💯Faster✅💯Lesser✅Detailed Explaination🎯Stack🔥Height🧠Step-by-Step Explaination✅Python🐍Java🍵
\uD83D\uDE80 Hi, I\'m Mohammed Raziullah Ansari, and I\'m excited to share solution to this question with detailed explanation:\n\n# \uD83C\uDFAFProblem Explain
# \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share solution to this question with detailed explanation:\n\n# \uD83C\uDFAFProblem Explaination: \n- We have a 2D array `matrix` filled with `0\'s` and `1\'s`.\n- We need to find the largest rectangle containing only 1\'s and return its `area`.\n# \uD83E\uDDE0Thinking Behind the Solution:\n\nLet\'s break down the solution process into two distinct sub-processes: **Processing Height Array** and **Max Area Calculation**.\n\n### 1\uFE0F\u20E3Processing Height Array\u2705:\n1. **Initialization**:\n   - Create an array `height` initialized with zeros, with length equal to the number of columns in the matrix. This array will represent the heights of bars in a histogram.\n\n2. **Processing Each Row**:\n   - For each row `currRow` in the 2D matrix:\n     - Traverse through each element of `currRow`.\n     - If the element is `1`, increment the corresponding index in the `height` array.\n     - If the element is `0`, reset the corresponding index in the `height` array to `0`.\n\n### 2\uFE0F\u20E3Max Area Calculation (Naive Approach)\uD83E\uDD13:\n- Now that we got `height` array assume that you are solving `Leetcode Problem 84. Largest Rectangle in Histogram` where we are given an array of integers `heights` representing the histogram\'s bar height where the width of each bar is 1.\n- We want to find and return the area of the largest rectangle in the histogram.\n\n**Approach**:\n   - Iterate over all possible pairs of bars (i, j) where i < j.\n   - For each pair (i, j), determine the minimum height `h` between the bars from index `i` to `j`.\n   - Calculate the area of the rectangle formed by this pair of bars, which is `area = h * (j - i + 1)`.\n   - Keep track of the maximum area found during these iterations.\n\n#### Code\uD83D\uDC69\uD83C\uDFFB\u200D\uD83D\uDCBB(TLE\u274C):\n\n\n```python []\nclass Solution:\n    def maximalRectangle(self, matrix: List[List[str]]) -> int:\n        if not matrix:\n            return 0\n        \n        rows, cols = len(matrix), len(matrix[0])\n        heights = [0] * (cols + 1)  # Include an extra element for easier calculation\n        max_area = 0\n        \n        for row in matrix:\n            for i in range(cols):\n                heights[i] = heights[i] + 1 if row[i] == \'1\' else 0\n            \n            # Calculate max area using histogram method\n            n = len(heights)  # Number of bars in the histogram\n\n            for i in range(n):\n                for j in range(i, n):\n                    # Determine the minimum height between bar i and bar j\n                    min_height = min(heights[k] for k in range(i, j + 1))\n                    # Calculate the area of the rectangle\n                    area = min_height * (j - i + 1)\n                    # Update maximum area if the current rectangle\'s area is larger\n                    if area > max_area:\n                        max_area = area\n\n        return max_area\n        \n```\n     \n```Java []\nclass Solution {\n    public int maximalRectangle(char[][] matrix) {\n        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)\n            return 0;\n        \n        int rows = matrix.length;\n        int cols = matrix[0].length;\n        int[] heights = new int[cols + 1]; // Include an extra element for easier calculation\n        int maxArea = 0;\n        \n        for (char[] row : matrix) {\n            for (int i = 0; i < cols; i++) {\n                heights[i] = (row[i] == \'1\') ? heights[i] + 1 : 0;\n            }\n            \n            // Calculate max area using histogram