<p>It is a sweltering summer day, and a boy wants to buy some ice cream bars.</p>

<p>At the store, there are <code>n</code> ice cream bars. You are given an array <code>costs</code> of length <code>n</code>, where <code>costs[i]</code> is the price of the <code>i<sup>th</sup></code> ice cream bar in coins. The boy initially has <code>coins</code> coins to spend, and he wants to buy as many ice cream bars as possible.&nbsp;</p>

<p><strong>Note:</strong> The boy can buy the ice cream bars in any order.</p>

<p>Return <em>the <strong>maximum</strong> number of ice cream bars the boy can buy with </em><code>coins</code><em> coins.</em></p>

<p>You must solve the problem by counting sort.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> costs = [1,3,2,4,1], coins = 7
<strong>Output:</strong> 4
<strong>Explanation: </strong>The boy can buy ice cream bars at indices 0,1,2,4 for a total price of 1 + 3 + 2 + 1 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> costs = [10,6,8,7,7,8], coins = 5
<strong>Output:</strong> 0
<strong>Explanation: </strong>The boy cannot afford any of the ice cream bars.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> costs = [1,6,3,1,2,5], coins = 20
<strong>Output:</strong> 6
<strong>Explanation: </strong>The boy can buy all the ice cream bars for a total price of 1 + 6 + 3 + 1 + 2 + 5 = 18.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>costs.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= costs[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= coins &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution:
    def maxIceCream(self, costs, coins):
        # Find the maximum cost in the costs array
        max_cost = max(costs)
        
        # Create a count array to store the frequency of each cost
        count = [0] * (max_cost + 1)
        for cost in costs:
            count[cost] += 1
        
        # Initialize the number of ice cream bars the boy can buy
        bars = 0
        
        # Iterate through the count array
        for cost in range(1, max_cost + 1):
            if count[cost] > 0:
                # Calculate how many bars of the current cost the boy can buy
                buy = min(count[cost], coins // cost)
                bars += buy
                coins -= cost * buy
                
                # If the boy has no more coins left, break the loop
                if coins == 0:
                    break
        
        return bars

# Example usage:
solution = Solution()
costs = [1,3,2,4,1]
coins = 7
print(solution.maxIceCream(costs, coins))  # Output: 4
30800