Problem Description

You have a grid with n + 2 horizontal bars and m + 2 vertical bars, creating a grid of 1×1 unit cells. The bars are numbered starting from 1.

You're given two arrays:

• hBars: contains indices of horizontal bars that can be removed
• vBars: contains indices of vertical bars that can be removed

All other bars (those not in these arrays) are fixed and cannot be removed.

When you remove bars, you can create holes in the grid. Your task is to find the maximum area of a square-shaped hole that can be created by removing some (or none) of the removable bars.

For example, if you can remove consecutive horizontal bars 2, 3, 4 and consecutive vertical bars 3, 4, 5, you could create a 3×3 square hole (since removing 3 consecutive bars creates a gap of size 3).

The key insight is that to create a square hole of side length k, you need to remove k-1 consecutive horizontal bars and k-1 consecutive vertical bars. The solution finds the longest sequence of consecutive removable bars in both directions, and the minimum of these two values determines the maximum square size possible.

The function returns the area of the largest square hole that can be created.

Intuition

Think about what happens when we remove bars from the grid. If we remove horizontal bar at position i, we're essentially merging the cells above and below that bar. If we remove consecutive horizontal bars at positions i, i+1, i+2, we create a vertical gap that spans multiple cells.

To create a square hole, we need the same size gap in both horizontal and vertical directions. For a square of side length s, we need to remove s-1 consecutive bars in each direction. Why s-1? Because removing one bar creates a gap of size 2 (merges 2 cells), removing two consecutive bars creates a gap of size 3, and so on.

The challenge becomes: what's the longest sequence of consecutive bars we can remove in each direction? This is similar to finding the longest consecutive sequence in an array of numbers.

For the horizontal bars in hBars, we need to find groups where the numbers are consecutive (like 2, 3, 4 or 5, 6, 7, 8). The same applies to vertical bars in vBars.

Once we find the longest consecutive sequence in both arrays, we have:

• Maximum horizontal gap = (longest consecutive sequence in hBars) + 1
• Maximum vertical gap = (longest consecutive sequence in vBars) + 1

Since we need a square, we take the minimum of these two values as our square's side length. The area is then the square of this side length.

The sorting step makes it easy to identify consecutive sequences - we just check if each element is exactly 1 more than the previous element.

Learn more about Sorting patterns.

Solution Approach

The solution implements a helper function f(nums) that finds the length of the longest consecutive increasing subsequence in an array, then adds 1 to get the gap size.

Step 1: Helper Function f(nums)

This function processes an array of bar indices:

1. Sort the array: nums.sort() arranges the bars in ascending order
2. Initialize counters:
   • ans = 1: tracks the maximum consecutive sequence length found
   • cnt = 1: tracks the current consecutive sequence length
3. Traverse the sorted array: For each element starting from index 1:
   • If nums[i] == nums[i-1] + 1: The current bar is consecutive to the previous one
     • Increment cnt to extend the current sequence
     • Update ans = max(ans, cnt) to track the longest sequence
   • Otherwise: The sequence is broken
     • Reset cnt = 1 to start counting a new sequence
4. Return the gap size: return ans + 1 (if we have k consecutive bars, we can create a gap of size k+1)

Step 2: Main Logic

1. Call f(hBars) to get the maximum horizontal gap size
2. Call f(vBars) to get the maximum vertical gap size
3. Take the minimum of these two values: min(f(hBars), f(vBars))
   • This ensures we have a square (equal width and height)
4. Return the square of this value to get the area: min(f(hBars), f(vBars)) ** 2

Example walkthrough:

• If hBars = [2, 3, 4, 6, 7] and vBars = [3, 4, 5, 6]
• For hBars: After sorting, we find sequences [2,3,4] (length 3) and [6,7] (length 2)
  • Maximum consecutive = 3, so gap size = 4
• For vBars: After sorting, we find sequence [3,4,5,6] (length 4)
  • Maximum consecutive = 4, so gap size = 5
• Minimum gap = min(4, 5) = 4
• Maximum square area = 4² = 16

Example Walkthrough

Let's walk through a concrete example with hBars = [2, 4, 3] and vBars = [3, 5, 4].

Step 1: Process horizontal bars (hBars = [2, 4, 3])

• Sort the array: [2, 3, 4]
• Initialize: ans = 1, cnt = 1
• Traverse from index 1:
  • At index 1: nums[1] = 3, nums[0] = 2
    • Is 3 == 2 + 1? Yes! They're consecutive
    • Increment cnt = 2
    • Update ans = max(1, 2) = 2
  • At index 2: nums[2] = 4, nums[1] = 3
    • Is 4 == 3 + 1? Yes! They're consecutive
    • Increment cnt = 3
    • Update ans = max(2, 3) = 3
• Return ans + 1 = 3 + 1 = 4
• Result: Removing 3 consecutive horizontal bars creates a gap of size 4

Step 2: Process vertical bars (vBars = [3, 5, 4])

• Sort the array: [3, 4, 5]
• Initialize: ans = 1, cnt = 1
• Traverse from index 1:
  • At index 1: nums[1] = 4, nums[0] = 3
    • Is 4 == 3 + 1? Yes! They're consecutive
    • Increment cnt = 2
    • Update ans = max(1, 2) = 2
  • At index 2: nums[2] = 5, nums[1] = 4
    • Is 5 == 4 + 1? Yes! They're consecutive
    • Increment cnt = 3
    • Update ans = max(2, 3) = 3
• Return ans + 1 = 3 + 1 = 4
• Result: Removing 3 consecutive vertical bars creates a gap of size 4

Step 3: Calculate maximum square area

• Maximum horizontal gap = 4
• Maximum vertical gap = 4
• Square side length = min(4, 4) = 4
• Maximum square area = 4² = 16

The grid visualization shows that by removing horizontal bars 2, 3, 4 and vertical bars 3, 4, 5, we create a 4×4 square hole with area 16.

