How do you solve Maximum Ice Cream Bars on LeetCode?

Maximum Ice Cream Bars (LeetCode #1833) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Buying cheaper items first maximizes quantity.

What is the time complexity of Maximum Ice Cream Bars?

The optimal solution for Maximum Ice Cream Bars runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, which is much more efficient than the brute force approach.

What is the brute force solution for Maximum Ice Cream Bars?

The brute force approach for Maximum Ice Cream Bars is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of ice cream bars to see how many can be bought within the given budget. This method is straightforward but inefficient due to its high time complexity. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Ice Cream Bars?

Maximum Ice Cream Bars uses the following concepts: Array, Greedy, Sorting, Counting Sort. The recommended patterns to study are: Greedy, Sorting, Array.

What are the interview tips for Maximum Ice Cream Bars?

Explain your thought process clearly. Discuss edge cases, like when coins are less than the cheapest item. Practice coding the solution without looking at references.

What are common mistakes when solving Maximum Ice Cream Bars?

Not sorting the costs before attempting to buy. Forgetting to check if the total cost exceeds the budget.

#1833

Maximum Ice Cream Bars

Medium

ArrayGreedySortingCounting SortGreedySortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal approach uses sorting and a greedy algorithm to ensure we buy the cheapest ice cream bars first, maximizing the number we can buy within the budget.

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Algorithm

3 steps

1Step 1: Sort the costs array in ascending order.
2Step 2: Initialize a count of ice cream bars and a total cost variable.
3Step 3: Iterate through the sorted costs, adding the cost of each bar to the total until the total exceeds the available coins.

solution.py13 lines

1# Full working Python code
2
3def maxIceCream(costs, coins):
4    costs.sort()
5    count = 0
6    total_cost = 0
7    for cost in costs:
8        if total_cost + cost <= coins:
9            total_cost += cost
10            count += 1
11        else:
12            break
13    return count

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, which is much more efficient than the brute force approach.

1Buying cheaper items first maximizes quantity.
2Sorting helps in efficiently managing budget.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.