<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 {

    pair<int, pair<int,int>> elFromMatrix(vector<vector<int>>& matrix, int r, int c) {
        return make_pair(matrix[r][c], make_pair(r, c));
    }

    int getIdx(pair<int,int> rc) {
        return rc.first*10000 + rc.second;
    }

public:
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        priority_queue<pair<int, pair<int,int>>> pq;
        unordered_set<int> visited;
        pq.push(make_pair(-1*matrix[0][0], make_pair(0,0)));
        --k;

        while(pq.size() > 0) {
            pair<int, pair<int,int>> el = pq.top();
            pq.pop();

            if(visited.find(getIdx(el.second)) != visited.end())
                continue;

            if(k == 0) {
                return -1*el.first;
            }

            if(el.second.first < matrix.size()-1) {
                auto ell = elFromMatrix(matrix, el.second.first+1, el.second.second);
                ell.first = -1*ell.first;

                //if(visited.find(getIdx(ell.second)) != visited.end())
                    pq.push(ell);
            }

            if(el.second.second < matrix.size()-1) {
                auto ell = elFromMatrix(matrix, el.second.first, el.second.second+1);
                ell.first = -1*ell.first;

                //if(visited.find(getIdx(ell.second)) != visited.end())
                    pq.push(ell);
            }

            visited.insert(getIdx(el.second));
            --k;
        }   

        return 0; 
    }
};
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