<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>

# 1.24/1.30/1.40/0/0

# basically it is the XOR of the square, foriming from  0,0 to corner contianing i,j
# a,b,c,..
# d,e,f,..

# a, a^b, a^b^c...
# a^d,a^b^d^e, a^b^c^d^e^f

# XOR above, XOR next to, then XOR diagonally 

# f ^ (a^b^c) ^ (a^b^d^e) ^ (a^b) = 

# runtime = going through each cell o(m*n)
# XOR op = 0(3) = O(1)
# O(m*n)

import heapq
class Solution:
    def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
        m,n = len(matrix), len(matrix[0])
        xorVals = [[0]*n for x in range(m)]
        xorHeap = []
        for i,row in enumerate(matrix):
            for j,cell in enumerate(row):
                val = matrix[i][j]
                if i > 0 :
                    val = val ^ xorVals[i-1][j]
                if j > 0:
                    val = val ^ xorVals[i][j-1]
                if i > 0 and j > 0:
                    val = val ^ xorVals[i-1][j-1]
                xorVals[i][j] = val
                heapq.heappush(xorHeap, -val)
        # print(xorVals)
        while (k > 1):
            heapq.heappop(xorHeap)
            k-=1

        return -heapq.heappop(xorHeap)
62900