<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) {

        // Approach-1: (Using Prefix Sum + MinHeap)
        // T.C. --> O(m*n*logk)
        // S.C. --> O(2*m*n) + O(k) => O(m*n)
        
        int m = matrix.size(), n = matrix[0].size();

        vector<vector<int>> upper_xor(m, vector<int>(n)), left_xor(m, vector<int>(n));

        priority_queue<int, vector<int>, greater<int>> pq;

        for(int i=0; i<m; ++i)
        {
            for(int j=0; j<n; ++j)
            {
                int xor_up = (i-1 < 0) ? 0 : upper_xor[i-1][j];
                int xor_left = (j-1 < 0 ) ? 0 : left_xor[i][j-1];
                int curr_xor = (matrix[i][j] ^ xor_up ^ xor_left);

                upper_xor[i][j] = curr_xor;
                left_xor[i][j] = (matrix[i][j] ^ xor_left);

                pq.push(curr_xor);

                if(pq.size() > k) pq.pop();
            }
        }

        return pq.top();
    }
};
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