<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 {
private: 
    map<int, vector<int>> hm;
public:
    int sumOfDigits(int n) {
        int sum = 0;
        while (n > 0) {
            int rem = n % 10;
            n = n / 10;
            sum += rem;
        }
        return sum;
    }

    int countLargestGroup(int n) {
        for (int i = 1; i <= n; ++i) {
            int sum = sumOfDigits(i);
            hm[sum].push_back(i);
        }    
        int count = 0;
        int maxSize = 0;
        for (const auto &it : hm) {
            vector<int> nums = it.second;
            int currSize = nums.size();
            if (currSize == maxSize) {
                ++count;
            }
            else if (currSize > maxSize) {
                maxSize = currSize;
                count = 1;
            }
        }

        return count;
    }
};
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