Maximum Ice Cream Bars
Medium
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Topics:
ArraysGreedy Algorithms
It is a sweltering summer day, and a boy wants to buy some ice cream bars.
There are n ice cream bars. You are given an array costs of length n, where costs[i] is the price of the i<sup>th</sup> ice cream bar in coins. The boy initially has coins coins to spend, and he wants to buy as many ice cream bars as possible.
Note: The boy can buy the ice cream bars in any order.
Return the maximum number of ice cream bars the boy can buy with coins coins.
You must solve the problem by counting sort.
Example 1:
Input: costs = [1,3,2,4,1], coins = 7
Output: 4
Explanation: The boy can buy ice cream bars at indices 0,1,2,4 for a total price of 1 + 3 + 2 + 1 = 7.
Example 2:
Input: costs = [10,6,8,7,7,8], coins = 5
Output: 0
Explanation: The boy cannot afford any of the ice cream bars.
Example 3:
Input: costs = [1,6,3,1,2,5], coins = 20
Output: 6
Explanation: The boy can buy all the ice cream bars for a total price of 1 + 6 + 3 + 1 + 2 + 5 = 18.
Constraints:
costs.length == n1 <= n <= 10<sup>5</sup>1 <= costs[i] <= 10<sup>5</sup>1 <= coins <= 10<sup>8</sup>
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the possible ranges for the values in the `costs` array and for the `coins` value? Are they all non-negative integers?
- Can the `costs` array be empty?
- If I don't have enough coins to buy any ice cream bars, what should I return (0, null, or throw an exception)?
- Are the ice cream bar costs distinct, or can there be duplicates?
- Is the order of the ice cream bars returned important?
Brute Force Solution
Approach
Imagine you have a certain amount of money and a list of ice cream bars with different prices. The brute force approach is like trying every possible combination of ice cream bars you could buy with your money. We want to find the largest number of ice cream bars we can purchase without exceeding our budget.
Here's how the algorithm would work step-by-step:
- Start by trying to buy zero ice cream bars.
- Then try to buy only the first ice cream bar, then the second, then the third, and so on, each time checking if you can afford it.
- Next, try buying pairs of ice cream bars: the first and second, the first and third, the second and third, and so on, making sure you don't spend too much money.
- Continue trying all possible combinations of three ice cream bars, four ice cream bars, and so on, always verifying that the total cost does not exceed your budget.
- For each combination you can afford, count the number of ice cream bars.
- Compare the number of ice cream bars in each affordable combination, and choose the combination with the most ice cream bars. That's the maximum number you can buy.
Code Implementation
def max_ice_cream_bars_brute_force(costs, money):
max_ice_cream_count = 0
# Iterate through all possible combinations of ice cream bars.
for i in range(1 << len(costs)):
current_cost = 0
current_ice_cream_count = 0
for j in range(len(costs)):
# Check if the j-th ice cream bar is included in the current combination.
if (i >> j) & 1:
current_cost += costs[j]
current_ice_cream_count += 1
# If the total cost is within the budget,
# update the maximum ice cream count.
if current_cost <= money:
max_ice_cream_count = max(max_ice_cream_count, current_ice_cream_count)
return max_ice_cream_countBig(O) Analysis
Time Complexity
O(2^n) – The brute force approach involves generating all possible subsets of the ice cream prices. For an input array of size n, there are 2^n possible subsets. For each subset, we calculate the total cost to determine if it is within the given coins budget. Therefore, the time complexity is dominated by the generation of all subsets, leading to O(2^n).
Space Complexity
O(N!) – The brute force approach, as described, explores all possible combinations of ice cream bars. This implicitly involves generating and storing subsets of the input list of ice cream prices. In the worst case, where we have to consider all possible combinations, the number of combinations grows factorially with the number of ice cream bars, N. Although the algorithm description does not explicitly state that these combinations are stored in an array or list, the implicit step of checking the costs of combinations entails storing each combination to calculate the cumulative cost. Thus, the space complexity is O(N!).
Optimal Solution
Approach
The most efficient way to figure out the maximum number of ice cream bars you can buy is to buy the cheapest ones first. This way, you use your money wisely and get the most for your budget. Essentially, prioritize the lower cost options to maximize your purchase.
Here's how the algorithm would work step-by-step:
- First, figure out how much each ice cream bar costs and put them in order from least to most expensive.
- Then, start buying the cheapest ice cream bars one at a time.
- Keep track of how much money you've spent as you buy each ice cream bar.
- Stop buying ice cream bars when you run out of money.
- The number of ice cream bars you bought before running out of money is the maximum you can buy.
Code Implementation
def max_ice_cream(costs, coins):
costs.sort()
ice_cream_count = 0
money_spent = 0
# Iterate through the sorted costs to buy ice cream.
for cost in costs:
if money_spent + cost <= coins:
# Purchase if we have enough money.
money_spent += cost
ice_cream_count += 1
else:
# Stop when we can't afford more.
break
return ice_cream_countBig(O) Analysis
Time Complexity
O(n log n) – The dominant operation in this algorithm is sorting the costs array, which has a time complexity of O(n log n), where n is the number of ice cream bars. The subsequent loop iterates at most n times to buy ice cream bars, but its time complexity is overshadowed by the sorting step. Therefore, the overall time complexity is determined by the sorting algorithm.
Space Complexity
O(1) – The algorithm sorts the input array of costs in place, meaning it modifies the original array directly without using additional storage proportional to the input size. It then iterates through this sorted array, using a few variables to track the current cost and the number of ice cream bars purchased. These variables take up a constant amount of space, irrespective of the number of ice cream bars (N). Therefore, the auxiliary space complexity is O(1).
Edge Cases
| Case | How to Handle |
|---|---|
| Empty costs array | Return 0 since no ice cream bars can be bought. |
| Coins is zero | Return 0 since no ice cream bars can be bought with zero coins. |
| Costs array contains a single zero | Return 1 if coins are non-negative, otherwise 0, as a single zero-cost item can always be bought. |
| Costs array contains only negative values | Consider only the non-negative elements or throw an exception if negative values are not allowed per the problem description. |
| All costs are greater than coins | Return 0 since no ice cream bars can be purchased. |
| Coins is very large, and costs are mostly small | Ensure the data type used to track remaining coins doesn't overflow during subtractions; also consider optimizing the sorting/selection process if needed. |
| Costs array contains duplicate values | The sorting algorithm handles this correctly; items with the same cost are treated equally and considered in order. |
| Costs array is already sorted | The sorting algorithm should handle already sorted arrays efficiently (ideally O(n) or O(n log n)). |
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