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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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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Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemMaximal 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: 6
Explanation: 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;
    Stack stack = 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

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