Jobs

Interviews

Discussions

Find the Minimum Area to Cover All Ones I

Medium

Salesforce

0 views

Topics:

Arrays

You are given a 2D binary array grid. Find a rectangle with horizontal and vertical sides with the smallest area, such that all the 1's in grid lie inside this rectangle.

Return the minimum possible area of the rectangle.

Example 1:

Input: grid = [[0,1,0],[1,0,1]]

Output: 6

Explanation:

The smallest rectangle has a height of 2 and a width of 3, so it has an area of 2 * 3 = 6.

Example 2:

Input: grid = [[1,0],[0,0]]

Output: 1

Explanation:

The smallest rectangle has both height and width 1, so its area is 1 * 1 = 1.

Constraints:

1 <= grid.length, grid[i].length <= 1000
grid[i][j] is either 0 or 1.
The input is generated such that there is at least one 1 in grid.

Try coding it in LeetCode

Find the Minimum Area to Cover All Ones I

Medium

Salesforce

0 views

Topics:

Arrays

You are given a 2D binary array grid. Find a rectangle with horizontal and vertical sides with the smallest area, such that all the 1's in grid lie inside this rectangle.

Return the minimum possible area of the rectangle.

Example 1:

Input: grid = [[0,1,0],[1,0,1]]

Output: 6

Explanation:

The smallest rectangle has a height of 2 and a width of 3, so it has an area of 2 * 3 = 6.

Example 2:

Input: grid = [[1,0],[0,0]]

Output: 1

Explanation:

The smallest rectangle has both height and width 1, so its area is 1 * 1 = 1.

Constraints:

1 <= grid.length, grid[i].length <= 1000
grid[i][j] is either 0 or 1.
The input is generated such that there is at least one 1 in grid.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the grid (maximum number of rows and columns), and what is the expected range for these dimensions?
What should I return if the grid is empty or if it contains no 1s?
Is the grid guaranteed to be rectangular (i.e., all rows have the same number of columns)?
Are the grid values guaranteed to be only 0 or 1, or could there be other values?
Is there a possibility of integer overflow when calculating the area of the rectangle?

Brute Force Solution

Approach

The brute force strategy for this problem involves examining every possible rectangle that could cover all the ones. It essentially tests all rectangle combinations until it discovers the smallest one that works. This approach is simple to understand but not the most efficient.

Here's how the algorithm would work step-by-step:

Imagine every possible rectangle that you can draw on the grid.
For each rectangle, check if it covers all the 'ones' in the grid.
If a rectangle does not cover all the 'ones', discard it and move on to the next possible rectangle.
If a rectangle covers all the 'ones', remember the area of this rectangle.
After checking all possible rectangles, find the rectangle with the smallest area from the rectangles you remembered.
The area of this smallest rectangle is the answer.

Code Implementation

def find_minimum_area_brute_force(grid):
    if not grid or not grid[0]:
        return 0

    row_count = len(grid)
    column_count = len(grid[0])
    minimum_area = float('inf')

    # Iterate through all possible top-left corners
    for top_row in range(row_count):
        for left_column in range(column_count):

            # Iterate through all possible bottom-right corners
            for bottom_row in range(top_row, row_count):
                for right_column in range(left_column, column_count):

                    # Check if the current rectangle covers all ones
                    all_ones_covered = True
                    for row_index in range(row_count):
                        for column_index in range(column_count):
                            if grid[row_index][column_index] == 1:
                                if not (top_row <= row_index <= bottom_row and\
                                        left_column <= column_index <= right_column):
                                    all_ones_covered = False
                                    break
                        if not all_ones_covered:
                            break

                    # If all ones are covered, update minimum_area
                    if all_ones_covered:
                        current_area = (bottom_row - top_row + 1) * (right_column - left_column + 1)

                        # Keep track of smallest area
                        minimum_area = min(minimum_area, current_area)

    # Handle the case where no 1s are present in the grid
    if minimum_area == float('inf'):
        return 0

    return minimum_area

Big(O) Analysis

Time Complexity

O(m^6) – Let m be the larger dimension between the number of rows and columns in the grid. The brute force approach iterates through all possible rectangles. There are O(m^2) ways to choose the top-left corner of a rectangle and O(m^2) ways to choose the bottom-right corner, giving O(m^4) possible rectangles. For each rectangle, we iterate through all the cells in the grid which takes O(m^2) time to check if all ones are covered by the current rectangle. Therefore, the total time complexity is O(m^4 * m^2) = O(m^6).

