Maximal Rectangle

class Solution {
public:
    int maximumHistogramArea(vector<vector<char>>& grid) {
        if (grid.empty()) return 0;
        int rows = grid.size();
        int cols = grid[0].size();
        vector<int> minHeight(cols), maxRight(cols), maxLeft(cols);
        fill(maxRight.begin(), maxRight.end(), cols);
        int maxRectangleArea = 0;
        for (int i = 0; i < rows; i++) {
            int openLeft = 0, openRight = cols;
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1')
                    minHeight[j]++;
                else
                    minHeight[j] = 0;
            }
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1')
                    maxLeft[j] = max(maxLeft[j], openLeft);
                else {
                    maxLeft[j] = 0;
                    openLeft = j + 1;
                }
            }
            for (int j = cols - 1; j >= 0; j--) {
                if (grid[i][j] == '1')
                    maxRight[j] = min(maxRight[j], openRight);
                else {
                    maxRight[j] = cols;
                    openRight = j;
                }
            }
            for (int j = 0; j < cols; j++) {
                maxRectangleArea = max(maxRectangleArea, (maxRight[j] - maxLeft[j]) * minHeight[j]);
            }
        }
        return maxRectangleArea;
    }
};

This solution presents an efficient algorithm implemented in C++ to find the area of the largest rectangle filled only with '1's in a binary matrix. The key approach is to leverage the concept of histograms to compute maximum areas in each row of the matrix, integrating both left and right limits for each cell that contains a '1'.

• Begin by handling edge cases, returning 0 if the input grid is empty. Extract the number of rows and columns from the grid.
• Introduce vectors minHeight, maxRight, and maxLeft to store the minimum height up to the current row, the maximal right and left boundaries where the histogram extends without dropping down. Initialize maxRight with the number of columns (cols) as it represents the right limit for each bar in the histogram.
• Utilize a main loop to iterate through each row of the grid. Within this loop:
  • Update minHeight by incrementing for a '1' or resetting to 0 for a '0', representing the height of the histogram bar at that column.
  • Use a secondary loop from left to right to determine maxLeft, which is the farthest left that the current bar extends without interruption by a '0'. This loop also manages an openLeft index to keep track of openings.
  • A similar loop, from right to left, updates maxRight. This loop uses an openRight index to smoothly mark the right limit of the histogram bar when uninterrupted by a '0'.
  • After setting the boundaries via maxLeft and maxRight, calculate the possible rectangle area for each element in the current row. Update maxRectangleArea to ensure the maximum is retained after evaluating each row.
• Finally, the result stored in maxRectangleArea is the area of the largest rectangle, and it is returned.

This technique optimally identifies the maximal rectangular region row by row, balancing complexity and performance and ensuring an effective solution to the problem.

class Solution {
    public int maxRectangleSize(char[][] grid) {
        if (grid.length == 0) return 0;
        int rows = grid.length;
        int cols = grid[0].length;

        int[] minHeight = new int[cols]; 
        int[] maxLeft = new int[cols]; 
        int[] maxRight = new int[cols]; 

        Arrays.fill(maxRight, cols); 

        int largestArea = 0;
        for (int i = 0; i < rows; i++) {
            int currentLeft = 0, currentRight = cols;
            // update heights
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1') minHeight[j]++;
                else minHeight[j] = 0;
            }
            // update maxLeft
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1') maxLeft[j] = Math.max(maxLeft[j], currentLeft);
                else {
                    maxLeft[j] = 0;
                    currentLeft = j + 1;
                }
            }
            // update maxRight
            for (int j = cols - 1; j >= 0; j--) {
                if (grid[i][j] == '1') maxRight[j] = Math.min(maxRight[j], currentRight);
                else {
                    maxRight[j] = cols;
                    currentRight = j;
                }
            }
            // calculate max area
            for (int j = 0; j < cols; j++) {
                largestArea = Math.max(largestArea, (maxRight[j] - maxLeft[j]) * minHeight[j]);
            }
        }
        return largestArea;
    }
}

The provided solution in Java efficiently finds the maximal rectangle composed entirely of '1's in a binary matrix. Here's a clear breakdown of the approach employed:

