Overview

This problem asks you to find the smallest rectangle (bounding box) that encloses all painting tiles (#) placed on a grid, and calculate the perimeter of that rectangle.

Analysis

To enclose all painting tiles, we need to determine the following 4 values:

1. The row number of the topmost painting tile (\(r_{min}\))
2. The row number of the bottommost painting tile (\(r_{max}\))
3. The column number of the leftmost painting tile (\(c_{min}\))
4. The column number of the rightmost painting tile (\(c_{max}\))

Once these values are determined, the smallest rectangle enclosing all painting tiles has a row range from \(r_{min}\) to \(r_{max}\) and a column range from \(c_{min}\) to \(c_{max}\).

The height of this rectangle can be expressed as \(r_{max} - r_{min} + 1\), and the width as \(c_{max} - c_{min} + 1\). As stated in the problem, the perimeter can be calculated as \(2 \times (\text{height} + \text{width})\).

If no painting tiles exist, there is nothing to enclose, so we output \(0\) as instructed by the problem.

Algorithm

1. Initialize variables \(r_{min}, r_{max}, c_{min}, c_{max}\) to track the minimum and maximum positions of painting tiles.
2. Examine each cell \((r, c)\) of the grid in order.
3. If cell \((r, c)\) is a painting tile (#):
   • Update \(r_{min}\) with the minimum of itself and \(r\)
   • Update \(r_{max}\) with the maximum of itself and \(r\)
   • Update \(c_{min}\) with the minimum of itself and \(c\)
   • Update \(c_{max}\) with the maximum of itself and \(c\)
   • Set a flag indicating “a painting tile was found”
4. After checking all cells, if the flag is not set, output 0.
5. If the flag is set, calculate the perimeter from the obtained minimum and maximum values and output it.

Complexity

• Time complexity: \(O(H \times W)\)
  • Since we scan all cells of the grid exactly once, the time is proportional to the total number of cells. With \(H, W \leq 1000\), this results in at most about \(10^6\) operations, which runs sufficiently fast.
• Space complexity: \(O(H \times W)\)
  • Memory proportional to the total number of cells is used to store the input grid information.

Implementation Notes

• Efficient search: For each row, using find('#') (leftmost) and rfind('#') (rightmost) allows us to determine the column range more efficiently than checking each character individually.

• Initial value setting: Setting \(r_{min}\) and \(c_{min}\) to sufficiently large values, and \(r_{max}\) and \(c_{max}\) to sufficiently small values (or \(-1\), etc.) ensures smooth updates when the first tile is found.

• Exception handling: Handle the case where no painting tiles exist by checking something like if not found: to correctly output \(0\).

  Source Code

import sys

def solve():
    # Read all input data from standard input and split by whitespace
    try:
        input_data = sys.stdin.read().split()
    except EOFError:
        return
        
    if not input_data:
        return
    
    # H (number of rows) and W (number of columns) are the first two values
    try:
        H = int(input_data[0])
        W = int(input_data[1])
    except (ValueError, IndexError):
        return
    
    # The grid characters start from index 2 in input_data.
    # We join all parts to handle cases where rows might be split by whitespace.
    # This creates a single string containing all '#' and '.' characters in order.
    all_chars = "".join(input_data[2:])
    
    # Initialize the bounding box boundaries for '#' characters
    # r_min, r_max for row indices; c_min, c_max for column indices
    r_min, r_max = H, -1
    c_min, c_max = W, -1
    found = False
    
    # Iterate through the grid row by row
    for r in range(H):
        # Extract the r-th row from the combined string
        # Each row is expected to have exactly W characters.
        row = all_chars[r * W : (r + 1) * W]
        
        # Find the first and last occurrence of '#' in the current row
        first_idx = row.find('#')
        if first_idx != -1:
            found = True
            last_idx = row.rfind('#')
            
            # Update the global bounding box coordinates
            if r < r_min:
                r_min = r
            if r > r_max:
                r_max = r
            if first_idx < c_min:
                c_min = first_idx
            if last_idx > c_max:
                c_max = last_idx
    
    if not found:
        # If no painting tile '#' was found, the perimeter is 0
        print(0)
    else:
        # The number of rows and columns covered by the bounding box
        # height = r2 - r1 + 1, width = c2 - c1 + 1
        height = r_max - r_min + 1
        width = c_max - c_min + 1
        
        # The perimeter is 2 * (height + width)
        # This matches the formula: 2 * ((r2 - r1 + 1) + (c2 - c1 + 1))
        print(2 * (height + width))

if __name__ == '__main__':
    solve()

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This editorial was generated by gemini-3-flash-thinking.