<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 digSum(int k){
        int sum = 0;
        while(k){
            sum += k%10;
            k/=10;
        }
        return sum;
    }
    static bool compareByValue(const pair<int, int>& a, const pair<int, int>& b) {
        return a.second > b.second;
    }

    int countLargestGroup(int n) {
        map<int, int> group;
        for(int i=1; i<=n; i++){
            group[digSum(i)]++;
        }
        vector<pair<int, int>> vp(group.begin(), group.end());
        sort(vp.begin(), vp.end(), compareByValue);
        int s = vp[0].second;
        int count = 0;
        for(auto x: vp){
            if(x.second == s){
                count++;
            }
            else{
                break;
            }
        }
        return count;
    }
};
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