1. Initialize auxiliary arrays minHeight, maxLeft, and maxRight for calculations regarding the heights and boundaries of potential rectangles.
2. Use Arrays.fill(maxRight, cols); for initializing the maxRight array to the number of columns as the initial maximum right boundary.
3. Iterate through each row of the matrix and execute the following:
   • Update the height (minHeight) for each column to track the accumulated number of consecutive '1's vertically up to the current row.
   • Determine the maximum left boundary (maxLeft) for each column by comparing the current index boundary kept in currentLeft or by resetting upon encountering a '0'.
   • Similarly, identify the minimum right boundary (maxRight) by decrementing from the end of the row towards the start, adjusting boundaries, or reset upon encountering a '0'.
   • Calculate the maximum area rectangle possible for each column using the formula (maxRight[j] - maxLeft[j]) * minHeight[j] and update the maximum area found.
4. The maximum found area is finally returned.

This solution leverages dynamic programming principles where each cell contributes to the potential maximum rectangle in which it can be the bottom-right corner, thus efficiently reducing the complexity compared to brute force approaches. The adjustments of left and right boundaries during each row scan are crucial for accurately determining the possible widths of rectangles while the continuous update of height values serves to calculate potential areas dynamically.

#define maximum(x, y) ((x) > (y) ? (x) : (y))
#define minimum(x, y) ((x) < (y) ? (x) : (y))
int largestRectangle(char** grid, int numRows, int* numCols) {
    if (numRows == 0) return 0;
    int rows = numRows;
    int cols = numCols[0];
    int leftBound[cols];
    int rightBound[cols];
    int heights[cols];
    memset(heights, 0, sizeof(int) * cols);
    for (int i = 0; i < cols; i++) {
        leftBound[i] = 0;
        rightBound[i] = cols;
    }
    int maxArea = 0;
    for (int i = 0; i < rows; i++) {
        int leftLimit = 0, rightLimit = cols;
        for (int j = 0; j < cols; j++) {
            if (grid[i][j] == '1')
                heights[j]++;
            else
                heights[j] = 0;
        }
        for (int j = 0; j < cols; j++) {
            if (grid[i][j] == '1')
                leftBound[j] = maximum(leftBound[j], leftLimit);
            else {
                leftBound[j] = 0;
                leftLimit = j + 1;
            }
        }
        for (int j = cols - 1; j >= 0; j--) {
            if (grid[i][j] == '1')
                rightBound[j] = minimum(rightBound[j], rightLimit);
            else {
                rightBound[j] = cols;
                rightLimit = j;
            }
        }
        for (int j = 0; j < cols; j++)
            maxArea = maximum(maxArea, (rightBound[j] - leftBound[j]) * heights[j]);
    }
    return maxArea;
}

This solution involves finding the maximum rectangular area in a binary grid where '1' represents that the cell is filled, and '0' represents the cell is empty. The primary approach used in this C program leverages dynamic programming combined with the largest rectangle in histogram problem-solving technique.

The key components include:

• Intializing arrays to keep track of the left and right boundaries (leftBound, rightBound) and heights of potential rectangles (heights) at each column for each row.
• Iterating through each row, for each cell where the value is '1':
  • Update the heights array to count consecutive 1s vertically.
  • Adjust leftBound using the maximum of the current boundary and the last seen position of '0' on the left, ensuring the boundary is squeezed as closely as possible to the right around blocks of 1s.
  • Update rightBound using the minimum of the current boundary and the last seen position of '0' on the right, squeezing the boundary to the left around blocks of 1s.
• The area for any position (i, j) is calculated as (rightBound[j] - leftBound[j]) * heights[j], ensuring the largest possible rectangle made up solely of 1s that can originate from any particular starting position.
• Maintain a running tally of maxArea to hold the maximum area found so far.

Every time a 0 is encountered:

• The heights at that column is reset to 0 because a 0 breaks any rectangle that could have been formed involving cells above the 0.
• The left limit is reset because any new rectangle must start after this column.
• The right limit is reset in a similar volumetric manner for rectangles extending to the left.

