Introduction

In this blog post, we will explore how to solve the LeetCode problem "2141" 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 have n computers. You are given the integer n and a 0-indexed integer array batteries where the ith battery can run a computer for batteries[i] minutes. You are interested in running all n computers simultaneously using the given batteries. Initially, you can insert at most one battery into each computer. After that and at any integer time moment, you can remove a battery from a computer and insert another battery any number of times. The inserted battery can be a totally new battery or a battery from another computer. You may assume that the removing and inserting processes take no time. Note that the batteries cannot be recharged. Return the maximum number of minutes you can run all the n computers simultaneously. Example 1: Input: n = 2, batteries = [3,3,3] Output: 4 Explanation: Initially, insert battery 0 into the first computer and battery 1 into the second computer. After two minutes, remove battery 1 from the second computer and insert battery 2 instead. Note that battery 1 can still run for one minute. At the end of the third minute, battery 0 is drained, and you need to remove it from the first computer and insert battery 1 instead. By the end of the fourth minute, battery 1 is also drained, and the first computer is no longer running. We can run the two computers simultaneously for at most 4 minutes, so we return 4. Example 2: Input: n = 2, batteries = [1,1,1,1] Output: 2 Explanation: Initially, insert battery 0 into the first computer and battery 2 into the second computer. After one minute, battery 0 and battery 2 are drained so you need to remove them and insert battery 1 into the first computer and battery 3 into the second computer. After another minute, battery 1 and battery 3 are also drained so the first and second computers are no longer running. We can run the two computers simultaneously for at most 2 minutes, so we return 2. Constraints: 1 <= n <= batteries.length <= 105 1 <= batteries[i] <= 109

Explanation

Here's the solution to the problem, focusing on efficiency:

• Binary Search: The core idea is to use binary search to find the maximum possible time. We search the range from 0 to the sum of all battery times.
• Greedy Check: For each potential time mid in the binary search, we check if it's feasible to run all computers for mid minutes. We iterate through the batteries, and for each battery, if it's sufficient to power a computer for mid minutes, we simply move on. If not, we add the battery's power to a "spare" power pool. If the total spare power is enough to power the remaining computers for mid minutes, it's a feasible time.

• Optimization: The critical optimization is to avoid actually simulating the battery swapping. Instead, we leverage the aggregate spare power calculation, significantly improving performance.

• Runtime & Storage Complexity: The runtime complexity is O(N log(Sum)), where N is the number of batteries and Sum is the sum of all battery times. The storage complexity is O(1).

Code

    def maxRunTime(n: int, batteries: list[int]) -> int:
    """
    Finds the maximum number of minutes all n computers can run simultaneously.

    Args:
        n: The number of computers.
        batteries: A list of integers representing the battery runtimes.

    Returns:
        The maximum number of minutes.
    """

    total_power = sum(batteries)

    def is_possible(minutes):
        spare_power = 0
        for battery in batteries:
            if battery >= minutes:
                spare_power += battery - minutes

        return spare_power >= (n * minutes - total_power + spare_power) if (n * minutes - total_power + spare_power) >= 0 else True

    low = 0
    high = total_power // n
    ans = 0

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

    return ans