<p>Nearly everyone has used the <a href="https://en.wikipedia.org/wiki/Multiplication_table" target="_blank">Multiplication Table</a>. The multiplication table of size <code>m x n</code> is an integer matrix <code>mat</code> where <code>mat[i][j] == i * j</code> (<strong>1-indexed</strong>).</p>

<p>Given three integers <code>m</code>, <code>n</code>, and <code>k</code>, return <em>the </em><code>k<sup>th</sup></code><em> smallest element in the </em><code>m x n</code><em> multiplication table</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/multtable1-grid.jpg" style="width: 500px; height: 254px;" />
<pre>
<strong>Input:</strong> m = 3, n = 3, k = 5
<strong>Output:</strong> 3
<strong>Explanation:</strong> The 5<sup>th</sup> smallest number is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/multtable2-grid.jpg" style="width: 493px; height: 293px;" />
<pre>
<strong>Input:</strong> m = 2, n = 3, k = 6
<strong>Output:</strong> 6
<strong>Explanation:</strong> The 6<sup>th</sup> smallest number is 6.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= m * n</code></li>
</ul>

class Solution {
public:
    long long helper(int mid , int k , int m , int n)
    {
         long long cnt =0;
         for(int i = 0 ; i<m ; i++)
         {
             long long value = min(mid/(i+1) , n);
             cnt+=value;
         }
         if(cnt>=k)
         {
            return true ;
         }
         return false ;
    }
    int findKthNumber(int m, int n, int k) {
        long long lo = 1 ;
        long long hi = 1e15;
        long long ans = hi ;
        while(hi>=lo)
        {
             long long mid = (lo+hi)/2 ;
             if(helper(mid , k,m,n))
             {
                 ans=mid ;
                 hi=mid-1 ;
             }
             else 
             {
                 lo = mid+1 ;
             }
        }
        return ans ;
    }
};
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