Java documented code
\nclass Solution {\n    public int kthLargestValue(int[][] matrix, int k) {\n        \n        //This is basic checking done for the matrix.\n        int m = ma
```\nclass Solution {\n    public int kthLargestValue(int[][] matrix, int k) {\n        \n        //This is basic checking done for the matrix.\n        int m = matrix.length;\n        if(m==0)\n            return -1;\n        int n = matrix[0].length;\n        if(n==0)\n            return -1;\n        \n        //This is the dp matrix which will contain all the xor values\n        int[][] dp = new int[m][n];\n        //We use this list to keep all the integers\n         List<Integer> list = new ArrayList<>();\n        \n        //This is the base case . THe first element at (0,0) index doesnot have any other values to xor with. We add that element to the list\n        dp[0][0] = matrix[0][0];\n        list.add(dp[0][0]);\n        \n        //This is for the first column. Our required value is the dp[i-1][0]^matrix[i][0]. We do this for the whole first column\n        for(int i=1;i<m;i++){\n            dp[i][0] = dp[i-1][0] ^ matrix[i][0];\n            list.add(dp[i][0]);\n        }\n        //This is for the first row. Our required value is the dp[0][i-1]^matrix[0][i]. We do this for the whole first row\n       \n        for(int i=1;i<n;i++){\n            dp[0][i] = dp[0][i-1] ^ matrix[0][i];\n            list.add(dp[0][i]);\n        }\n        \n        \n        //For all the other elements the xor value would be dp[i-1][j]^dp[i][j-1]^dp[i-1][j-1]^matrix[i][j].We add the elements to the list.\n        for(int i=1;i<m;i++){\n            for(int j=1;j<n;j++){\n                dp[i][j] = dp[i-1][j]^dp[i][j-1]^dp[i-1][j-1]^matrix[i][j];\n                list.add(dp[i][j]);\n            }\n        }\n        \n        //AS we need the Kth largest element , as we have sorted it in increasing order the kth largest element will be m*n-k th element in the ascending order\n        Collections.sort(list);\n        \n        return list.get(m*n-k);\n        \n        \n    }\n}\n```
['Java']