method\n            int n = heights.length; // Number of bars in the histogram\n            \n            for (int i = 0; i < n; i++) {\n                for (int j = i, minHeight = Integer.MAX_VALUE; j < n; j++) {\n                    minHeight = Math.min(minHeight, heights[j]);\n                    int area = minHeight * (j - i + 1);\n                    maxArea = Math.max(maxArea, area);\n                }\n            }\n        }\n        \n        return maxArea;\n    }\n}\n\n```\n     \n```C++ []\nclass Solution {\npublic:\n    int maximalRectangle(vector<vector<char>>& matrix) {\n        if (matrix.empty() || matrix[0].empty())\n            return 0;\n        \n        int rows = matrix.size();\n        int cols = matrix[0].size();\n        vector<int> heights(cols + 1, 0); // Include an extra element for easier calculation\n        int maxArea = 0;\n        \n        for (const auto& row : matrix) {\n            for (int i = 0; i < cols; i++) {\n                heights[i] = (row[i] == \'1\') ? heights[i] + 1 : 0;\n            }\n            \n            // Calculate max area using histogram method\n            int n = heights.size(); // Number of bars in the histogram\n            \n            for (int i = 0; i < n; i++) {\n                for (int j = i, minHeight = INT_MAX; j < n; j++) {\n                    minHeight = min(minHeight, heights[j]);\n                    int area = minHeight * (j - i + 1);\n                    maxArea = max(maxArea, area);\n                }\n            }\n        }\n        \n        return maxArea;\n    }\n};\n```\n\n**Complexity**:\nThe naive approach examines all possible rectangles by iterating through each pair of bars in the histogram, resulting in a time complexity of O(n^3), where n is the number of bars. \n\n![Screenshot 2024-04-13 105011.png](https://assets.leetcode.com/users/images/253b83d7-1b98-401e-85db-f74b22828c68_1712985627.9949334.png)\n\n**Optimization**:\nTo optimize, identify and address inefficiencies in the brute-force method. This involves recognizing repetitive calculations, such as repeatedly finding the minimum height between pairs of bars. We can use techniques like dynamic programming (DP) or stack-based methods to streamline the process and reduce unnecessary recalculations, ultimately improving efficiency and scalability for larger datasets.\n\n### 2\uFE0F\u20E3Max Area Calculation (Using Stack-Based Approach)\uD83D\uDE0E:\n1. **Initialization**:\n   - Initialize a stack `stack` to keep track of indices of bars in the histogram.\n   - Initialize `max_area` to store the maximum rectangle area found.\n\n2. **Iterate Through Each Bar**:\n   - Traverse each bar\'s height in the `height` array from left to right.\n\n3. **Stack Operations**:\n   - For each bar at index `i`:\n     - While the stack is not empty and the current bar\'s height (`height[i]`) is less than the height of the bar represented by the index on top of the stack (`height[stack[-1]]`):\n       - Pop the top index `j` from the stack.\n       - Calculate the area with the popped bar\'s height:\n         - `h = height[j]`\n         - `w = i - stack[-1] - 1` (if stack is not empty), otherwise `w = i`\n         - `area = h * w`\n       - Update `max_area` with the maximum of `max_area` and `area`.\n\n4. **Final Maximum Area**:\n   - After processing all bars in the `height` array:\n     - Check if the stack is not empty:\n       - Pop each remaining index `j` from the stack and calculate the area similarly to step 3.\n\n5. **Return Result**:\n   - `max_area` will hold the value of the largest rectangle that can be formed within the histogram.\n\n\n# Let\'s walkthrough\uD83D\uDEB6\uD83C\uDFFB\u200D\u2642\uFE0F the implementation process with an example for better understanding\uD83C\uDFAF:\nLet\'s walk through the code step by step with the input `matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]`.