Formatted question description: https://leetcode.ca/all/1833.html
1833. Maximum Ice Cream Bars
Level
Medium
Description
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
Solution
Use a greedy approach. Sort the array costs
and calculate the sum of the smallest elements such that the sum is maximized and the sum does not exceed coins
. Return the maximum number of elements possible.

class Solution { public int maxIceCream(int[] costs, int coins) { Arrays.sort(costs); int count = 0; int length = costs.length; for (int i = 0; i < length; i++) { int cost = costs[i]; if (coins >= cost) { coins = cost; count++; } else break; } return count; } }

// OJ: https://leetcode.com/problems/maximumicecreambars/ // Time: O(NlogN) // Space: O(1) class Solution { public: int maxIceCream(vector<int>& A, int B) { sort(begin(A), end(A)); int ans = 0; for (int n : A) { if (B < n) break; ++ans; B = n; } return ans; } };

# 1833. Maximum Ice Cream Bars # https://leetcode.com/problems/maximumicecreambars/ class Solution: def maxIceCream(self, costs: List[int], coins: int) > int: n = len(costs) res = 0 costs.sort() for c in costs: if coins >= c: res += 1 coins = c else: break return res