Maximize Area of Square Hole in Grid

Maximize Area of Square Hole in Grid

What is this problem about?

You are given a grid made of horizontal and vertical bars. You can remove specific horizontal and vertical bars given to you in two separate arrays. When bars are removed, adjacent grid cells merge to form larger holes. Your goal is to determine the area of the maximum perfect square hole you can form by removing the allowed bars.

Why is this asked in interviews?

This is a geometry and array problem. Interviewers use it to see if candidates can decouple a 2D spatial problem into two independent 1D problems. Attempting to simulate the 2D grid and the hole merging physically is extremely complex and slow. The optimal solution requires recognizing that the width and height of the hole are determined independently by the longest contiguous sequence of removed bars in each direction.

Algorithmic pattern used

This problem relies on the Sorting and Contiguous Sequence pattern.

1. The width of a hole is determined by the number of consecutive vertical bars removed. If you remove 2 consecutive bars, the gap width is 3. If you remove $k$ consecutive bars, the gap is $k + 1$.
2. Sort the hBars array and the vBars array.
3. Find the longest sequence of consecutive integers in hBars. Let the length of this sequence be max_h. The max height gap is max_h + 1.
4. Find the longest sequence of consecutive integers in vBars. Let the length be max_v. The max width gap is max_v + 1.
5. Because the hole MUST be a perfect square, the side length is limited by the smaller of the two maximum gaps: side = min(max_h + 1, max_v + 1). The area is side * side.

Example explanation

Allowed removals: hBars = [3, 2, 4], vBars = [4, 2].

1. Sort hBars: [2, 3, 4].
   • They are completely consecutive! Length of consecutive sequence = 3.
   • Max height gap = $3 + 1 = 4$.
2. Sort vBars: [2, 4].
   • They are NOT consecutive. The longest consecutive sequence is just 1 (either [2] or [4]).
   • Max width gap = $1 + 1 = 2$. We can make a hole of height 4, but only width 2. To make a perfect square, we are bottlenecked by the width. Side length = min(4, 2) = 2. Max square area = $2 \times 2 = 4$.

Common mistakes candidates make

The most frequent mistake is assuming the array of bars is already sorted. If you iterate through hBars = [3, 2, 4] without sorting, you will calculate a max consecutive length of 1, resulting in completely wrong dimensions. Another mistake is calculating the area of the maximum rectangular hole instead of restricting it to a square. You must apply the min() function to the dimensions before squaring.

Interview preparation tip

For the Maximize Area of Square Hole coding problem, always look for opportunities to decouple 2D problems. If operations on rows do not affect operations on columns, solve the rows as a 1D array problem, solve the columns as a 1D array problem, and then mathematically combine the two independent results at the very end.