Problem Description

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.

Intuition

To solve this problem, we need to find the boundaries of a rectangle that encompass all the 1s in the grid. The idea is to traverse the grid and determine the minimum and maximum rows and columns where 1 appears. These boundaries are:

• (x1, y1) representing the top-left corner of the rectangle.
• (x2, y2) representing the bottom-right corner of the rectangle.

Solution 1: Find Minimum and Maximum Boundaries

We traverse through each element of the grid. For each 1 we encounter:

• Update x1 and y1 to be the minimum row and column indices, respectively.
• Update x2 and y2 to be the maximum row and column indices, respectively.

Once we have the boundaries, the area of the rectangle can be calculated as:

[ \text{area} = (x2 - x1 + 1) \times (y2 - y1 + 1) ]

This formula accounts for the number of rows and columns covered by the rectangle. By iterating through the grid once, this approach efficiently finds the smallest rectangle encompassing all 1s.

Solution Approach

The implementation of the solution is straightforward. We will use a single pass over the grid to calculate the boundaries of the rectangle.

• Initialize x1 and y1 to a very large value (inf in Python) and x2 and y2 to a very small value (-inf in Python). This sets up the initial boundaries.

• Iterate over each element in the grid using nested loops. The outer loop iterates over rows with an index i, and the inner loop iterates over columns with an index j.

• When a 1 is found at grid[i][j], update the boundaries:

  • x1 = min(x1, i): Keep the smallest row index where a 1 appears.
  • y1 = min(y1, j): Keep the smallest column index where a 1 appears.
  • x2 = max(x2, i): Keep the largest row index where a 1 appears.
  • y2 = max(y2, j): Keep the largest column index where a 1 appears.

• Once all elements have been processed, calculate the area of the rectangle using the formula:

  [ \text{area} = (x2 - x1 + 1) \times (y2 - y1 + 1) ]

The key patterns used here involve:

• Initialization for boundary tracking using inf and -inf.
• A simple traversal pattern to check each element, updating boundary values accordingly.
• Calculation of the area from the derived boundary indices, ensuring all 1s are enclosed.

This approach efficiently finds the minimum possible area containing all 1s using a single traversal of the grid, with a time complexity of O(n * m) where n is the number of rows and m is the number of columns.

Example Walkthrough

Let's consider an example grid:

grid = [
  [0, 0, 1, 0],
  [0, 1, 1, 0],
  [0, 0, 0, 0]
]

To find the smallest rectangle that encompasses all 1s, we initialize our boundary indicators:

• x1 = inf, y1 = inf for the top-left corner.
• x2 = -inf, y2 = -inf for the bottom-right corner.

Now, let's iterate through the grid to update these boundaries.

1. Row 0, Column 2: We find a 1.

   • Update boundaries:
     • x1 = min(inf, 0) = 0 (smallest row index)
     • y1 = min(inf, 2) = 2 (smallest column index)
     • x2 = max(-inf, 0) = 0 (largest row index)
     • y2 = max(-inf, 2) = 2 (largest column index)

2. Row 1, Column 1: We find another 1.

   • Update boundaries:
     • x1 = min(0, 1) = 0
     • y1 = min(2, 1) = 1
     • x2 = max(0, 1) = 1
     • y2 = max(2, 1) = 2

3. Row 1, Column 2: Another 1.

   • Update boundaries:
     • x1 = min(0, 1) = 0
     • y1 = min(1, 2) = 1
     • x2 = max(1, 1) = 1
     • y2 = max(2, 2) = 2

Once we have completed the grid traversal, the boundaries are fixed: x1 = 0, y1 = 1, x2 = 1, and y2 = 2.

Finally, calculate the area of the rectangle:

[ \text{area} = (x2 - x1 + 1) \times (y2 - y1 + 1) = (1 - 0 + 1) \times (2 - 1 + 1) = 2 \times 2 = 4 ]

Therefore, the minimum possible area of the rectangle that contains all 1s in the grid is 4.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(m * n), where m is the number of rows and n is the number of columns in grid. This is because the code iterates through each element of the grid once.

The space complexity of the code is O(1) since the space used does not depend on the size of grid and only a constant amount of space is needed for the variables x1, y1, x2, and y2.

Learn more about how to find time and space complexity quickly.