5 Best Ways to Find Maximum Ice Cream Bars in Python

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💡 Problem Formulation: You’re given an array of ice cream bars’ costs and an amount of coins. The goal is to find out the maximum number of ice cream bars you can buy with those coins. For example, if the input array is [1,3,2,4,1], and the amount of coins is 7, the output should be 4 bars: buying the bars with costs 1, 2, 1, and 3.

Method 1: Sorting and Iteration

This method involves first sorting the array of ice cream bar costs in ascending order and then iterating through the array, incrementally deducting the cost from the total amount of coins until you can no longer afford an ice cream bar. It’s efficient and straightforward.

Here’s an example:

def max_ice_cream_bars(costs, coins):
    costs.sort()
    bars = 0
    for cost in costs:
        if cost > coins:
            break
        coins -= cost
        bars += 1
    return bars

print(max_ice_cream_bars([1,3,2,4,1], 7))

Output: 4

This function sorts the cost of ice cream bars and then iterates through the sorted list, reducing the coins accordingly and counting the number of bars purchased until the coins run out.

Method 2: Using a Min Heap

Maximizing the number of ice cream bars can be approached by continuously buying the cheapest bar available. A min-heap—or a priority queue—provides an efficient structure to access the minimum element each time. This Python method leverages the heapq module.

Here’s an example:

import heapq

def max_ice_cream_bars(costs, coins):
    heapq.heapify(costs)
    bars = 0
    while costs and coins >= costs[0]:
        coins -= heapq.heappop(costs)
        bars += 1
    return bars

print(max_ice_cream_bars([1,3,2,4,1], 7))

Output: 4

The heapq.heapify function turns the list into a min-heap, after which the minimum elements are continuously popped (and the coins reduced accordingly) to calculate the maximum number of ice cream bars that can be bought.

Method 3: Counting Sort for Small Fixed Costs

When the cost of ice cream bars is within a small fixed range (e.g., 1 to 100), counting sort can be a viable and efficient solution. It counts the occurrence of each cost and then iteratively consumes the bars starting with the cheapest.

Here’s an example:

def max_ice_cream_bars(costs, coins):
    count = [0] * (max(costs) + 1)
    for cost in costs:
        count[cost] += 1
    
    bars = 0
    for cost, frequency in enumerate(count):
        if coins >= cost:
            buy = min(frequency, coins // cost)
            bars += buy
            coins -= cost * buy
        else:
            break
    return bars

print(max_ice_cream_bars([1,3,2,4,1], 7))

Output: 4

This method uses a list to count the frequency of each cost and then iterates through the list to determine how many ice cream bars can be bought at each cost level until there are no more coins left.

Method 4: Dynamic Programming (Knapsack Approach)

For those interested in comprehensive programming solutions, the knapsack problem approach can be adapted to solve this. It works well when dealing with a large number of bars with varied costs. However, it’s more complex and may not be as efficient as other methods for smaller inputs.

Here’s an example:

def max_ice_cream_bars(costs, coins):
    dp = [0] * (coins + 1)
    for cost in costs:
        for i in range(coins, cost-1, -1):
            dp[i] = max(dp[i], dp[i-cost] + 1)
    return dp[coins]

print(max_ice_cream_bars([1,3,2,4,1], 7))

Output: 4

Dynamic programming is used to build up an array where each index represents the number of bars that can be bought with that many coins. The array is updated by considering each ice cream bar’s cost and maximizing the number at each step.

Bonus One-Liner Method 5: Functional Python

The task can elegantly be solved by employing Python’s functional abilities. This one-liner uses a generator expression within the sum() function and provides an ultra-compact solution suitable for cases where code brevity is a priority.

Here’s an example:

max_ice_cream_bars = lambda costs, coins: sum(coins >= (coin := sorted(costs)[bar]) and (coins := coins - coin) for bar in range(len(costs)))

print(max_ice_cream_bars([1,3,2,4,1], 7))

Output: 4

This compact lambda function sorts the list of costs and runs a generator expression that cumulatively subtracts the cost of each bar from the total number of coins, keeping track of the number of bars that can be bought.

Summary/Discussion

  • Method 1: Sorting and Iteration. Strengths: Easy-to-understand, efficient for small datasets. Weaknesses: Sort operation can become costly with very large datasets.
  • Method 2: Using a Min Heap. Strengths: Minimizes cost look-up time, efficient for large datasets. Weaknesses: Slightly more complex to understand and implement.
  • Method 3: Counting Sort for Small Fixed Costs. Strengths: Extremely efficient for datasets with small and predictable costs. Weaknesses: Not suitable for costs that are not within a small fixed range.
  • Method 4: Dynamic Programming. Strengths: Offers a precise and comprehensive solution. Weaknesses: Overkill for simple problems and less efficient for smaller datasets.
  • Bonus Method 5: One-Liner Functional Python. Strengths: Code brevity. Weaknesses: Potentially difficult to read and understand.