<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>

#include <bits/stdc++.h>
using namespace std;
#define lop(i, n) for (long long i = 0; i < n; i++)
#define lopr(i, n) for (long long i = n - 1; i >= 0; i--)
#define loop(i, a, b) for (long long i = a; i < b; i++)

auto init = []()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();

class Solution
{
public:
    int maxIceCream(vector<int> &costs, int coins)
    {
        vector<long long> pref(costs.size() + 1, 0);
        sort(costs.begin(), costs.end());
        for (long long i = 1; i <= costs.size(); ++i)
        {
            pref[i] = pref[i - 1] + costs[i - 1];
        }
        return upper_bound(pref.begin(), pref.end(), coins) - pref.begin() - 1;
    }
};
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