Problem Description

You have a rectangular field with corners at coordinates (1, 1) and (m, n). The field contains some internal fences:

• Horizontal fences run from (hFences[i], 1) to (hFences[i], n)
• Vertical fences run from (1, vFences[i]) to (m, vFences[i])

The field is also surrounded by boundary fences that cannot be removed:

• Two horizontal boundaries: from (1, 1) to (1, n) and from (m, 1) to (m, n)
• Two vertical boundaries: from (1, 1) to (m, 1) and from (1, n) to (m, n)

Your task is to find the maximum area of a square field that can be created by removing some (or none) of the internal fences. The square must be formed by selecting two horizontal lines (either boundaries or fences) and two vertical lines (either boundaries or fences) such that the distance between the horizontal lines equals the distance between the vertical lines.

Return the maximum square area modulo 10^9 + 7. If no square field can be formed, return -1.

For example, if you can create a square with side length d, the area would be d^2. You need to find the largest possible d where:

• The distance d exists between some pair of horizontal lines (considering both boundaries and fences)
• The same distance d exists between some pair of vertical lines (considering both boundaries and fences)

Intuition

To form a square field, we need to find two horizontal lines that are distance d apart and two vertical lines that are also distance d apart. The key insight is that we don't need to track which specific fences we remove - we only care about what distances are achievable.

Think of it this way: if we have horizontal fences at positions [2, 5, 7] plus boundaries at 1 and m, we can create various distances by choosing any two of these positions. For instance, choosing positions 2 and 7 gives us a distance of 5. Similarly for vertical fences.

The crucial observation is that a square with side length d can be formed if and only if:

1. Distance d can be achieved between some pair of horizontal lines (including boundaries)
2. The same distance d can be achieved between some pair of vertical lines (including boundaries)

This naturally leads to a set intersection approach:

• Calculate all possible distances between pairs of horizontal lines (including boundaries at 1 and m)
• Calculate all possible distances between pairs of vertical lines (including boundaries at 1 and n)
• Find the common distances that appear in both sets
• The largest common distance gives us the maximum square side length

Why does this work? Because if distance d appears in both sets, it means we can select two horizontal lines d units apart AND two vertical lines d units apart, forming a square of side length d. The area would then be d^2.

The beauty of this approach is that it reduces a potentially complex geometric problem to a simple set operation - finding the intersection of two sets of distances and taking the maximum value.

Solution Approach

The solution implements the intuition using a helper function and set operations:

Step 1: Generate All Possible Distances

The helper function f(nums: List[int], k: int) takes a list of fence positions and a boundary value k:

• First, it extends the list with boundary positions [1, k] to include the field boundaries
• Sorts the combined list to have positions in order
• Uses combinations(nums, 2) to generate all pairs of positions
• Calculates the distance b - a for each pair (a, b) where a < b (guaranteed by combinations on sorted list)
• Returns a set containing all unique distances

Step 2: Calculate Horizontal and Vertical Distances

• Call f(hFences, m) to get all possible horizontal distances hs
  • This includes distances between any two horizontal lines (fences + boundaries at positions 1 and m)
• Call f(vFences, n) to get all possible vertical distances vs
  • This includes distances between any two vertical lines (fences + boundaries at positions 1 and n)

Step 3: Find Common Distances

• Use set intersection hs & vs to find distances that appear in both sets
• These common distances represent possible square side lengths
• Use max(hs & vs, default=0) to find the largest common distance
  • The default=0 handles the case when there's no intersection (no square possible)

Step 4: Calculate the Result

• If ans (maximum common distance) is non-zero:
  • Calculate the square area as ans**2
  • Return the result modulo 10**9 + 7
• If ans is 0 (no common distances found):
  • Return -1 to indicate no square field is possible

The time complexity is O(p^2 + q^2) where p and q are the total number of horizontal and vertical lines respectively (including boundaries). The space complexity is O(p^2 + q^2) for storing all possible distances in the sets.

Example Walkthrough

Let's walk through a small example where m = 6, n = 7, hFences = [3], and vFences = [4].

Step 1: Identify all horizontal lines

• Boundary lines at positions: 1 and 6
• Fence lines at position: 3
• All horizontal lines: [1, 3, 6]

Step 2: Calculate all possible horizontal distances

• Distance between positions 1 and 3: 3 - 1 = 2
• Distance between positions 1 and 6: 6 - 1 = 5
• Distance between positions 3 and 6: 6 - 3 = 3
• Set of horizontal distances: {2, 3, 5}

Step 3: Identify all vertical lines

• Boundary lines at positions: 1 and 7
• Fence lines at position: 4
• All vertical lines: [1, 4, 7]

Step 4: Calculate all possible vertical distances

• Distance between positions 1 and 4: 4 - 1 = 3
• Distance between positions 1 and 7: 7 - 1 = 6
• Distance between positions 4 and 7: 7 - 4 = 3
• Set of vertical distances: {3, 6}

