Buy the Most Bars Using Counting Sort

### Problem Each price in `costs` belongs to one ice-cream bar. With `coins` coins, buy as many bars as possible in any order and return the greatest count. Use counting sort, as required by the source, to process prices from least to greatest. ```cpp class Solution { public: int maxIceCream(vector<int>& costs, int coins); }; ``` ### Examples **Example 1:** ```text costs = [2,2,5,1], coins = 5 Output: 3 Explanation: Buy prices 1, 2, and 2. ``` **Example 2:** ```text costs = [4,6,8], coins = 3 Output: 0 Explanation: Even the cheapest bar exceeds the budget. ``` ### Constraints - `n == costs.length`, with `1 <= n <= 10^5`. - `1 <= costs[i] <= 10^5`. - `1 <= coins <= 10^8`. - The solution must use counting sort. ### Clarifying Questions A bar is indivisible and may be purchased at most once. Equal-priced bars are distinct available items. ```hint Inspect the structure Do equal-priced bars remain separate purchases, and can a single bar be bought more than once? ``` ### Reasoning checkpoints - Retain the explicit counting-sort requirement. - Prove that the selected bars are affordable and no larger count can be purchased. ### Optional extensions How does the algorithm’s cost depend on the maximum price as well as the number of bars?