<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> mp;
        for(int i=1;i<=n;i++)
        {
            if(i<=9)
            {
                mp[i]++;
                continue;
            }
            else
            {
                int a=i;
                int sum=0;
                while(a>0)
                {
                    sum+=a%10;
                    a=a/10;
                }
                mp[sum]++;
            }
                
        }

        vector<pair<int,int>> res;

        for(auto x:mp)
            res.push_back({x.second,x.first});

        sort(res.rbegin(),res.rend());
        int a=res[0].first,count=0;
        
        for(int i=0;i<res.size();i++)
        {
            if(res[i].first==a)
                count++;
            else
                break;
        }
        return count;
        
    }
};
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