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Open coding challenge: Bottle sets

This open coding challenge is designed to develop both programming and data analysis skills. Anyone can attempt their own solution, the objective is to pass with 100% accuracy all test cases. While efficienty in coding and analysis is a plus, the primary aim is accuracy in the solution.

Introduction

In the following sections, you will find the instructions to solve the challenge and the problem description, which outlines the rules, objective and constrains. This challenge format was inspired by the Turing platform, where similar problems are proposed, combining data analysis skills along with coding, and the user has to solve it within a strict time limit, just 45 minutes! Here, you can take it easy and solve it at your own pace, or challenge yourself by setting a timer. Either way... have fun!

Instructions

Files

The Github repository includes several files:

• 'Challenge_Description.txt': this file includes the problem description, task, examples and constrains. You will also find them further in this website.

• 'codes/challenge.py': this Python file is where you have to implement your solution. Only modify the section marked as '# Implement your code here'. The rest of the script runs your solution against the test cases.

• 'codes/test_cases.py': this Python script evaluates your solution using hidden test cases and returns your success rate (e.g. "20% of the test cases were successful"). You don't need to use this file.

• 'codes/FM_solution.py': this file contains my own solution. Later I'll explain how to run it if you want to test it and play around.

In order to take this challenge, download the 'codes' folder to your computer, and follow the instructions below.

Running your solution

Prerequisites (Python):

Linux system: Ensure Python is installed. For assistance, refer to how to install python on Linux.

Windows system: Install a Python interpreter. For assistance, refer to Python releases for Windows and also Using Python on Windows.

MacOS system: Ensure Python is installed. For assistance, refer to how to install Python on Mac 👈🏽.

General instructions:

Open a terminal (Linux), PowerShell (Windows) or IDLE Shell (Mac) in the same directory as the python files and run:

python challenge.py

This line will return your solution's success rate on the test cases.

Running my solution (optional)

If you want to run my own solution, open the 'codes/test_cases.py' file, comment the first line (put a hash # at the beggining) and uncomment the second line (remove the hash #). After the changes, the first line without any hash should be:

from FM_solution import optim_sets

You can then revert these changes to evaluate your solution again.

Problem description, task and constrains

Once you are familiar with the instructions, you can start the challenge. If you are setting a timer, start it as soon as you begin reading the following section.

^{Conceptual image, AI-generated using the prompt: "Two shelves with crystal bottles of many sizes and shapes, with an alchemist theme".}

Description

You are a seller of liquid products, which can be stored in bottles of any size. Recently, you received an unusual order: a client requests 'as many sets of bottles as possible, where each set consists of one or two bottles, and all sets must have the same total volume'. The volume, measured in litres, can be any integer value, e.g. 1L or 5L, what matters is that you maximize the number of sets, as the mysterious client will pay you a very high amount per set, regardless of the liquid volume.

Task

Design an algorithm that takes as input a list of bottle capacities (in litres) available in your store, and returns the maximum number of sets that can be made. Each set must have one or two bottles, and the total volume of each set must be the same across all sets.

Examples

Input	Output	Explanation
[1,1,2,2,5,5]	3	Three sets with 2L each: [1,1] + [2] + [2]
[1,1,3,5]	2	Two sets with 1L each: [1] + [1]
[1,2,5,6,10,11,20]	3	Three sets with 11L each: [1,10] + [5,6] + [11]
[12]	1	Single set with 12L: [12]
[1,1,5,5]	2	Either two sets of 1L: [1] + [1]; two sets of 5L: [5] + [5]; or two sets of 6L: [1,5] + [1,5]

Constrains:

• The input list contains integer values within the range 1 to 100.
• The input list includes 1 to 100 elements.

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My solution

Below, I explain my solution for this challenge, including the overall analysis and step-by-step breakdown of the code.

Overall stragegy

The "Bottle sets" challenge is based on a relatively simple mathematical problem, or at least simple to enunciate. There are many ways to build a solution, ranging from brute force, by trying all possibilities within the input list, to elegant mathematical formulae that provide an exact description. Although the later seems the best approach, there is a competition between the time invested in the solution (up to 45 minutes if you take that challenge) and its elegance. As a consequence, my solution combines some analysis and general rules with a bit of brute force. While not the most efficient algorithm, this balance produces a functioning and simple code.

Unique bottle configurations

After some analysis, I arrived to the (good) conclusion that for any given capacity value $C$, the bottles in the input list $L$ can be arranged in a unique configuration using sets having up to 2 bottles. Let me explain first the opposite case, and imagine that the rules are different and the sets can include up to 3 bottles:

Example for rule "Sets can have up to 3 bottles"

Input list: [1,1,2,2,3,4].
Capacity for all sets: $C=4$.
Configuration A (sets): [2,2] + [3,1] + [4].
Configuration B (sets): [1,1,2] + [4].

