Java easy and concise solution PriorityQueue
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```\nclass Solution {\n    public int kthLargestValue(int[][] matrix, int k) {\n        PriorityQueue<Integer> pq = new PriorityQueue<>();\n        int m = matrix.length;\n        int n = matrix[0].length;\n\n        int[][] ans = new int[m+1][n+1];\n\n        for(int i=0;i<m;i++){\n            for(int j=0;j<n;j++){\n                ans[i+1][j+1] = matrix[i][j]^ans[i+1][j]^ans[i][j+1]^ans[i][j];\n                pq.offer(ans[i+1][j+1]);\n                if(pq.size()>k)pq.poll();\n            }\n        }\n        return pq.peek();\n    }\n}\n```
['Java']