<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>

#include <bits/stdc++.h>
#define ll long long 
#define nl '\n'
class Solution {
public:
    int kthLargestValue(vector<vector<int>>& matrix,int k) {
        int m = matrix.size();
        int n = matrix[0].size();
        vector<vector<int>>xorMatrix(m,vector<int>(n, 0));
     priority_queue<int,vector<int>,greater<int>>minHeap;
        // Compute the XOR matrix
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(i == 0 && j == 0){
                    xorMatrix[i][j] = matrix[i][j];
                }else if (i == 0){
                    xorMatrix[i][j] = xorMatrix[i][j-1] ^ matrix[i][j];
                }else if (j == 0){
                    xorMatrix[i][j] = xorMatrix[i-1][j] ^ matrix[i][j];
                }else{
                    xorMatrix[i][j] = xorMatrix[i-1][j] ^ xorMatrix[i][j-1] ^ xorMatrix[i-1][j-1] ^ matrix[i][j];
                }
                // Maintain a min-heap of the k largest values
                if(minHeap.size() < k){
                    minHeap.push(xorMatrix[i][j]);
                }else if(xorMatrix[i][j] > minHeap.top()){/// maximum xormatrix[i][j] value in the top of heap
                    minHeap.pop();
                    minHeap.push(xorMatrix[i][j]);
                }
            }
        }
        return minHeap.top();
    }

//Matrix:        XOR Matrix:
// 5  2           5  7
// 1  6           4  4
};
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