Dynamic programming approach
\n# Approach\n Describe your approach to solving the problem. \nThe current approach iterates over each row and column, updating the matrix values and then sort
\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe current approach iterates over each row and column, updating the matrix values and then sorting for each row. To make it more efficient, we can keep track of the height of consecutive ones for each cell in a separate array. This eliminates the need to sort for each row, improving the time complexity.\n\nThe updated approach calculates the heights for each column, maintains an array of heights, and sorts it once. Then, it iterates over the rows, updating the heights array and computing the maximum submatrix sum. This avoids redundant sorting and should result in a slight improvement in efficiency.\n\n# Complexity\n- Time complexity: \n     O(m*nlog n)\n \n\n- Space complexity:\n     O(n)\n\n# Code\n```\nclass Solution {\n    public int largestSubmatrix(int[][] matrix) {\n        int m = matrix.length;\n        int n = matrix[0].length;\n        int ans = 0;\n        int[] heights = new int[n];\n        for (int i = 0; i < m; i++) {\n            for (int j = 0; j < n; j++) {\n                if (matrix[i][j] == 0) {\n                    heights[j] = 0;\n                } else {\n                    heights[j] += matrix[i][j];\n                }\n            }\n            int[] sortedHeights = Arrays.copyOf(heights, n);\n            Arrays.sort(sortedHeights);\n            for (int k = 0; k < n; k++) {\n                ans = Math.max(ans, sortedHeights[k] * (n - k));\n            }\n        }\n        return ans;\n    }\n}\n```
['Java']