<p>Given an <code>n x n</code> <code>matrix</code> where each of the rows and columns is sorted in ascending order, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest element in the matrix</em>.</p>

<p>Note that it is the <code>k<sup>th</sup></code> smallest element <strong>in the sorted order</strong>, not the <code>k<sup>th</sup></code> <strong>distinct</strong> element.</p>

<p>You must find a solution with a memory complexity better than <code>O(n<sup>2</sup>)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
<strong>Output:</strong> 13
<strong>Explanation:</strong> The elements in the matrix are [1,5,9,10,11,12,13,<u><strong>13</strong></u>,15], and the 8<sup>th</sup> smallest number is 13
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[-5]], k = 1
<strong>Output:</strong> -5
</pre>

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

<ul>
	<li><code>n == matrix.length == matrix[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 300</code></li>
	<li><code>-10<sup>9</sup> &lt;= matrix[i][j] &lt;= 10<sup>9</sup></code></li>
	<li>All the rows and columns of <code>matrix</code> are <strong>guaranteed</strong> to be sorted in <strong>non-decreasing order</strong>.</li>
	<li><code>1 &lt;= k &lt;= n<sup>2</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>Could you solve the problem with a constant memory (i.e., <code>O(1)</code> memory complexity)?</li>
	<li>Could you solve the problem in <code>O(n)</code> time complexity? The solution may be too advanced for an interview but you may find reading <a href="http://www.cse.yorku.ca/~andy/pubs/X+Y.pdf" target="_blank">this paper</a> fun.</li>
</ul>

class Solution {
public:

    int n, k;
    int countNums(int x, vector<vector<int>>& arr){
        int top=0, bottom=n-1,mid, count1=0;
        while(top<=bottom){
            mid=top+(bottom-top)/2;
            if(arr[mid][0]<=x) top=mid+1;
            else{
                if(mid>0 && arr[mid-1][0]<=x){
                top=mid;
                break;
                } 
                bottom=mid-1;
            }
        }
        for(int i=0;i<top;i++){
            int left=0, right=n-1;
            while(left<=right){
                mid=left+(right-left)/2;
                if(arr[i][mid]<=x) left=mid+1;
                else{
                    if(mid>0 && arr[i][mid-1]<=x){
                    left=mid;
                    break;  
                    } 
                    right=mid-1;
                }
            }
            count1+=left;
        }
        return count1;
    }

    int kthSmallest(vector<vector<int>>& matrix, int k1) {
        k=k1, n=matrix.size();
        vector<vector<int>> arr=matrix;
        int left=arr[0][0], right=arr[n-1][n-1], mid;
        while(left<=right){
            mid=left+(right-left)/2;
            if(countNums(mid, arr)<k) left=mid+1;
            else{
                if(countNums(mid-1, arr)<k) break;
                right=mid-1;
            }
        }
        return mid;      
    }
};
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