It's clear that Configuration A, with 3 sets, is better than Configuration B, with only 2 sets. The takeaway message here is that using up to 3 bottles per set, the configuration of bottles for any capacity $C$ may not be unique. Therefore, it's necessary to first find all possible configurations and then choose the best one.

Now consider the actual rule: sets can have up to 2 bottles. For a single bottle $L_i$ with capacity $C_i$, there are three possible cases when a total capacity value $C$ is fixed for all sets:

1. If $C_i=C$, the bottle $L_i$ makes a single-bottle set.
2. If $C_i>C$, the bottle $L_i$ is useless.
3. If $C_i<C$, the bottle $L_i$ needs another bottle $L_j$ with capacity $C_j = C - C_i$.

Since this applies to all bottles in the input list, single bottles with capacity $C_i=C$ will form simple sets, while the other bottles will be arranged in sets $(C_i,C_j)$ that add $C_i+C_j=C$. The conclusion is: for any capacity $C$, there is a unique configuration of bottles.

My algorithm makes use of this valuable information: it finds the unique configuration for each capacity $C$, counts the number of sets, and chooses the highest count as the final answer. Therefore, the main question becomes how to efficiently analyze all possible capacities...

The simplest algorithm

To set a baseline, meaning a basic code that just works, the simplest algorithm may analyze all capacity values ranging from the smallest capacity in the input list $C_{min} = \min(L)$ to the largest sum $C_{max} = \max(C_i+C_j)$ for any two bottles $(L_i,L_j)$, one by one.

Example for simple algorithm analyzing a full range of capacity values

Input list: [2,2,2,4,4,8,8,8].
Minimum capacity: $C_{min}=2$.
Maximum capacity: $C_{max}=8+8=16$.
Possible capacities: [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16].

As you can imagine from the long list of possible capacities, this algorithm is highly inefficient as it analyzes values that are not even feasible for the input list.

A refined algorithm

A better approach is to analyze only feasible capacity values, using both single- and two-bottle sets.

Example of feasible capacity values

Input list: [2,2,2,4,4,8,8,8].
Single-bottle set capacities: [2,4,8].
Two-bottle set capacities: [4,6,8,10,12,16].
Possible capacities: [2,4,6,8,10,12,16].

In this example, the list of possible capacities is complete and none of the values are expendable. Notice that the values $C=4$ and $C=8$ appear in both single- and two-bottle lists, as they can be reached in both ways.

Code: step by step

Now that I've explained my reasoning, I first introduce a pseudo-code that outlines the main steps of the algorithm, and then explain all the details of the full code.

Pseudo-code

The following coding block is the simplified version of my original code, in which the main steps are described and fictional functions are called, such as "count_bottles". Later, in the full description, this auxiliar functions will be replaced by a more complex structure.

def optim_set(stock):
  """
  From the input {stock} list, return the optimal number...
  of one- and two-bottle sets that can be made with a...
  common total capacity value.
  """ 
  # 1. Count bottles for each capacity value, store them in dictionary with [key = capacity], [value = number of bottles]
  groups = {C: count_bottles(stock,C) for C in stock}

  # 2. Make a first guess for optimal number of sets using only single-bottle sets:
  best_N = max(groups.values())

  # 3. Identify the capacities of all two-bottle sets:
  C_sum = [C_i+C_j for (C_i,C_j) in groups]

  # 4. For values in C_sum, count sets and improve {N_best} if possible:
  for C in C_sum:
    n_sets = count_sets(C,groups)
    best_N = max(best_N,n_sets)

  # Finally, return optimal number of sets:
  return best_N

Full description of the original code

Let's break down the full code, step by step. For clarity, I will display segments of the code sequentially, along with their explanations. You can view the original file codes/FM_solution.py in the Github repository.

def optim_sets(stock):
  # 1. Identify sub-groups with equal values:
  vals = list(set(stock)) # Unique values
  groups = {val: stock.count(val) for val in vals} # key=capacity, value=counts  
#...

Explanation:

The first line identifies all unique capacity values from the input list $\color{teal}{\text{stock}}$ by converting it into a set, then back to a list, $\color{teal}{\text{vals}}$, for future convenience.

