1] / speed
        return time <= hoursBefore

    low = 0
    high = n - 1
    ans = -1

    while low <= high:
        mid = (low + high) // 2
        if check(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    return ans