<p>You are given an integer <code>n</code>.</p>

<p>Each number from <code>1</code> to <code>n</code> is grouped according to the sum of its digits.</p>

<p>Return <em>the number of groups that have the largest size</em>.</p>

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

<pre>
<strong>Input:</strong> n = 13
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 9 groups in total, they are grouped according sum of its digits of numbers from 1 to 13:
[1,10], [2,11], [3,12], [4,13], [5], [6], [7], [8], [9].
There are 4 groups with largest size.
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 groups [1], [2] of size 1.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int sum(int n)
    {
        if(n==0) return 0;
        else return n%10 + sum(n/10);
    }
    int countLargestGroup(int n) {
        vector<int> count (37,0); 
        int max = 0;
        for(int i=1; i<=n; i++) ++count[sum(i)];
        // for(int i=0; i<37; i++) cout<<i<<":"<<count[i]<<" ";
        sort(count.begin(), count.end());
        int ans = 0;
        for(int i=36; i>-1; i--) 
        {
            if(count[i]==count[36]) ans++;
            else break;
        }
        return ans;
    }
};
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