C# Solution for Largest Submatrix With Rearrangements Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThis solution maintains a count of consecutive 1s vertically in the input matrix and th
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis solution maintains a count of consecutive 1s vertically in the input matrix and then sorts each row to find the maximum area of the submatrix containing only 1s after rearranging the columns optimally.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tIterate through the matrix starting from the second row.\n2.\tFor each cell containing \u20181\u2019, add the value of the cell above it. This operation accumulates the count of consecutive 1s vertically.\n3.\tSort each row of the matrix.\n4.\tTraverse each row and calculate the maximum area by considering the maximum consecutive 1s in a row, multiplied by its width.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tThe iteration through the matrix to update the counts of consecutive 1s vertically takes O(m * n) time.\n\u2022\tSorting each row takes O(n * log n) time, and since it\u2019s performed for each of the \u2018m\u2019 rows, the overall sorting takes O(m * n * log n) time.\n\u2022\tFinding the maximum area by traversing the sorted rows takes O(m * n) time.\nHence, the overall time complexity is O(m * n * log n), dominated by the sorting step.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tThe space complexity is O(1) because the operations are performed in place on the input matrix.\n\u2022\tAdditionally, there\u2019s some extra space used for variables and temporary storage, but they don\u2019t significantly impact the overall space complexity.\n\n# Code\n```\npublic class Solution {\n    public int LargestSubmatrix(int[][] matrix) {\n        int m = matrix.Length;\n        int n = matrix[0].Length;\n\n        // Update the matrix to maintain the count of consecutive 1s vertically\n        for (int i = 1; i < m; i++) {\n            for (int j = 0; j < n; j++) {\n                if (matrix[i][j] == 1) {\n                    matrix[i][j] += matrix[i - 1][j];\n                }\n            }\n        }\n\n        // Sort each row to find the maximum area\n        int maxArea = 0;\n        for (int i = 0; i < m; i++) {\n            Array.Sort(matrix[i]);\n            for (int j = n - 1; j >= 0; j--) {\n                maxArea = Math.Max(maxArea, matrix[i][j] * (n - j));\n            }\n        }\n\n        return maxArea;\n    }\n}\n```
['C#']