SIMPLE MIN-HEAP C++ SOLUTION
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n#include <bits/stdc++.h>\n#define ll long long \n#define nl \'\\n\'\nclass Solution {\npublic:\n    int kthLargestValue(vector<vector<int>>& matrix,int k) {\n        int m = matrix.size();\n        int n = matrix[0].size();\n        vector<vector<int>>xorMatrix(m,vector<int>(n, 0));\n     priority_queue<int,vector<int>,greater<int>>minHeap;\n        // Compute the XOR matrix\n        for(int i = 0; i < m; i++){\n            for(int j = 0; j < n; j++){\n                if(i == 0 && j == 0){\n                    xorMatrix[i][j] = matrix[i][j];\n                }else if (i == 0){\n                    xorMatrix[i][j] = xorMatrix[i][j-1] ^ matrix[i][j];\n                }else if (j == 0){\n                    xorMatrix[i][j] = xorMatrix[i-1][j] ^ matrix[i][j];\n                }else{\n                    xorMatrix[i][j] = xorMatrix[i-1][j] ^ xorMatrix[i][j-1] ^ xorMatrix[i-1][j-1] ^ matrix[i][j];\n                }\n                // Maintain a min-heap of the k largest values\n                if(minHeap.size() < k){\n                    minHeap.push(xorMatrix[i][j]);\n                }else if(xorMatrix[i][j] > minHeap.top()){/// maximum xormatrix[i][j] value in the top of heap\n                    minHeap.pop();\n                    minHeap.push(xorMatrix[i][j]);\n                }\n            }\n        }\n        return minHeap.top();\n    }\n\n//Matrix:        XOR Matrix:\n// 5  2           5  7\n// 1  6           4  4\n};\n```
['C++']