<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:
    def countLargestGroup(self, n: int) -> int:

        dict_ = {}

        for i in range(1, n+1):
            num = i
            sum_ = 0
            while i > 0:
                sum_ += i%10
                i = i // 10
            dict_[sum_] = dict_.get(sum_, []) + [num]
        max_ = -1
        for v in dict_.values():
            max_ = max(max_, len(v))
        cnt = 0
        for v in dict_.values():
            if len(v) == max_:
                cnt += 1
        return cnt
17000