CF 1147B - Chladni Figure

Rating: 1900
Tags: brute force, strings
Solve time: 1m 2s
Verified: yes

Solution

Problem Understanding

The problem gives a rectangular grid of characters representing a pattern on a metal plate, where each cell contains either a . (empty) or a * (filled). The task is to find the smallest rectangle that contains all the * characters. This rectangle is described by its topmost row, bottommost row, leftmost column, and rightmost column, using 1-based indices.

The input first gives the number of rows and columns, followed by the grid itself. The output should be four integers defining the bounding rectangle of all stars.

Because the grid can be as large as 500x500, any approach that examines every possible rectangle explicitly is infeasible. A naive brute-force that considers every pair of top/bottom and left/right coordinates to check if it contains all stars would require up to (500^2)^2 operations, roughly 6.25×10^10, which is far too slow for a typical 2-second limit.

Edge cases to be careful about include grids with a single *, grids where all * are in a single row or column, and grids where * touch the boundaries. For example, a grid like:

3 4
*...
....
....

should return 1 1 1 1. A careless approach that assumes multiple rows or columns will fail.

Approaches

The brute-force approach would scan every possible rectangle, compute whether it contains all stars, and track the smallest such rectangle. This works because checking a rectangle is straightforward, but it fails when the grid is large: with n rows and m columns, the operation count is O(n^2 * m^2), which is too slow for n=m=500.

The key observation is that we do not need to check every rectangle. Each star only contributes to the extremes of the rectangle. The topmost row of the rectangle is the minimum row index containing a *, the bottommost row is the maximum row index containing a *, the leftmost column is the minimum column index with a *, and the rightmost column is the maximum column index with a *. Scanning the grid once and recording these extrema is sufficient.

This reduces the problem from a four-nested-loop brute-force to a single O(n*m) scan.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2 * m^2)	O(1)	Too slow
Optimal	O(n*m)	O(1)	Accepted

Algorithm Walkthrough

Initialize four variables to track the rectangle boundaries. Set top to a large value (larger than any possible row), bottom to -1, left to a large value, and right to -1. These represent uninitialized extremes.
Iterate over each row i from 0 to n-1. For each row, iterate over each column j from 0 to m-1. If the cell (i,j) contains *, update the extremes:

Set top to min(top, i).
Set bottom to max(bottom, i).
Set left to min(left, j).
Set right to max(right, j).

After scanning the grid, the rectangle containing all stars has top row top+1, bottom row bottom+1, left column left+1, and right column right+1. The +1 accounts for 1-based indexing.

Why it works: The variables top, bottom, left, right always hold the extreme indices of stars seen so far. Because we scan the entire grid, no star is missed. By taking the minima and maxima of coordinates, we guarantee the rectangle exactly bounds all stars. There is no risk of missing a boundary because each * is considered individually.

Python Solution

import sys
input = sys.stdin.readline

def main():
    n, m = map(int, input().split())
    grid = [input().strip() for _ in range(n)]

    top, bottom = n, -1
    left, right = m, -1

    for i in range(n):
        for j in range(m):
            if grid[i][j] == '*':
                top = min(top, i)
                bottom = max(bottom, i)
                left = min(left, j)
                right = max(right, j)

    print(top + 1, left + 1, bottom + 1, right + 1)

if __name__ == "__main__":
    main()

Each part of this solution corresponds directly to the algorithm walkthrough. Initializing extremes ensures we can compare each * correctly. The double loop ensures every cell is checked, preventing missed stars. The final addition of +1 converts from 0-based Python indexing to the 1-based output required by the problem.

Worked Examples

Sample input 1:

3 4
....
.*..
....

i	j	grid[i][j]	top	bottom	left	right
1	1	.	3	-1	4	-1
1	2	.	3	-1	4	-1
2	2	*	1	1	2	2

Output: 2 2 2 2. The table shows that after encountering the only *, all extremes collapse to that cell.

Sample input 2:

4 5
*....
.....
..*..
.....

i	j	grid[i][j]	top	bottom	left	right
0	0	*	0	0	0	0
2	2	*	0	2	0	2

Output: 1 1 3 3. Both stars are included in the rectangle. The table confirms that the algorithm correctly extends boundaries when a new star appears.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n*m)	We scan each cell exactly once.
Space	O(n*m)	We store the grid; only four integers extra are used.

With n, m up to 500, the number of operations is 250,000, well within a 2-second time limit. Memory usage is minimal.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    main()
    return output.getvalue().strip()

# provided sample
assert run("3 4\n....\n.*..\n....\n") == "2 2 2 2", "sample 1"

# single row
assert run("1 5\n..*..\n") == "1 3 1 3", "single row"

# single column
assert run("5 1\n.\n*\n.\n*\n.\n") == "2 1 4 1", "single column"

# all stars
assert run("2 2\n**\n**\n") == "1 1 2 2", "all stars"

# one star at corner
assert run("3 3\n*..\n...\n...\n") == "1 1 1 1", "corner"

# stars in diagonal
assert run("3 3\n*..\n.*.\n..*\n") == "1 1 3 3", "diagonal"

Test input	Expected output	What it validates
1x5 with center star	1 3 1 3	single row
5x1 with two stars	2 1 4 1	single column
2x2 all stars	1 1 2 2	rectangle spans entire grid
3x3 corner star	1 1 1 1	boundary condition
3x3 diagonal stars	1 1 3 3	rectangle expands with multiple extremes

Edge Cases

For a single * anywhere, the algorithm sets top, bottom, left, and right to the same value, producing a rectangle of size 1x1. For a star at the top-left corner, top and left remain zero, and the final +1 converts to 1-based indices. For a star at the bottom-right, bottom and right are correctly updated as the maximum indices. In all cases, the algorithm correctly bounds all stars without overshooting or undershooting.

The algorithm correctly handles grids where all stars lie in a single row or single column because minima and maxima converge to the actual positions. This avoids pitfalls where a naive approach might assume a minimum size rectangle larger than necessary.