\n- Iteration 1:\n```\nCurrent Row Array: [\'1\', \'0\', \'1\', \'0\', \'0\']\nHeight Array: [1, 0, 1, 0, 0]\nCurrent Histogram:\n# #  \nMax Area: 1\n```\n- Iteration 2:\n```\nCurrent Row Array: [\'1\', \'0\', \'1\', \'1\', \'1\']\nHeight Array: [2, 0, 2, 1, 1]\nCurrent Histogram:\n# #  \n# ###\nMax Area: 3\n```\n- Iteration 3:\n```\nCurrent Row Array: [\'1\', \'1\', \'1\', \'1\', \'1\']\nHeight Array: [3, 1, 3, 2, 2]\nCurrent Histogram:\n# #  \n# ###\n#####\nMax Area: 6\n```\n- Iteration 4:\n```\nCurrent Row Array: [\'1\', \'0\', \'0\', \'1\', \'0\']\nHeight Array: [4, 0, 0, 3, 0]\nCurrent Histogram:\n#    \n#  # \n#  # \n#  # \nMax Area: 6\n```\n\n# Code\uD83D\uDC68\uD83C\uDFFB\u200D\uD83D\uDCBB:\n```Python []\nclass Solution:\n    def maximalRectangle(self, matrix: List[List[str]]) -> int:\n        if not matrix:\n            return 0\n        \n        rows, cols = len(matrix), len(matrix[0])\n        heights = [0] * (cols + 1)  # Include an extra element for easier calculation\n        max_area = 0\n        \n        for row in matrix:\n            for i in range(cols):\n                heights[i] = heights[i] + 1 if row[i] == \'1\' else 0\n            \n            # Calculate max area using histogram method\n            stack = []\n            for i in range(len(heights)):\n                while stack and heights[i] < heights[stack[-1]]:\n                    h = heights[stack.pop()]\n                    w = i if not stack else i - stack[-1] - 1\n                    max_area = max(max_area, h * w)\n                stack.append(i)\n        \n        return max_area               \n```\n```Java []\nclass Solution {\n    public int maximalRectangle(char[][] matrix) {\n        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)\n            return 0;\n\n        int rows = matrix.length;\n        int cols = matrix[0].length;\n        int[] heights = new int[cols + 1]; // Include an extra element for easier calculation\n        int maxArea = 0;\n\n        for (char[] row : matrix) {\n            for (int i = 0; i < cols; i++) {\n                heights[i] = (row[i] == \'1\') ? heights[i] + 1 : 0;\n            }\n\n            // Calculate max area using stack-based method\n            Stack<Integer> stack = new Stack<>();\n            for (int i = 0; i < heights.length; i++) {\n                while (!stack.isEmpty() && heights[i] < heights[stack.peek()]) {\n                    int h = heights[stack.pop()];\n                    int w = stack.isEmpty() ? i : i - stack.peek() - 1;\n                    maxArea = Math.max(maxArea, h * w);\n                }\n                stack.push(i);\n            }\n        }\n\n        return maxArea;\n    }\n}\n\n```\n```C++ []\nclass Solution {\npublic:\n    int maximalRectangle(vector<vector<char>>& matrix) {\n        if (matrix.empty() || matrix[0].empty())\n            return 0;\n\n        int rows = matrix.size();\n        int cols = matrix[0].size();\n        vector<int> heights(cols + 1, 0); // Include an extra element for easier calculation\n        int maxArea = 0;\n\n        for (const auto& row : matrix) {\n            for (int i = 0; i < cols; i++) {\n                heights[i] = (row[i] == \'1\') ? heights[i] + 1 : 0;\n            }\n\n            // Calculate max area using stack-based method\n            stack<int> stk;\n            for (int i = 0; i < heights.size(); i++) {\n                while (!stk.empty() && heights[i] < heights[stk.top()]) {\n                    int h = heights[stk.top()];\n                    stk.pop();\n                    int w = stk.empty() ? i : i - stk.top() - 1;\n                    maxArea = max(maxArea, h * w);\n                }\n                stk.push(i);\n            }\n        }\n\n        return maxArea;\n    }\n};\n\n```\n
['Array', 'Stack', 'Matrix', 'Monotonic Stack', 'C++', 'Java', 'Python3']