<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) {
        int maxSize = INT_MIN;
        int count = 0;
        unordered_map<int, vector<int>>mp;
        for(int i = 1; i <= n; i++) {
            int sum = 0;
            int num = i;
            while(num != 0)
            {
                int dig = num % 10;
                sum += dig;
                num = num/10;
            }
            mp[sum].push_back(num);
            maxSize = max(maxSize, static_cast<int>(mp[sum].size()));
        }
        for(auto it = mp.begin(); it != mp.end(); it++) {
            if(it->second.size() == maxSize)
                count++;
        }
        return count;
    }
};
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