Description Link to heading
1833. Maximum Ice Cream Bars (Medium)
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
Solution Link to heading
Greedy algorithm: the optimal way is to buy the cheapest ice-cream each time, we can use a priority queue to simulate the process.
Code Link to heading
class Solution {
public:
int maxIceCream(vector<int> &costs, int coins) {
priority_queue<int, vector<int>, std::greater<int>> pq; // small top stack
for (int &price : costs) {
pq.push(price);
}
int cnt = 0;
while (!pq.empty() && coins >= pq.top()) {
cnt++;
coins -= pq.top();
pq.pop();
}
return cnt;
}
};