Maximum Running Time of N Computers

Problem Statement

In this problem, you have a certain number of computers n and a list of batteries each with a distinct capacity (in minutes). Each battery's capacity is represented by its respective value in the array batteries. Your main goal is to determine how long you can run all n computers simultaneously.

You must begin by assigning one battery to each computer. However, as the simulation progresses, you can interchange these batteries amongst computers or replace a used battery with another. This replacement process is instantaneous and can be performed repeatedly without any time constraint. It's essential to note that batteries are non-rechargeable; once their capacity is exhausted, they cannot be reused.

Your task is to calculate the longest time all computers can run concurrently without any of them shutting down due to a lack of battery power.

Examples

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

Approach and Intuition

From the problem statement and examples provided, we can infer a strategy:

1. Initial setup: At the beginning, distribute the batteries strategically among the computers. An initial interpretation might suggest putting higher battery capacities on computers that have higher needs or splitting them as evenly as possible.

2. Simulation of operation:

   • Start all computers at the same time with their assigned batteries.
   • As any battery runs out, replace it with another from the pool if available.

3. Optimization using max operation time: The ideal approach involves finding a balance such that each computer can operate for as long as possible. This often means swapping out batteries between computers to ensure that no computer is idle while there are still batteries available with remaining capacity.

4. Simulation continues until all batteries are depleted: The operation goes on until it's no longer possible to run all computers simultaneously.

The examples provide clear scenarios where strategic distribution and timely swapping of batteries ensure maximum operational time. These scenarios involve:

• Calculating ongoing usage to balance out battery lifetimes across computers.
• Monitoring battery usage in real-time to swiftly replace depleted units.

Therefore, this method leverages the practice of simultaneously balancing battery usage and keeping all computers operational, ensuring a maximized operational period given the constraints and resources available.

Solutions

class Solution {
    public long maximumRuntime(int count, int[] cells) {
        long totalPower = 0;
        for (int energy : cells)
            totalPower += energy;
        long low = 1, high = totalPower / count;
            
        while (low < high) {
            long mid = high - (high - low) / 2;
            long surplus = 0;
                
            for (int energy : cells)
                surplus += Math.min(energy, mid);
    
            if (surplus >= (long)(count * mid))
                low = mid;
            else
                high = mid - 1;
        }
        return low;
    }
}