Step 5: Find common distances

• Horizontal distances: {2, 3, 5}
• Vertical distances: {3, 6}
• Intersection: {2, 3, 5} ∩ {3, 6} = {3}
• Maximum common distance: 3

Step 6: Calculate the result

• Maximum square side length: 3
• Maximum square area: 3² = 9
• Return: 9 % (10⁹ + 7) = 9

This means we can form a square field with side length 3. For instance:

• Using horizontal lines at positions 3 and 6 (distance = 3)
• Using vertical lines at positions 1 and 4 (distance = 3)
• Or using vertical lines at positions 4 and 7 (distance = 3)

The key insight is that we don't need to track which specific fences to remove - we only need to find matching distances in both horizontal and vertical directions to form a square.

Solution Implementation

Time and Space Complexity

The time complexity is O(h^2 + v^2), where h is the length of hFences and v is the length of vFences.

Breaking down the analysis:

• The function f is called twice - once for horizontal fences and once for vertical fences
• Inside function f, after extending the list with 2 elements (boundaries), we have h+2 horizontal positions and v+2 vertical positions
• Sorting takes O((h+2)log(h+2)) for horizontal and O((v+2)log(v+2)) for vertical
• The combinations operation generates all pairs from the sorted list, which creates C(h+2, 2) = (h+2)(h+1)/2 pairs for horizontal and C(v+2, 2) = (v+2)(v+1)/2 pairs for vertical
• Computing differences for all pairs takes O(h^2) for horizontal and O(v^2) for vertical
• The set intersection operation hs & vs takes O(min(|hs|, |vs|)) which is bounded by O(h^2 + v^2)
• Finding the maximum takes linear time in the size of the intersection

Since O(h^2) and O(v^2) dominate the sorting operations, the overall time complexity is O(h^2 + v^2).

The space complexity is O(h^2 + v^2):

• The set hs stores up to (h+2)(h+1)/2 differences, which is O(h^2)
• The set vs stores up to (v+2)(v+1)/2 differences, which is O(v^2)
• The intersection creates a new set with size at most min(|hs|, |vs|)
• Other variables use constant space

Therefore, the total space complexity is O(h^2 + v^2).

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

Common Pitfalls

1. Integer Overflow When Calculating Square Area

The most critical pitfall in this problem is calculating max_side_length ** 2 without considering integer overflow. Since the field dimensions can be as large as 10^9, the maximum possible side length is also up to 10^9. When squaring this value, we get 10^18, which exceeds the typical 32-bit integer limit and can cause overflow issues in some programming languages.

Incorrect approach:

# This might cause issues in languages with fixed integer sizes
area = max_side_length ** 2
return area % MOD

Correct approach:

# Apply modulo during calculation to prevent overflow
area = pow(max_side_length, 2, MOD)  # Built-in modular exponentiation
# Or explicitly handle large numbers
area = (max_side_length * max_side_length) % MOD

2. Forgetting to Include Boundary Fences

A common mistake is only considering the internal fences and forgetting that the field boundaries at positions 1 and m (for horizontal) and 1 and n (for vertical) are also valid lines for forming squares.

Incorrect approach:

# Only using internal fences
distances = set()
for i in range(len(hFences)):
    for j in range(i + 1, len(hFences)):
        distances.add(hFences[j] - hFences[i])

Correct approach:

# Include boundaries with internal fences
all_positions = hFences + [1, m]  # Add boundaries
all_positions.sort()
# Then calculate distances from all positions

3. Not Handling Empty Intersection Properly

When there are no common distances between horizontal and vertical possibilities, the intersection will be empty. Using max() on an empty set will raise a ValueError.

Incorrect approach:

max_side_length = max(horizontal_distances & vertical_distances)  # Raises error if empty

Correct approach:

max_side_length = max(horizontal_distances & vertical_distances, default=0)
# Or check explicitly:
common_distances = horizontal_distances & vertical_distances
if not common_distances:
    return -1
max_side_length = max(common_distances)

4. Inefficient Distance Calculation

Using nested loops without optimization or using combinations incorrectly can lead to inefficiency or errors.

Less efficient approach:

# Creating unnecessary intermediate lists
from itertools import combinations
all_pairs = list(combinations(all_positions, 2))
distances = set([b - a for a, b in all_pairs])

More efficient approach:

# Direct set construction without intermediate list
distances = {b - a for a, b in combinations(sorted(all_positions), 2)}

5. Not Sorting Positions Before Distance Calculation

If positions aren't sorted, you might calculate negative distances or need additional checks.

Incorrect approach:

# Without sorting, might get negative distances
for a, b in combinations(all_positions, 2):
    distances.add(abs(b - a))  # Need abs() to handle order

Correct approach:

all_positions.sort()  # Sort first
for a, b in combinations(all_positions, 2):
    distances.add(b - a)  # No need for abs() since b > a is guaranteed