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Maximize Area of Square Hole in Grid

Difficulty: Medium

Problem Description

You are given two integers, n and m, and two integer arrays, hBars and vBars. The grid has n + 2 horizontal and m + 2 vertical bars, creating 1 x 1 unit cells. You can remove some of the bars in hBars and some of the bars in vBars. Return an integer denoting the maximum area of a square-shaped hole in the grid after removing some bars (possibly none).

Key Insights

  • The grid is defined by the bars, where removing a bar increases the potential size of the square hole.
  • The maximum square side length is determined by the minimum distance between the remaining horizontal and vertical bars.
  • The problem can be reduced to finding gaps between the bars to determine the largest continuous space.

Space and Time Complexity

Time Complexity: O(hBars.length + vBars.length)
Space Complexity: O(1)

Solution

To solve the problem, we can follow these steps:

  1. Sort the arrays hBars and vBars to facilitate gap calculation.
  2. Calculate the maximum possible gaps between consecutive bars in both arrays.
  3. The largest square hole can be formed by the smallest gap among the maximum gaps calculated from both horizontal and vertical bars.
  4. The area of the square hole is then the side length squared.

The main data structures used are arrays (for hBars and vBars) and the algorithm primarily involves sorting and linear scans to find gaps.

Code Solutions

def maxSquareArea(n, m, hBars, vBars):
    # Add the boundaries of the grid
    hBars = [0] + sorted(hBars) + [n + 1]
    vBars = [0] + sorted(vBars) + [m + 1]
    
    # Find the maximum gaps between horizontal bars
    max_h_gap = max(hBars[i] - hBars[i - 1] for i in range(1, len(hBars)))
    
    # Find the maximum gaps between vertical bars
    max_v_gap = max(vBars[i] - vBars[i - 1] for i in range(1, len(vBars)))
    
    # The side length of the largest square is the minimum of the two max gaps
    max_side = min(max_h_gap, max_v_gap)
    
    # Area of the square
    return max_side * max_side
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