Space Complexity

O(1) – The provided brute force algorithm iterates through all possible rectangles and checks if each covers all ones. The algorithm only needs to store the minimum area found so far. No auxiliary data structures dependent on the input grid's size (N, where N represents the number of cells in the grid) are used to store intermediate rectangle combinations or grid states. Therefore, the space complexity is constant.

Optimal Solution

Approach

To find the smallest rectangle covering all the '1's, we don't need to check every possible rectangle. Instead, we use the positions of the '1's to find the boundaries of the smallest possible rectangle directly. We can do this by finding the highest, lowest, leftmost, and rightmost '1' and using these to define the corners.

Here's how the algorithm would work step-by-step:

Find the topmost row that contains a '1'. This will be the top edge of the rectangle.
Find the bottommost row that contains a '1'. This will be the bottom edge of the rectangle.
Find the leftmost column that contains a '1'. This will be the left edge of the rectangle.
Find the rightmost column that contains a '1'. This will be the right edge of the rectangle.
Now that you have the top, bottom, left, and right edges, you know the boundaries of the smallest rectangle that contains all the '1's.
Calculate the area of the rectangle using its width and height (which you get from the boundaries you found).

Code Implementation

def find_minimum_area(grid):
    top_row = -1
    bottom_row = -1
    leftmost_column = -1
    rightmost_column = -1

    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                top_row = row_index
                break
        if top_row != -1:
            break

    for row_index in range(number_of_rows - 1, -1, -1):
        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                bottom_row = row_index
                break
        if bottom_row != -1:
            break

    for column_index in range(number_of_columns):
        for row_index in range(number_of_rows):
            if grid[row_index][column_index] == 1:
                leftmost_column = column_index
                break
        if leftmost_column != -1:
            break

    for column_index in range(number_of_columns - 1, -1, -1):
        for row_index in range(number_of_rows):
            if grid[row_index][column_index] == 1:
                rightmost_column = column_index
                break
        if rightmost_column != -1:
            break

    # Handle edge case where no 1s exist.
    if top_row == -1:
        return 0

    # Calculate the area based on extreme points found
    height = bottom_row - top_row + 1

    width = rightmost_column - leftmost_column + 1

    return height * width

Big(O) Analysis

Time Complexity

O(m * n) – The algorithm iterates through all the rows (m) and columns (n) of the matrix to find the topmost, bottommost, leftmost, and rightmost '1'. Finding the topmost row involves iterating through the rows until a row with a '1' is found. Similarly, finding the bottommost, leftmost, and rightmost involves iterating through the rows and columns, respectively, in the worst-case scenario going through all elements. Therefore, each of these operations takes O(m * n) time because in the worst case, we scan every element in the grid. Consequently, the overall time complexity is O(m * n).

Space Complexity

O(1) – The algorithm uses a fixed number of integer variables to store the topmost row, bottommost row, leftmost column, and rightmost column containing a '1'. The number of such variables doesn't depend on the size of the input matrix. Thus, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 immediately as there are no 1s to cover.
Grid with no 1s	Return 0 since the minimum area to cover nothing is zero.
Grid with only one 1	Return 1 as the minimum rectangle covering one 1 is a 1x1 rectangle.
Grid with all 1s	The algorithm should correctly calculate the area of the entire grid in this case.
Grid with one row or one column	The algorithm should correctly identify the min/max indices of the 1s in the row/column and calculate area.
Large grid dimensions potentially causing integer overflow when calculating the area.	Use long data type for intermediate calculations to prevent overflow, and handle cases where even `long` might overflow based on problem constraints.
Grid with 1s scattered far apart	The algorithm should find the extreme coordinates (min/max row/col) correctly regardless of 1s distribution.
Grid exceeding memory limits (very large dimensions)	Consider using an iterative approach or divide-and-conquer to reduce memory footprint if memory is a significant concern for very large grids.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the grid (maximum number of rows and columns), and what is the expected range for these dimensions?
What should I return if the grid is empty or if it contains no 1s?
Is the grid guaranteed to be rectangular (i.e., all rows have the same number of columns)?
Are the grid values guaranteed to be only 0 or 1, or could there be other values?
Is there a possibility of integer overflow when calculating the area of the rectangle?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine every possible rectangle that you can draw on the grid.
For each rectangle, check if it covers all the 'ones' in the grid.
If a rectangle does not cover all the 'ones', discard it and move on to the next possible rectangle.
If a rectangle covers all the 'ones', remember the area of this rectangle.
After checking all possible rectangles, find the rectangle with the smallest area from the rectangles you remembered.
The area of this smallest rectangle is the answer.

