Introduction

In this blog post, we will explore how to solve the LeetCode problem "2187" 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 array time where time[i] denotes the time taken by the ith bus to complete one trip. Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus. You are also given an integer totalTrips, which denotes the number of trips all buses should make in total. Return the minimum time required for all buses to complete at least totalTrips trips. Example 1: Input: time = [1,2,3], totalTrips = 5 Output: 3 Explanation: - At time t = 1, the number of trips completed by each bus are [1,0,0]. The total number of trips completed is 1 + 0 + 0 = 1. - At time t = 2, the number of trips completed by each bus are [2,1,0]. The total number of trips completed is 2 + 1 + 0 = 3. - At time t = 3, the number of trips completed by each bus are [3,1,1]. The total number of trips completed is 3 + 1 + 1 = 5. So the minimum time needed for all buses to complete at least 5 trips is 3. Example 2: Input: time = [2], totalTrips = 1 Output: 2 Explanation: There is only one bus, and it will complete its first trip at t = 2. So the minimum time needed to complete 1 trip is 2. Constraints: 1 <= time.length <= 105 1 <= time[i], totalTrips <= 107

Explanation

• Binary Search: Use binary search to find the minimum time required. The search space is between the minimum possible time (minimum element in time array) and the maximum possible time (maximum element in time array multiplied by totalTrips).
  • Check Feasibility: For a given time t (midpoint of the binary search), calculate how many trips each bus can complete within that time. Sum up the trips completed by all buses and check if the total number of trips is greater than or equal to totalTrips.
  • Adjust Search Space: Based on whether the current time t is feasible (i.e., enough trips are completed), adjust the binary search space to either search for a smaller time or a larger time.

• Runtime Complexity: O(n log(m)), where n is the length of the time array and m is the search space (max(time) totalTrips). *Storage Complexity: O(1).

Code

    def minimumTime(time, totalTrips):
    """
    Finds the minimum time required for all buses to complete at least totalTrips trips.

    Args:
        time (list of int): Time taken by each bus to complete one trip.
        totalTrips (int): Total number of trips required.

    Returns:
        int: Minimum time required.
    """

    def is_possible(t):
        trips = 0
        for bus_time in time:
            trips += t // bus_time
        return trips >= totalTrips

    left = 1
    right = min(time) * totalTrips
    ans = right

    while left <= right:
        mid = (left + right) // 2
        if is_possible(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans