Problem Description

In this LeetCode problem, we're tasked with finding the largest possible square-shaped hole that can be created in a grid by removing certain bars. The grid consists of n + 2 horizontal bars and m + 2 vertical bars, forming 1 x 1 unit cells. The bars are 1-indexed for reference.

To create a hole, we have the option of removing horizontal bars at positions listed in hBars, and vertical bars at positions listed in vBars. Each array contains distinct positions, and they fall within certain ranges—hBars values are within [2, n + 1], and vBars within [2, m + 1].

Our goal is to return the maximum area of the square hole. To visualize this, imagine removing consecutive bars to create the largest possible square void in the grid's structure.

Intuition

Firstly, we need to understand that the maximum square hole we can make in the grid is limited by the number of consecutive bars we can remove either horizontally or vertically. Therefore, the key to this problem is figuring out the longest sequence of consecutive bar numbers in the hBars and vBars that we’re permitted to take out.

Let's consider any sequence of removable bars. The longest consecutive run of bar numbers in either direction will ultimately determine the maximum size of the square-shaped hole we can achieve. For instance, if we can remove three consecutive horizontal bars and two consecutive vertical bars, the largest square hole we can create is 2 x 2, since the square cannot exceed the smaller side.

The solution approach here is quite clever. We sort each bar array to make it easier to find consecutive sequences. Then, using a helper function f(nums), we traverse through the sorted array and count the length of the longest increasing consecutive subsequence. This function returns the longest length plus one (to account for the extra space at the end of the last bar, as a square hole need one more unit space to be formed).

After calculating this for both hBars and vBars, the side length of the largest possible square hole is the minimum of these two lengths. Since square area is equal to the side length squared, we finally return the side length squared (calculated minimum) as the answer.

Learn more about Sorting patterns.

Solution Approach

The implementation relies on a straightforward but effective sorting and counting approach.

Here's a step-by-step breakdown:

1. Sort both arrays, hBars and vBars. Sorting brings all consecutive sequences into adjacent positions in the array, which makes it easy to count them in the next step.

2. Define a helper function f(nums) which takes a sorted list of bar positions. Initialize two variables: ans and cnt to 1. Here, ans will store the length of the longest consecutive sequence we find, and cnt is a counter for the current consecutive sequence.

3. Loop through the nums list starting from the second element (indexed at 1), and compare each element with its predecessor.

   • If the current element is exactly one more than the previous (nums[i] == nums[i - 1] + 1), it means that we've found consecutive bars. Increment cnt to extend the current consecutive sequence. Update ans to hold the maximum value between ans and the current cnt.

   • If the current element is not consecutive, reset cnt to 1 because we've hit a break in the sequence, and we'll need to start counting a new sequence from this point.

4. When the loop finishes, add 1 to the final ans. This is because the square hole size is actually one unit larger than the number of bars removed (the last bar removed implies there is one extra unit of open space that completes the square).

5. Apply the helper function to both hBars and vBars to get the size of the longest consecutive sequence for both horizontal and vertical bars.

6. The side length of the largest square hole we can create is the minimum of these two sizes. This is because the shape has to be a square, therefore, it is limited by the shorter side.

7. Calculate the area of the largest square hole by squaring the side length found in the previous step. The square of the minimum consecutive sequence length provides us with the maximum square hole area after removing the bars, which is the required output.

Using the steps from the approach above, the code neatly encapsulates the logic within one main function maximizeSquareHoleArea and one helper function f. The use of sorting and a single pass counting mechanism ensures an efficient solution, with the overall time complexity being dominated by the sorting step, which is O(n log n) for each array.

Example Walkthrough

To illustrate the solution approach, let's walk through a small example. Suppose we have the following input:

• n + 2 = 5 horizontal bars (indexed from 1 to 5)
• m + 2 = 5 vertical bars (indexed from 1 to 5)
• hBars = [3, 4]
• vBars = [2, 4]

Step 1: Sort both arrays, hBars = [3, 4], vBars = [2, 4].

The arrays are already sorted in this example, so we can proceed directly to the next step.

Step 2: We will use a helper function f(nums) to find the length of the longest consecutive sequence.

Step 3: Let's apply function f(nums) on hBars:

• We start with ans = 1 and cnt = 1.
• Loop through hBars, starting from the second element:
  • Compare hBars[1] to hBars[0]. Since 4 == 3 + 1, they are consecutive, so increase cnt to 2.
  • Since we are at the end of the array, we finish the loop with ans being the value of cnt, which is 2.

Step 4: Add 1 to the final ans, yielding 2 + 1 = 3. This means horizontally, we can fit a square hole of side length 3.

Step 5: Repeat steps 3 and 4 for vBars:

• cnt starts at 1, and ans starts at 1.
• Comparing vBars[1] to vBars[0], we see that 4 != 2 + 1, so these are not consecutive.
• The loop ends with ans still equal to 1.

Step 6: Add 1 to the final ans, yielding 1 + 1 = 2. This means vertically, the largest square hole we can form has a side length of 2.

Step 7: The comparison of the two sizes (3 and 2) shows that the maximum square hole we can create is limited by the vertical bars, which allow for a side length of 2.

Step 8: The area of the square is the side length squared, so in this example, the largest square hole we can create has an area of 2 x 2 = 4.

Following this approach, the maximizeSquareHoleArea function would return the value 4 as the answer for this example.

Solution Implementation

Time and Space Complexity

The function maximizeSquareHoleArea includes two calls to the helper function f, once for hBars and once for vBars. The complexity analysis will be the same for both calls since the operations performed by f are identical for both lists.

Time Complexity

The time complexity is dominated by the sorting operation inside the function f. Since Python's sort method typically uses Timsort, which has an average and worst-case time complexity of O(n log n) where n is the length of the list being sorted. The subsequent loop runs in linear time, O(n), since it iterates through the sorted list once.

Therefore, the time complexity of the function f is O(n log n) + O(n) which simplifies to O(n log n) because the n log n term dominates.

Since we call the function f twice, once for hBars and once for vBars, and assuming n refers to the longer of the two lists for worst-case analysis, then the overall time complexity remains O(n log n) since these two calls are sequential, not nested.

Space Complexity

The space complexity of the function f is O(n) because of the space required to sort the list. The sort operation may require additional space to hold elements temporarily during the sort. The variables ans and cnt use constant space and do not scale with input size.

Overall, since we consider the larger of the two lists hBars and vBars, the total space complexity of the maximizeSquareHoleArea function is O(n), where n is the length of the larger input list.

Learn more about how to find time and space complexity quickly using problem constraints.