Maximal Rectangle

Description:

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 formed by the 1s in the middle two columns and bottom three rows.

Example 2:

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

Example 3:

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

---

Approaches:

1. Monotonic Stack (Histogram Approach)

This problem is an extension of the "Largest Rectangle in Histogram" problem. We can treat each row of the matrix as the base of a histogram.

1. Create a heights array to store the accumulated height of consecutive 1s for each column.
2. Iterate through each row. If a cell is 1, we increment its corresponding height. If it is 0, we reset the height to 0.
3. For each row's updated heights array, apply the monotonic stack approach to find the largest rectangle area.
4. Maintain a max_area variable to keep track of the maximum area found across all rows.

Interactive Visualizer

Click the cells below to toggle between 0 and 1, then calculate to see the maximal rectangle.

• Time Complexity: $O(R \times C)$ where $R$ is the number of rows and $C$ is the number of columns. We visit each cell a constant number of times while updating heights and processing the stack.
• Space Complexity: $O(C)$ because we use an array of size $C + 1$ to store the heights and a stack that can grow up to size $C$.

Solutions:

C++

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) return 0;
        
        int cols = matrix[0].size();
        vector<int> heights(cols + 1, 0);
        int maxArea = 0;
        
        for (const auto& row : matrix) {
            for (int i = 0; i < cols; i++) {
                heights[i] = row[i] == '1' ? heights[i] + 1 : 0;
            }
            
            vector<int> stack = {-1};
            for (int i = 0; i <= cols; i++) {
                while (stack.back() != -1 && heights[i] < heights[stack.back()]) {
                    int h = heights[stack.back()];
                    stack.pop_back();
                    int w = i - 1 - stack.back();
                    maxArea = max(maxArea, h * w);
                }
                stack.push_back(i);
            }
        }
        
        return maxArea;
    }
};

Java

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

Python

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

JavaScript

/**
 * @param {character[][]} matrix
 * @return {number}
 */
const maximalRectangle = function(matrix) {
    if (matrix.length === 0 || matrix[0].length === 0) return 0;
    
    const cols = matrix[0].length;
    const heights = new Array(cols + 1).fill(0);
    let maxArea = 0;
    
    for (let row of matrix) {
        for (let i = 0; i < cols; i++) {
            heights[i] = row[i] === '1' ? heights[i] + 1 : 0;
        }
        
        const stack = [-1];
        for (let i = 0; i <= cols; i++) {
            while (stack[stack.length - 1] !== -1 && heights[i] < heights[stack[stack.length - 1]]) {
                const h = heights[stack.pop()];
                const w = i - 1 - stack[stack.length - 1];
                maxArea = Math.max(maxArea, h * w);
            }
            stack.push(i);
        }
    }
    
    return maxArea;
};