Code Implementation

def find_minimum_area_brute_force(grid):
    if not grid or not grid[0]:
        return 0

    row_count = len(grid)
    column_count = len(grid[0])
    minimum_area = float('inf')

    # Iterate through all possible top-left corners
    for top_row in range(row_count):
        for left_column in range(column_count):

            # Iterate through all possible bottom-right corners
            for bottom_row in range(top_row, row_count):
                for right_column in range(left_column, column_count):

                    # Check if the current rectangle covers all ones
                    all_ones_covered = True
                    for row_index in range(row_count):
                        for column_index in range(column_count):
                            if grid[row_index][column_index] == 1:
                                if not (top_row <= row_index <= bottom_row and\
                                        left_column <= column_index <= right_column):
                                    all_ones_covered = False
                                    break
                        if not all_ones_covered:
                            break

                    # If all ones are covered, update minimum_area
                    if all_ones_covered:
                        current_area = (bottom_row - top_row + 1) * (right_column - left_column + 1)

                        # Keep track of smallest area
                        minimum_area = min(minimum_area, current_area)

    # Handle the case where no 1s are present in the grid
    if minimum_area == float('inf'):
        return 0

    return minimum_area

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Find the topmost row that contains a '1'. This will be the top edge of the rectangle.
Find the bottommost row that contains a '1'. This will be the bottom edge of the rectangle.
Find the leftmost column that contains a '1'. This will be the left edge of the rectangle.
Find the rightmost column that contains a '1'. This will be the right edge of the rectangle.
Now that you have the top, bottom, left, and right edges, you know the boundaries of the smallest rectangle that contains all the '1's.
Calculate the area of the rectangle using its width and height (which you get from the boundaries you found).

Code Implementation

def find_minimum_area(grid):
    top_row = -1
    bottom_row = -1
    leftmost_column = -1
    rightmost_column = -1

    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                top_row = row_index
                break
        if top_row != -1:
            break

    for row_index in range(number_of_rows - 1, -1, -1):
        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                bottom_row = row_index
                break
        if bottom_row != -1:
            break

    for column_index in range(number_of_columns):
        for row_index in range(number_of_rows):
            if grid[row_index][column_index] == 1:
                leftmost_column = column_index
                break
        if leftmost_column != -1:
            break

    for column_index in range(number_of_columns - 1, -1, -1):
        for row_index in range(number_of_rows):
            if grid[row_index][column_index] == 1:
                rightmost_column = column_index
                break
        if rightmost_column != -1:
            break

    # Handle edge case where no 1s exist.
    if top_row == -1:
        return 0

    # Calculate the area based on extreme points found
    height = bottom_row - top_row + 1

    width = rightmost_column - leftmost_column + 1

    return height * width

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 immediately as there are no 1s to cover.
Grid with no 1s	Return 0 since the minimum area to cover nothing is zero.
Grid with only one 1	Return 1 as the minimum rectangle covering one 1 is a 1x1 rectangle.
Grid with all 1s	The algorithm should correctly calculate the area of the entire grid in this case.
Grid with one row or one column	The algorithm should correctly identify the min/max indices of the 1s in the row/column and calculate area.
Large grid dimensions potentially causing integer overflow when calculating the area.	Use long data type for intermediate calculations to prevent overflow, and handle cases where even `long` might overflow based on problem constraints.
Grid with 1s scattered far apart	The algorithm should find the extreme coordinates (min/max row/col) correctly regardless of 1s distribution.
Grid exceeding memory limits (very large dimensions)	Consider using an iterative approach or divide-and-conquer to reduce memory footprint if memory is a significant concern for very large grids.