This efficient calculation at each step ensures that the maximum rectangle of 1’s is determined with optimum boundary conditions without having to resort to brute-force checking of every possible rectangle in the grid.

var largestRectangleArea = function (grid) {
    if (grid.length === 0) return 0;
    let row = grid.length;
    let col = grid[0].length;
    let leftLimit = Array(col).fill(0);
    let rightLimit = Array(col).fill(col);
    let heights = Array(col).fill(0);
    let maximumArea = 0;
    for (let i = 0; i < row; i++) {
        let leftBoundary = 0, rightBoundary = col;
        for (let j = 0; j < col; j++) {
            if (grid[i][j] === "1") heights[j]++;
            else heights[j] = 0;
        }
        for (let j = 0; j < col; j++) {
            if (grid[i][j] === "1") leftLimit[j] = Math.max(leftLimit[j], leftBoundary);
            else {
                leftLimit[j] = 0;
                leftBoundary = j + 1;
            }
        }
        for (let j = col - 1; j >= 0; j--) {
            if (grid[i][j] === "1") rightLimit[j] = Math.min(rightLimit[j], rightBoundary);
            else {
                rightLimit[j] = col;
                rightBoundary = j;
            }
        }
        for (let j = 0; j < col; j++) {
            maximumArea = Math.max(maximumArea, (rightLimit[j] - leftLimit[j]) * heights[j]);
        }
    }
    return maximumArea;
};

Solve the "Maximal Rectangle" problem using the provided JavaScript function largestRectangleArea. The function determines the area of the largest rectangle containing only '1's in a 2D binary matrix.

• Understand the problem by noting that it requires finding the maximum area of a continuous series of '1's in a grid that is potentially rectangular.
• The function initializes dimensions (row and col), along with arrays leftLimit, rightLimit, and heights to track constraints for each column during matrix traversal.
• Loop through each row to dynamically update the height of series of '1's using the heights array.
• For each element in the row, update leftLimit to hold the leftmost valid starting index for a rectangle, and rightLimit to hold the rightmost valid ending index.
• While traversing each column in the grid, use leftLimit and rightLimit to determine permissible boundaries of the rectangle and calculate potential areas.
• Continuously update maximumArea if a larger area is found.

The function iterates over the grid with a complexity roughly O(n*m) where n and m are the number of rows and columns of the grid, respectively. It efficiently calculates maximum rectangle area by combining principles from histograms and dynamic programming.

class Solution:

    def largestRectangle(self, grid: List[List[str]]) -> int:
        if not grid:
            return 0

        rows = len(grid)
        columns = len(grid[0])

        left_bound = [0] * columns  # left boundary for maximum rectangle
        right_bound = [columns] * columns  # right boundary for maximum rectangle
        heights = [0] * columns

        max_area = 0

        for i in range(rows):

            current_left, current_right = 0, columns
            # Compute height
            for j in range(columns):
                if grid[i][j] == "1":
                    heights[j] += 1
                else:
                    heights[j] = 0
            # Compute left bounds
            for j in range(columns):
                if grid[i][j] == "1":
                    left_bound[j] = max(left_bound[j], current_left)
                else:
                    left_bound[j] = 0
                    current_left = j + 1
            # Compute right bounds
            for j in range(columns - 1, -1, -1):
                if grid[i][j] == "1":
                    right_bound[j] = min(right_bound[j], current_right)
                else:
                    right_bound[j] = columns
                    current_right = j
            # Compute maximum area
            for j in range(columns):
                max_area = max(max_area, heights[j] * (right_bound[j] - left_bound[j]))

        return max_area

You will solve the problem of finding the maximum rectangular area in a grid filled with binary values ('0' and '1') using Python. The given code defines the largestRectangle function, which accepts a grid of binary values, and through efficient computations, determines the largest rectangle that can be formed from connected '1's.

• Initialize variables for the number of rows and columns within the grid. Create arrays to track the heights of consecutive '1's, as well as the left and right boundaries of potential rectangles.
• Iterate over each row in the grid to update heights for each column based on the presence of '1's. If '0' is encountered, reset height to zero.
• Calculate left boundaries by marking the start of a new potential rectangle whenever '1' follows a '0'.
• Calculate right boundaries in a reverse manner starting from the last column towards the first, determining the end of a rectangle span.
• For each column within a row, calculate the potential area formed between left and right boundaries and update maximum found area if the current is larger.
• Return the computed maximum area.

The solution utilizes dynamic programming by updating left and right boundary markers and heights progressively to optimize the performance, avoiding redundant recalculations. Make sure you understand how dynamic boundaries and heights influence rectangle calculation in a grid. This approach ensures that each cell contributes to a potential maximal rectangle efficiently.