**π‘ Problem Formulation:** Imagine a swimming competition with final match results stored in a list or array, where each element represents a swimmer’s time. In a typical swimming match, the swimmer with the lowest timing wins. This article demonstrates different methods (in Python) to determine the number of swimmers who finished at a winning timeβclarifying that there could be a tie for first place. For instance, with input `[59.80, 59.80, 60.00]`

, the desired output should be `2`

as the top two swimmers have tied for the win.

## Method 1: Using Basic Iteration

This method involves looping through all the swimmer times and comparing each time to find the smallest one. Once the winning time is found, iterate again to count how many swimmers have this winning time. This method is straightforward and does not require importing additional modules.

Here’s an example:

swimmer_times = [59.80, 59.80, 60.00] min_time = min(swimmer_times) winners = sum(time == min_time for time in swimmer_times) print(winners)

Output: `2`

In the code snippet above, we first find the minimum time using `min(swimmer_times)`

, then use list comprehension to count how many times match this minimum time with `sum(time == min_time for time in swimmer_times)`

. The final output gives us the number of swimmers who won the match.

## Method 2: Using the Counter Class from collections Module

The Counter class from Python’s collections module can count the frequency of each time in the swimmer times list. We can then select the most common time(s) and determine if there are multiple winners.

Here’s an example:

from collections import Counter swimmer_times = [59.80, 59.80, 60.00] time_counts = Counter(swimmer_times) winners = time_counts[min(swimmer_times)] print(winners)

Output: `2`

By using the Counter class, we create a dictionary that maps swimmer times to their frequency. We then directly access the count of the lowest time to get the number of winners, which is efficient and straightforward.

## Method 3: Sorting the List

This technique sorts the times list and then counts how many swimmers have the first time in the sorted list, under the assumption that the lowest time wins.

Here’s an example:

swimmer_times = [59.80, 59.80, 60.00] swimmer_times.sort() winning_time = swimmer_times[0] winners = swimmer_times.count(winning_time) print(winners)

Output: `2`

The snippet uses Python’s list sorting method `sort()`

, which rearranges the swimmer times in ascending order. The number of swimmers with the winning time is then counted with `swimmer_times.count(winning_time)`

.

## Method 4: Using numpy’s unique and argsort Functions

For datasets with a high volume of swimmer times, numpy can provide more efficient array operations. Using numpyβs `unique()`

and `argsort()`

functions, we can determine the number of swimmers sharing the lowest time.

Here’s an example:

import numpy as np swimmer_times = np.array([59.80, 59.80, 60.00]) sorted_indices = np.argsort(swimmer_times) winners = np.sum(swimmer_times == swimmer_times[sorted_indices[0]]) print(winners)

Output: `2`

With numpy, `argsort()`

gives us indices that would sort an array, and we use these to find winners. We use boolean indexing to count the frequencies of the winning time, resulting in an efficient and concise approach for large datasets.

## Bonus One-Liner Method 5: Using List Comprehension and min Function

A brief but potent one-liner can combine the min function with a list comprehension to provide the count of swimmers who won the match.

Here’s an example:

swimmer_times = [59.80, 59.80, 60.00] winners = len([time for time in swimmer_times if time == min(swimmer_times)]) print(winners)

Output: `2`

This one-liner creates a new list that contains only the occurrences of the winning time and then takes the length of that list to determine the number of winners, which is a quick and elegant solution.

## Summary/Discussion

**Method 1:**Basic Iteration. Simple and clear. Potentially inefficient for large datasets due to double iteration.**Method 2:**Using Counter Class. More Pythonic and efficient, especially for larger data sets with many repeating values.**Method 3:**Sorting the List. Easy to implement but requires sorting the entire list, which might not be optimal for large datasets.**Method 4:**Using numpy’s Functions. The most efficient for large numerical datasets with faster array operations.**Method 5:**Bonus One-Liner. Extremely concise but computes the minimum multiple times, which might be less efficient.