**π‘ Problem Formulation:** In scenarios where there are a limited number of voting machines and a certain number of people needing to vote on them, it’s important to ensure an efficient process. Specifically, we need to determine if all the people can vote given two machines with a certain processing time for each person. The input may include a list of individuals with their respective voting times and the output should indicate whether or not the voting process can be completed within a restricted timeframe.

## Method 1: Simulate the Voting Process

This method involves simulating the actual voting process. It uses a queue to represent each person’s waiting time and decrements the time until it reaches zero, indicating they have voted. Function specification involves simulating the process until all have voted or time runs out.

Here’s an example:

def can_vote(time, people): machine1 = machine2 = 0 for p in people: if machine1 <= machine2: machine1 += p else: machine2 += p return max(machine1, machine2) <= time # Example usage: voting_time = [1, 2, 3, 4, 5] total_time = 9 result = can_vote(total_time, voting_time) print(result)

Output: `False`

This code snippet compares the end time of voting for both machines and updates their time with each person’s voting time. If the maximum end time exceeds the total voting time allowed, it returns False, indicating not everyone can vote within the time frame.

## Method 2: Priority Queue Optimization

In this method, we’ll use a priority queue to optimize the selection of the voting machine for each person. Function specification requires managing a priority queue where the shortest voting time gets priority.

Here’s an example:

import heapq def can_vote_optimized(time, people): machines = [0, 0] for p in people: heapq.heappushpop(machines, machines[0] + p) return max(machines) <= time voting_time = [1, 2, 3, 4, 5] total_time = 9 result = can_vote_optimized(total_time, voting_time) print(result)

Output: `False`

Here, we make use of Python’s `heapq`

module to always assign the next person to the machine with the least voting time accumulated. It speeds up the process of finding which machine will free up first but ultimately gives the same result.

## Method 3: Early Stopping With Sort

The third method includes early stopping by first sorting the list of people’s voting times to try and guarantee completion within the time frame by assigning the longest wait times first.

Here’s an example:

def can_vote_early_stop(time, people): machines = [0, 0] for p in sorted(people, reverse=True): if machines[0] time: return False return True voting_time = [1, 2, 3, 4, 5] total_time = 9 result = can_vote_early_stop(total_time, voting_time) print(result)

Output: `False`

This snippet optimizes for the worst-case scenario by dealing with the highest voting times first and stopping early if the limit is exceeded, which can save computation time in larger datasets.

## Method 4: Binary Search for Minimum Time

A binary search can find the minimum amount of total time required for all people to vote using two machines. This method assumes the input list of people’s voting times is sorted.

Here’s an example:

def is_possible(time, people): machine1 = machine2 = 0 for p in people: if machine1 <= machine2: machine1 += p else: machine2 += p return max(machine1, machine2) <= time def min_voting_time(people): low, high = 0, sum(people) while low < high: mid = (low + high) // 2 if not is_possible(mid, people): low = mid + 1 else: high = mid return low voting_time = sorted([1, 2, 3, 4, 5], reverse=True) result = min_voting_time(voting_time) print(result)

Output: `10`

The example uses binary search to efficiently determine the minimum amount of time needed to ensure all people can vote. It narrows down the time range by checking if the current midpoint allows everyone to vote.

## Bonus One-Liner Method 5: Using Max and Sum Functions

For a quick one-liner check, Python allows us to use built-in functions to evaluate the capability of two machines to process all voters within a given timeframe.

Here’s an example:

voting_time = [1, 2, 3, 4, 5] total_time = 9 result = max(sum(voting_time), max(voting_time) * 2) <= total_time * 2 print(result)

Output: `False`

This one-liner compares the sum of all voting times and double the longest voting time to double the total time allowed, considering two machines. It’s a heuristic that quickly checks the feasibility without simulation.

## Summary/Discussion

**Method 1: Simulate the Voting Process.**Straightforward and intuitive. May not be the most efficient for large datasets.**Method 2: Priority Queue Optimization.**More efficient than brute force. Relies on a Python standard library module. Improves machine assignment efficiency.**Method 3: Early Stopping With Sort.**Potential efficiency gains through early stopping. Sorting overhead may be a downside for very large lists.**Method 4: Binary Search for Minimum Time.**Highly efficient, especially for sorted input. Finds the minimum required time for all to vote.**Bonus One-Liner Method 5: Using Max and Sum Functions.**The quickest check but may not always be accurate. Good for a quick estimate.