#2943

Medium

LeetCode Problem Workspace

Maximize Area of Square Hole in Grid

Learn how to maximize the area of a square-shaped hole by selectively removing horizontal and vertical bars efficiently using sorting.

array sorting

Problem Statement

You are given a grid formed by n + 2 horizontal bars and m + 2 vertical bars, creating 1x1 cells. Two integer arrays, hBars and vBars, represent removable horizontal and vertical bars respectively. Each bar is indexed starting from 1, and only the bars in hBars and vBars can be removed. Your goal is to remove some bars to create the largest possible square-shaped hole in the grid.

Return the area of the largest square-shaped hole after removing any subset of removable bars. Consider all gaps between remaining bars and choose removals that maximize the square side length. All values in hBars and vBars are distinct, and constraints ensure efficient computation is possible with sorting and array processing.

Examples

Example 1

Input: n = 2, m = 1, hBars = [2,3], vBars = [2]

Output: 4

The left image shows the initial grid formed by the bars. The horizontal bars are [1,2,3,4] , and the vertical bars are [1,2,3] . One way to get the maximum square-shaped hole is by removing horizontal bar 2 and vertical bar 2.

Example 2

Input: n = 1, m = 1, hBars = [2], vBars = [2]

Output: 4

To get the maximum square-shaped hole, we remove horizontal bar 2 and vertical bar 2.

Example 3

Input: n = 2, m = 3, hBars = [2,3], vBars = [2,4]

Output: 4

One way to get the maximum square-shaped hole is by removing horizontal bar 3, and vertical bar 4.

Constraints

1 <= n <= 109
1 <= m <= 109
1 <= hBars.length <= 100
2 <= hBars[i] <= n + 1
1 <= vBars.length <= 100
2 <= vBars[i] <= m + 1
All values in hBars are distinct.
All values in vBars are distinct.

Solution Approach

Sort Bars Separately

Sort hBars and vBars independently to easily compute gaps between consecutive bars. Sorting ensures that you can evaluate all possible continuous spaces without missing any potential square area.

Compute Maximum Gaps

Iterate through the sorted bars to calculate the maximum distance between consecutive horizontal bars and vertical bars. Include the fixed outer bars to ensure edge gaps are considered. The largest gap in each direction represents the possible side length of a square-shaped hole.

Determine Maximum Square Area

The maximum area is the square of the smaller of the largest horizontal gap and largest vertical gap. This guarantees the square fits within both dimensions of the grid, and no other combination of removals can yield a larger square.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(hBars.length log hBars.length + vBars.length log vBars.length) due to sorting. Space complexity is O(1) additional beyond the input arrays, as only gap calculations are needed.

What Interviewers Usually Probe

Sorting arrays is expected before evaluating gaps.
Focus on consecutive distances between bars rather than all combinations.
Clarify edge cases where the largest square is at the grid boundary.

Common Pitfalls or Variants

Common pitfalls

Forgetting to include the gaps at the edges of the grid outside the removable bars.
Assuming removing all bars always gives the largest square instead of computing actual gaps.
Ignoring the distinct nature of hBars and vBars and indexing from 1.

Follow-up variants

Compute largest rectangular hole instead of square-shaped hole using similar gap calculations.
Handle cases where some bars cannot be removed by marking them as fixed obstacles in the array.
Maximize square area with weighted bars where removing each bar has a cost.

FAQ

What is the key pattern for solving Maximize Area of Square Hole in Grid?

The key pattern is Array plus Sorting, where sorting the horizontal and vertical bars allows gap calculation to determine the largest square.

Why do we need to consider the outer fixed bars?

The outer fixed bars define the grid boundary, and ignoring them can lead to underestimating the maximum square area.

Can removing all bars always maximize the square area?

Not always; only specific removals that create the largest consecutive gaps produce the maximum square side length.

What is the time complexity for this approach?

Sorting hBars and vBars leads to O(hBars.length log hBars.length + vBars.length log vBars.length), which is efficient given the constraints.

How does GhostInterview assist with this problem?

It automates sorting, identifies critical gaps, and calculates the maximum square area efficiently without manual iteration over combinations.

terminal

Solution

Solution 1: Sorting

The problem essentially asks us to find the length of the longest consecutive increasing subsequence in the array, and then add $1$.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

class Solution:
    def maximizeSquareHoleArea(
        self, n: int, m: int, hBars: List[int], vBars: List[int]
    ) -> int:
        def f(nums: List[int]) -> int:
            nums.sort()
            ans = cnt = 1
            for i in range(1, len(nums)):
                if nums[i] == nums[i - 1] + 1:
                    cnt += 1
                    ans = max(ans, cnt)
                else:
                    cnt = 1
            return ans + 1

        return min(f(hBars), f(vBars)) ** 2

Continue Practicing

#2948 make lexicographically smallest array by swapping elements #2952 minimum number of coins to be added #2933 high access employees