<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, vector<int>> mp;
        int sum = 0;
        for(int i = 1; i <= n; i++){
            if(i < 10){
                mp[i].push_back(i);
            }else{
                int temp = i;
                while(temp != 0){
                    sum += temp%10;
                    temp = temp/10;
                }
                mp[sum].push_back(i);
                sum = 0;
            }
        }
        vector<int> ans;
        for(auto x : mp){
            ans.push_back(x.second.size());
        }
        sort(ans.rbegin(),ans.rend());
        int maxLen = INT_MIN;
        int count = 0;
        for(auto x : ans){
            if(x > maxLen){
                maxLen = x;
                count++;
            }else if(x == maxLen){
                count++;
            }
        }
        return count;
    }
};
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