✔ Simple C++ Solution
\n int kthLargestValue(vector<vector<int>>& matrix, int k) {\n        priority_queue<int,vector<int>,greater<int>> min_heap;\n       \n        for(int i=0;i<mat
```\n int kthLargestValue(vector<vector<int>>& matrix, int k) {\n        priority_queue<int,vector<int>,greater<int>> min_heap;\n       \n        for(int i=0;i<matrix.size();i++){\n            for(int j=0;j<matrix[i].size();j++){\n                if(i>0 && j>0) matrix[i][j]= (matrix[i-1][j]^matrix[i][j-1])^matrix[i][j]^matrix[i-1][j-1];\n                else if(i>0 && j<=0) matrix[i][j]=matrix[i-1][j]^matrix[i][j];\n                else if(i<=0 && j>0) matrix[i][j]=matrix[i][j-1]^matrix[i][j];\n              \n                min_heap.push(matrix[i][j]);\n            if(min_heap.size()>k)\n                    min_heap.pop();\n            }\n        }\n        \n    \n        return min_heap.top();\n    }\n```
['C', 'Heap (Priority Queue)']