The second line creates the dictionary, $\color{teal}{\text{groups}}$, where each key represents a unique capacity, and its value is the count of occurrences in the $\color{teal}{\text{stock}}$ list. I chose to use the built-in method count(), because I prefer to avoid importing libraries and build a more transparent code. Alternatively, I may use the Counter object from the collections library:

# Equivalent method (not in original code):
from collections import Counter
groups = Counter(stock)

Let's continue with the breakdown:

#...
  # 2. First guess with single-bottle sets:
  best_N = max(groups.values())
#...

Explanation:

This step just selects the highest count within the $\color{teal}{\text{groups}}$ dictionary, providing an initial guess for the optimal number of sets.

#...
  # 3. Identify the capacities of all two-bottle sets:
  sum_vals = [] # Initiate
  for i in range(len(vals)):
    for j in range(i,len(vals)):
      sum_vals.append(vals[i]+vals[j])
  sum_vals = list(set(sum_vals))
#...

Explanation:

The list $\color{teal}{\text{sum_vals}}$ collects all possible capacities formed by combining two bottles. The first line just initiates the list, followed by a double loop that iterates through each pair of elements ($i$,$j$) in $\color{teal}{\text{vals}}$ and adds their sum to $\color{teal}{\text{sum_vals}}$.

In the double loop, using $j$ starting from $i$ ensures that each pair is considered only once, avoiding equivalent results. On the other hand, $(i,i)$ pairs are included because the $\color{teal}{\text{stock}}$ list may include bottles with the same capacity.

Once the loop is finished, the list is converted into a set to avoid duplicated values, and back into a list for future convenience.

#...
  # 4. Count two-bottle sets and update first guess if needed:
  for sum_val in sum_vals:
    check_vals = []
    n_sum = groups[sum_val] if sum_val in groups else 0
    for val_i in groups:
      val_j = sum_val-val_i
      if val_i not in check_vals:
        check_vals += [val_i,val_j]
        if val_i == val_j:
          n_sum += int(groups[val_i]/2)
        elif val_i != val_j and val_j in groups:
          n_sum += min([groups[val_i],groups[val_j]])
    best_N = max([best_N,n_sum])

  return best_N

Explanation:

This final block updates the initial guess $\color{teal}{\text{best_N}}$ by considering both one- and two-bottle sets. The main loop processes each capacity value in $\color{teal}{\text{sum_vals}}$, by defining a counter, $\color{teal}{\text{n_sum}}$, that accumulates the number of sets for the current capacity, starting with the single-bottle counts if available.

Instead of directly iterating through the $\color{teal}{\text{stock}}$ list, the code uses the dictionary $\color{teal}{\text{groups}}$, which summarizes the count of bottles for each capacity. For a fixed capacity $C$ common to all sets, a single key $C_i$ from $\color{teal}{\text{groups}}$ has several options:

1. $C_i = C$, then all bottles with capacity $C_i$ will be used as single-bottle sets (initialization of $\color{teal}{\text{n_sum}}$).

2. $C_i > C$, then all bottles with capacity $C_i$ won't be used.

3. $C_i = C/2$, then each pair of bottles with capacity $C_i$ will be used as a two-bottle set. Their contribution to $\color{teal}{\text{n_sum}}$ is determined by halving their count and rounding down (for the odd case).

4. $C_i < C$ & $C_i \neq C/2$, then each bottle with capacity $C_i$ needs another bottle with the complement capacity $C_j = C - C_i$. If there is at least one bottle with capacity $C_j$, the number of two-bottle sets with capacity $C$ is limited by the smallest group size of the two capacities ($C_i,C_j$), and the counter $\color{teal}{\text{n_sum}}$ is updated accordingly.

The code handles these cases by iterating through keys in $\color{teal}{\text{groups}}$, namely $\color{teal}{\text{val_i}}$. To prevent double counting, the loop keeps track of the used capacities by a list $\color{teal}{\text{check_list}}$, which registers the current capacity $\color{teal}{\text{val_i}}$ and its complement $\color{teal}{\text{val_j}}$ in each step.

After processing all possible sets for a given capacity, the resulting counter $\color{teal}{\text{n_sum}}$ is compared with $\color{teal}{\text{N_best}}$, and the larger value is retained as the new optimal count ($\color{teal}{\text{N_best}}$). When all capacities in $\color{teal}{\text{sum_vals}}$ have been evaluated, the function returns $\color{teal}{\text{N_best}}$, which represents the optimal number of sets for the input $\color{teal}{\text{stock}}$ list.

Final comments and invitation to collaborate

My solution passes 100% of the test cases, meaning it works fine. I'm sure there is room for improvement, so any feedback is very welcome. Feel free to reach out via LinkedIn or email (fertmeneses@gmail.com).

Also, I invite you to join the GitHub project as a collaborator to share your own solutions or suggest improvements.

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