Introduction

In this blog post, we will explore how to solve the LeetCode problem "1833" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

It is a sweltering summer day, and a boy wants to buy some ice cream bars. At the store, there are n ice cream bars. You are given an array costs of length n, where costs[i] is the price of the ith 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 == n 1 <= n <= 105 1 <= costs[i] <= 105 1 <= coins <= 108

Explanation

Here's a solution that uses counting sort to determine the maximum number of ice cream bars a boy can buy.

• Counting Sort: Utilize counting sort because the cost values are constrained within a reasonable range (1 to 10^5). This allows us to efficiently count the occurrences of each price.
• Greedy Approach: Iterate through the sorted prices (implicitly sorted by the counting sort). Buy as many ice cream bars of the current price as possible without exceeding the available coins.

• Accumulate Count: Keep track of the total number of ice cream bars bought.

• Runtime Complexity: O(n + k), where n is the number of ice cream bars and k is the range of cost values (1 to 10^5).

• Storage Complexity: O(k), to store the count array.

Code

    def maxIceCream(costs, coins):
    """
    Calculates the maximum number of ice cream bars a boy can buy.

    Args:
        costs: A list of integers representing the cost of each ice cream bar.
        coins: An integer representing the total coins the boy has.

    Returns:
        An integer representing the maximum number of ice cream bars the boy can buy.
    """

    max_cost = 100000  # Maximum possible cost of an ice cream bar
    counts = [0] * (max_cost + 1)

    # Count the occurrences of each cost
    for cost in costs:
        counts[cost] += 1

    ice_cream_bars = 0
    for cost in range(1, max_cost + 1):
        if coins >= cost:
            num_affordable = min(counts[cost], coins // cost)
            ice_cream_bars += num_affordable
            coins -= num_affordable * cost
        else:
            break  # No more affordable ice cream bars

    return ice_cream_bars