5 Best Ways to Find the Closest Dessert Cost in Python

💡 Problem Formulation: Finding the dessert cost closest to a budget within a list of prices is a common programming problem. For instance, given a list of dessert costs [3.50, 4.90, 2.70, 5.80] and a budget of 5.00, the program should output 4.90 as it is the nearest cost under the budget.

Method 1: Brute Force

This method involves iterating through each dessert cost in the list, comparing them to the budget, and keeping track of the closest cost below the said budget. It’s a straightforward approach with no additional libraries required.

Here’s an example:

dessert_costs = [3.50, 4.90, 2.70, 5.80]
budget = 5.00
closest_cost = dessert_costs[0]

for cost in dessert_costs:
    if abs(budget - cost) < abs(budget - closest_cost) and cost <= budget:
        closest_cost = cost

print(closest_cost)

Output:

In this snippet, we initialize closest_cost with the first dessert cost in the list. We then loop through each cost, updating closest_cost if a closer cost under the budget is found. This approach is easy to understand and implement but may not be the most efficient for larger lists.

Method 2: Sorting and Binary Search

By first sorting the list of costs and then performing binary search to find the closest cost under the budget, we can improve efficiency for large datasets. This uses Python’s built-in algorithms for sorting and searching.

Here’s an example:

import bisect

dessert_costs = [3.50, 4.90, 2.70, 5.80]
budget = 5.00

# Sort the list of costs
sorted_costs = sorted(dessert_costs)

# Find the position to insert the budget, then move left for closest cost under budget
position = bisect.bisect_right(sorted_costs, budget) - 1
closest_cost = sorted_costs[position] if position >= 0 else None

print(closest_cost)

Output:

After sorting the dessert costs, we use bisect_right from the bisect module to find the index where the budget would be inserted in the list. We then step one index to the left to obtain the largest cost that does not exceed the budget. This method is more efficient but requires the costs to be sorted first.

Method 3: Minimize Absolute Difference

This method uses a minimization approach to find the dessert cost with the smallest absolute difference from the budget that is still less than or equal to the budget. It combines the efficiency of a single-pass solution with Python’s built-in min function.

Here’s an example:

dessert_costs = [3.50, 4.90, 2.70, 5.80]
budget = 5.00

closest_cost = min((cost for cost in dessert_costs if cost <= budget), key=lambda x: (budget - x))

print(closest_cost)

Output:

Using a generator expression, we filter dessert costs that are under the budget. We pass this generator along with a key function to the min function to find the value with the smallest difference from the budget. This method is concise and requires no preliminary sorting.

Method 4: Using NumPy for Large Datasets

For large datasets, using NumPy’s vectorized operations can significantly speed up the search. NumPy has built-in functions that allow for efficient array comparisons and effectively finding the nearest number under a certain threshold.

Here’s an example:

import numpy as np

dessert_costs = np.array([3.50, 4.90, 2.70, 5.80])
budget = 5.00

# Filter costs that are under the budget and find the closest cost
under_budget = dessert_costs[dessert_costs <= budget]
closest_cost = under_budget[np.argmin(np.abs(under_budget - budget))]

print(closest_cost)

Output:

This snippet uses NumPy arrays to keep track of dessert costs. After filtering the costs under the budget, we use np.argmin to find the index of the closest cost. This method can handle larger datasets more efficiently due to NumPy’s optimized performance.

Bonus One-Liner Method 5: Using a Comprehension

Python’s list comprehensions provide a compact way to combine filtering and minimization to find the closest cost. This one-liner approach is stylish and highly readable for those familiar with comprehensions.

Here’s an example:

dessert_costs = [3.50, 4.90, 2.70, 5.80]
budget = 5.00

closest_cost = min([cost for cost in dessert_costs if cost <= budget], default=None, key=lambda x: budget - x)

print(closest_cost)

Output:

The list comprehension filters dessert costs under the budget. The min function then finds the minimum element based on the smallest difference from the budget. The default=None parameter ensures that if there are no desserts under the budget, None is returned. This method is highly concise and perfect for smaller lists.

Summary/Discussion

• Method 1: Brute Force. Simple and straightforward. No extra libraries required, but not the most efficient for large datasets.
• Method 2: Sorting and Binary Search. Efficient, especially for large datasets, and leverages Python’s built-in sorting and binary search features. Requires sorting the list first, which can be expensive for very large lists.
• Method 3: Minimize Absolute Difference. A one-pass solution that’s efficient and doesn’t require sorting. Best suited for situations where budget adherence is strict and datasets are reasonably sized.
• Method 4: Using NumPy for Large Datasets. Excellent for handling large datasets. Utilizes NumPy’s optimized array operations, but requires NumPy to be installed.
• Bonus One-Liner Method 5: Using a Comprehension. Elegant and concise, ideal for quick scripts or smaller datasets. Might not be as intuitive for those unfamiliar with list comprehensions.