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In this post, you will find the solution for the **Maximal Rectangle** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Maximal Rectangle– LeetCode Problem

Maximal Rectangle– LeetCode Problem

**Problem:**

Given a `rows x cols`

binary `matrix`

filled with `0`

‘s and `1`

‘s, find the largest rectangle containing only `1`

‘s and return *its area*.

**Example 1:**

Input:matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]Output:6Explanation:The maximal rectangle is shown in the above picture.

**Example 2:**

Input:matrix = [["0"]]Output:0

**Example 3:**

Input:matrix = [["1"]]Output:1

**Constraints:**

`rows == matrix.length`

`cols == matrix[i].length`

`1 <= row, cols <= 200`

`matrix[i][j]`

is`'0'`

or`'1'`

.

Maximal Rectangle– LeetCode Solutions

Maximal Rectangle in C++:

class Solution { public: int maximalRectangle(vector>& matrix) { if (matrix.empty()) return 0; int ans = 0; vector hist(matrix[0].size()); for (const auto& row : matrix) { for (int i = 0; i < row.size(); ++i) hist[i] = row[i] == '0' ? 0 : hist[i] + 1; ans = max(ans, largestRectangleArea(hist)); } return ans; } private: int largestRectangleArea(const vector & heights) { int ans = 0; stack stack; for (int i = 0; i <= heights.size(); ++i) { while (!stack.empty() && (i == heights.size() || heights[stack.top()] > heights[i])) { const int h = heights[stack.top()]; stack.pop(); const int w = stack.empty() ? i : i - stack.top() - 1; ans = max(ans, h * w); } stack.push(i); } return ans; } };

Maximal Rectangle in Java:

class Solution { public int maximalRectangle(char[][] matrix) { if (matrix.length == 0) return 0; int ans = 0; int[] hist = new int[matrix[0].length]; for (char[] row : matrix) { for (int i = 0; i < row.length; ++i) hist[i] = row[i] == '0' ? 0 : hist[i] + 1; ans = Math.max(ans, largestRectangleArea(hist)); } return ans; } private int largestRectangleArea(int[] heights) { int ans = 0; Stackstack = new Stack<>(); for (int i = 0; i <= heights.length; ++i) { while (!stack.isEmpty() && (i == heights.length || heights[stack.peek()] > heights[i])) { final int h = heights[stack.pop()]; final int w = stack.isEmpty() ? i : i - stack.peek() - 1; ans = Math.max(ans, h * w); } stack.push(i); } return ans; } }

Maximal Rectangle in Python:

class Solution: def maximalRectangle(self, matrix: List[List[str]]) -> int: if not matrix: return 0 ans = 0 hist = [0] * len(matrix[0]) def largestRectangleArea(heights: List[int]) -> int: ans = 0 stack = [] for i in range(len(heights) + 1): while stack and (i == len(heights) or heights[stack[-1]] > heights[i]): h = heights[stack.pop()] w = i - stack[-1] - 1 if stack else i ans = max(ans, h * w) stack.append(i) return ans for row in matrix: for i, num in enumerate(row): hist[i] = 0 if num == '0' else hist[i] + 1 ans = max(ans, largestRectangleArea(hist)) return ans