1833 - Maximum Ice Cream Bars (Medium)

Problem Link

https://leetcode.com/problems/maximum-ice-cream-bars/

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.

Return the maximum number of ice cream bars the boy can buy with coins coins.

Note: The boy can buy the ice cream bars in any order.

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 <= 10^5
• 1 <= costs[i] <= 10^5
• 1 <= coins <= 10^8

Approach 1: Frequency

Time Complexity: $O(n + m)$ _{(where n: length of costs, m: maximum value among costs)} Space Complexity: $O(m)$ _{(where m: maximum value among costs)}

Python

Written by @hirotake111

class Solution:
    def maxIceCream(self, costs: List[int], coins: int) -> int:
        # Fill in the list of frequency
        # Each index is cost of icecream - so the length must be max(costs) + 1
        freq = [0] * (max(costs) + 1)
        answer = 0

        # Count each cost's frequency
        for cost in costs:
            freq[cost] += 1

        for cost, amount in enumerate(freq):
            # If frequency is 0, skip it
            if freq[cost] == 0:
                continue
            # If cost * amount is less than coins,
            # simply decrease the coins by cost * amount
            if cost * amount <= coins:
                coins -= cost * amount
                answer += amount
                continue
            # At this point we can't buy amount * cost
            # So coins // cost should be the amount of icecream we can buy
            answer += coins // cost
            # And don't forget to exit loop (as we can't buy icecreams anymore)
            break

        return answer