<p>You are given a 2D <code>matrix</code> of size <code>m x n</code>, consisting of non-negative integers. You are also given an integer <code>k</code>.</p>

<p>The <strong>value</strong> of coordinate <code>(a, b)</code> of the matrix is the XOR of all <code>matrix[i][j]</code> where <code>0 &lt;= i &lt;= a &lt; m</code> and <code>0 &lt;= j &lt;= b &lt; n</code> <strong>(0-indexed)</strong>.</p>

<p>Find the <code>k<sup>th</sup></code> largest value <strong>(1-indexed)</strong> of all the coordinates of <code>matrix</code>.</p>

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

<pre>
<strong>Input:</strong> matrix = [[5,2],[1,6]], k = 1
<strong>Output:</strong> 7
<strong>Explanation:</strong> The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[5,2],[1,6]], k = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[5,2],[1,6]], k = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>0 &lt;= matrix[i][j] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= m * n</code></li>
</ul>

class Solution {
public:
    int kthLargestValue(vector<vector<int>>& matrix, int k) {
        // note that at index (i,j)
        // A[i][j]=matrix[i][j] ^ A[i-1][j] ^ A[i][j-1] ^ A[i-1][j-1] wen i,j>0
        // if i==0 then A[i][j]=matrix[i][j] ^ A[i][j-1]
        // if j==0 then A[i][j]= matrix[i][j] ^ A[i-1][j]
        int m=matrix.size();
        int n=matrix[0].size();
        vector<vector<int>> A(m,vector<int>(n,0));
        priority_queue<int,vector<int>,greater<int>> pq; // i'll use a min Heap of size K
        pq.push(matrix[0][0]);
        A[0][0]=matrix[0][0];
        for(int i=1;i<m;i++){
            A[i][0]=matrix[i][0] ^ A[i-1][0];
            pq.push(A[i][0]);
            if(pq.size()>k) pq.pop();
        }
        for(int j=1;j<n;j++){
            A[0][j]=matrix[0][j] ^ A[0][j-1];
            pq.push(A[0][j]);
            if(pq.size()>k) pq.pop();
        }
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                A[i][j]=matrix[i][j] ^ A[i-1][j] ^ A[i][j-1] ^ A[i-1][j-1];
                pq.push(A[i][j]);
                if(pq.size()>k) pq.pop();
            }
        }
        // for(int i=0;i<m;i++){
        //     for(int j=0;j<n;j++){
        //         cout<<A[i][j]<<"   ";
        //     }
        //     cout<<endl;
        // }
        return pq.top();
    }
};
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