Solution Implementation

Time and Space Complexity

The time complexity is O(h × log h + v × log v), where h is the length of hBars and v is the length of vBars. This is because:

• The function f is called twice - once for hBars and once for vBars
• Inside function f, the main operation is sorting which takes O(n × log n) for an array of length n
• The subsequent loop through the sorted array takes O(n) time
• For hBars: sorting takes O(h × log h) and iteration takes O(h)
• For vBars: sorting takes O(v × log v) and iteration takes O(v)
• Total: O(h × log h + h + v × log v + v) = O(h × log h + v × log v)

Since typically h and v are of similar magnitude and can be generalized as n, the time complexity can be simplified to O(n × log n).

The space complexity is O(1) if we consider the sorting is done in-place (Python's sort modifies the list in-place). The only additional space used is for a few variables (ans, cnt, i) which are constant space. However, if we consider the space used by the sorting algorithm itself (Python's Timsort uses up to O(n) auxiliary space in worst case), then the space complexity is O(n), where n represents the maximum of the lengths of hBars and vBars.

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

Common Pitfalls

1. Misunderstanding the Gap Size Calculation

Pitfall: A common mistake is thinking that removing k consecutive bars creates a gap of size k, when it actually creates a gap of size k+1.

Why it happens: When you remove bars at positions 2, 3, and 4 (3 bars), you create a hole that spans from position 1 to position 5, which is 4 units wide.

Wrong approach:

def find_max_consecutive_gap(bars):
    bars.sort()
    max_consecutive = 1
    current_consecutive = 1
  
    for i in range(1, len(bars)):
        if bars[i] == bars[i - 1] + 1:
            current_consecutive += 1
            max_consecutive = max(max_consecutive, current_consecutive)
        else:
            current_consecutive = 1
  
    return max_consecutive  # ❌ Missing the +1

Correct approach:

def find_max_consecutive_gap(bars):
    bars.sort()
    max_consecutive = 1
    current_consecutive = 1
  
    for i in range(1, len(bars)):
        if bars[i] == bars[i - 1] + 1:
            current_consecutive += 1
            max_consecutive = max(max_consecutive, current_consecutive)
        else:
            current_consecutive = 1
  
    return max_consecutive + 1  # ✅ Add 1 for the actual gap size

2. Forgetting to Handle Empty Arrays

Pitfall: Not considering the case when hBars or vBars is empty, which would cause the function to return incorrect results.

Wrong approach:

def find_max_consecutive_gap(bars):
    bars.sort()  # ❌ If bars is empty, this still works but...
    max_consecutive = 1
    current_consecutive = 1
  
    for i in range(1, len(bars)):  # Never enters loop if empty
        # ...
  
    return max_consecutive + 1  # Returns 2 even with no removable bars!

Correct approach:

def find_max_consecutive_gap(bars):
    if not bars:  # ✅ Handle empty array
        return 1  # No bars to remove means gap size of 1
  
    bars.sort()
    max_consecutive = 1
    current_consecutive = 1
  
    for i in range(1, len(bars)):
        if bars[i] == bars[i - 1] + 1:
            current_consecutive += 1
            max_consecutive = max(max_consecutive, current_consecutive)
        else:
            current_consecutive = 1
  
    return max_consecutive + 1

3. Not Sorting the Input Arrays

Pitfall: Forgetting to sort the bars array before looking for consecutive sequences, which would miss valid consecutive bars that aren't adjacent in the original array.

Wrong approach:

def find_max_consecutive_gap(bars):
    # ❌ No sorting - bars might be [3, 1, 2, 5, 4]
    max_consecutive = 1
    current_consecutive = 1
  
    for i in range(1, len(bars)):
        if bars[i] == bars[i - 1] + 1:  # Won't find [1,2,3] sequence!
            current_consecutive += 1
            max_consecutive = max(max_consecutive, current_consecutive)
        else:
            current_consecutive = 1
  
    return max_consecutive + 1

Example where this fails:

• Input: hBars = [3, 1, 2] (unsorted)
• Without sorting: Would not detect the consecutive sequence [1, 2, 3]
• With sorting: Correctly identifies [1, 2, 3] as consecutive

4. Using Maximum Instead of Minimum for Square Dimension

Pitfall: Taking the maximum of horizontal and vertical gaps instead of the minimum, forgetting that a square needs equal dimensions.

Wrong approach:

max_square_side = max(max_horizontal_gap, max_vertical_gap)  # ❌ Creates rectangle, not square
return max_square_side ** 2

Correct approach:

max_square_side = min(max_horizontal_gap, max_vertical_gap)  # ✅ Ensures square shape
return max_square_side ** 2