<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 countLargestGroup(int&n) {
        unordered_map<int, int>rem;
        for(; n > 0; --n){
            string s = to_string(n);
            int val = 0;
            for(char&c:s){
                val+= (c - '0');
            }
            ++rem[val];
        }
        vector<int>sizes(rem.size());
        for(auto&p:rem){
            sizes[n++] = p.second;
        }
        sort(sizes.rbegin(), sizes.rend());
        n = 1;
        for(; n < sizes.size(); ++n){
            if(sizes[n] != sizes[n-1]) return n;
        }
        return n;
    }
};
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