Java Solution for Maximal Rectangle Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe above solution uses a dynamic programming approach to calculate the height of the l
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe above solution uses a dynamic programming approach to calculate the height of the largest rectangle that can be formed with the current cell at the bottom of the rectangle. It then applies a modified version of the algorithm for finding the maximum area of a histogram on each row of the matrix to find the maximum area of a rectangle that can be formed using that row as the bottom edge.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach first creates a new matrix height where height[i][j] represents the height of the largest rectangle that can be formed with cell (i,j) as the bottom cell. If matrix[i][j] is \'0\', then height[i][j] is set to 0, since no rectangle can be formed with a bottom cell of \'0\'. Otherwise, height[i][j] is set to height[i-1][j] + 1 if i > 0, since we can extend a rectangle from the cell above it, or to 1 if i == 0, since the current cell is the topmost cell of the rectangle.\n\nOnce the height matrix has been constructed, the solution applies the modified histogram algorithm to each row of the height matrix to find the maximum area of a rectangle that can be formed using that row as the bottom edge. The modified histogram algorithm maintains a stack of indices of increasing heights of bars seen so far. When a bar with height smaller than the bar at the top of the stack is encountered, the bars in the stack are popped and their maximum area is calculated based on the height of the popped bar and the width of the rectangle, which is the difference between the current index and the index of the bar at the top of the stack (or the entire length of the stack if it is empty). The maximum area seen so far is updated if the current area is greater than it. Finally, the function returns the maximum area of a rectangle that can be formed using the heights of the bars in the given input array.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(m * n)\nThe code iterates over every cell in the input matrix once to calculate the height array, where m and n are the dimensions of the input matrix. It then calls the maxAreaInHist method once for each row in the matrix, which has a time complexity of O(n) for each row. Thus, the total time complexity is O(m * n) for the two nested loops and the maxAreaInHist method.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(m * n)\nThe code creates a 2D array height with m rows and n+1 columns to store the height of the histogram. Each row of the height array has n+1 elements, which increases the space complexity by an extra n elements per row. Thus, the space complexity is O(m * (n+1)), which can be simplified to O(m * n) asymptotically. Additionally, the code uses a stack to store indices, which can have a worst-case space complexity of O(n) if all elements are strictly decreasing in height\n\n# Code\n```\nclass Solution {\n    public int maximalRectangle(char[][] matrix) {\n        int m = matrix.length;\n        int n = m == 0 ? 0 : matrix[0].length;\n        int[][] height = new int[m][n + 1];\n        int maxArea = 0;\n        for (int i = 0; i < m; i++) \n        {\n            for (int j = 0; j < n; j++) \n            {\n                if (matrix[i][j] == \'0\') \n                {\n                    height[i][j] = 0;\n                } \n                else \n                {\n                    height[i][j] = i == 0 ? 1 : height[i - 1][j] + 1;\n                }\n            }\n        }\n        for (int i = 0; i < m; i++) \n        {\n            int area = maxAreaInHist(height[i]);\n            if (area > maxArea) \n            {\n                maxArea = area;\n            }\n        }\n        return maxArea;\n    }\n    private int maxAreaInHist(int[] height) {\n        Stack<Integer> stack = new Stack<Integer>();\n        int i = 0;\n        int max = 0;\n        while (i < height.length) \n        {\n            if (stack.isEmpty() || height[stack.peek()] <= height[i]) \n            {\n                stack.push(i++);\n            } \n            else \n            {\n                int t = stack.pop();\n                max = Math.max(max, height[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));\n            }\n        }\n        return max;\n    }\n\n}\n```
['Java']