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.&nbsp;

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.

&nbsp;
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.


&nbsp;
Constraints:


	costs.length == n
	1 &lt;= n &lt;= 105
	1 &lt;= costs[i] &lt;= 105
	1 &lt;= coins &lt;= 108

class Solution:
    def maxIceCream(self, costs: List[int], coins: int) -> int:
        costs.sort()
        i= 0
        for price in costs:
            if price<= coins:
                i+= 1
                coins-= price
            else:
                break
        return i
class Solution {
    public int maxIceCream(int[] costs, int coins) {

        //Greedy Approach
        //a. sort cost in increasing order

        Arrays.sort(costs);

        int count = 0;
        for(int cost : costs){

            //b. check remainig coin is greater or equal than cuurent ice - cream cost
            if(coins - cost >= 0) {
                coins -= cost;
                count++;
            }
        }

        return count;
    }
}
class Solution {
public:
    int maxIceCream(vector<int>& costs, int coins) {
        int n=costs.size();

        sort(costs.begin(),costs.end());
        
        int i=0;
        for(;i<n && coins>=costs[i];i++){
            coins-=costs[i];
        }
        
        return i;
    }
};
var maxIceCream = function(costs, coins) {
  costs.sort((a, b) => a - b);
  let count = 0;

  for (let i = 0; i < costs.length; i++) {
    if (costs[i] <= coins) {
      count++;
      coins -= costs[i]
    } else {
      break; // a small optimization, end the loop early if coins go down to zero before we reach end of the length of costs.
    }
  }
  return count;
};
Maximum Ice Cream Bars