2975. Maximum Square Area by Removing Fences From a Field

Problem Description

This problem presents us with a large rectangular field, with the corners at (1, 1) and (m, n). Throughout this field, there are certain fences placed horizontally and vertically. A horizontal fence extends from (hFences[i], 1) to (hFences[i], n), while a vertical fence extends from (1, vFences[i]) to (m, vFences[i]). These fences partition the field into smaller sections. Our goal is to find the maximum area of a square that can be obtained by removing some of these fences. It is important to note that the boundary fences surrounding the field are immovable.

We need to return the area of the largest square that can be formed, modulo 10^9 + 7. If it's not possible to form any square, we return -1.

The challenge lies in identifying potential squares and calculating the maximum possible area that can be obtained by removing fences, while taking into account the fixed field boundaries.

Intuition

The solution approach uses enumeration and set theory. We start by identifying all possible distances between any two horizontal fences as well as between any two vertical fences. Every distinct distance represents a potential side length of a square.

In order to calculate these distances efficiently, we use a helper function f that adds the boundary fences (1, k) to the list of fences, sorts the list, and then returns a set containing every possible distance between two fences. Why use a set? Because we're interested in distinct distances – duplicate distances won't lead to larger squares.

Once we have sets for horizontal distances (hs) and vertical distances (vs), we seek the intersection of these sets. The intersection represents the side lengths of squares that can be formed both horizontally and vertically. The answer can only be a length that is common to both, because a square needs equal sides.

We find the maximum value in the intersection of these sets to get the largest possible side length of a square. The answer is the square of this length, modulo 10^9 + 7 as required by the problem. If there are no common distances (ans is 0), it is impossible to form a square and thus we return -1.

Solution Approach

The solution involves two main stages: calculating all possible distances between fences and finding the maximum square area using those distances.

Here's how the solution is implemented step-by-step:

1. Helper Function f:

   • This function takes a list of fence positions (nums) and the maximum extent of the field (k), which would be m for horizontal and n for vertical fences.
   • It first adds the boundary positions, 1 and k, to the list of fence positions to consider the entire field.
   • The next step is to sort the extended list of positions. Sorting is crucial because we want to find all possible pair-wise distances, and distances are positive and undirected—so sorting helps in enumerating these systematically.
   • Finally, using combinations from Python's itertools module, the function generates all pairs of fence positions and calculates their differences, which are the potential side lengths of squares.
   • It returns a set of these calculated distances to ensure each distance is unique.

2. Finding Common Distances:

   • The f function is called twice, once for hFences with m and once for vFences with n, returning sets hs and vs respectively.
   • The intersection of these sets (hs & vs) gives us all potential side lengths that are common to both horizontal and vertical directions—this is a key part because for a square, the lengths must be the same in both directions.

3. Determining the Largest Square:

   • From the set intersection, we find the maximum value (max(hs & vs, default=0)) which represents the side length of the largest possible square.
   • If ans (the maximum length) is non-zero, we calculate the area by squaring ans and then take the modulus with 10**9 + 7 to comply with the problem's requirement for large numbers.
   • If ans is zero, which means no common distances were found, it indicates that forming a square is impossible, and thus the function returns -1.

Algorithmically, this approach is efficient because it avoids brute force checks for every possible square by leveraging mathematical set operations to identify valid square side lengths. The use of sorting and combination generation is a blend of pattern recognition and combinatorics, which makes the solution elegant and efficient. The dealing with a large number modulo is a common practice in programming contests to prevent integer overflow and is handled with a simple modulus operation.

Example Walkthrough

Let's consider a small example to illustrate the solution approach:

Suppose we have a field with the corners at (1, 1) and (5, 6) (m = 5, n = 6). There are horizontal fences at positions 2 and 3 (hFences = [2, 3]) and vertical fences at positions 2, 4 and 5 (vFences = [2, 4, 5]).

We would like to know the area of the largest square that can be formed by possibly removing fences.

1. Calculate Horizontal Distances:

   • Apply the helper function f to hFences with k = 5.
   • Add the boundary fences, so the modified fence list is [1, 2, 3, 5].
   • After sorting, we have the sorted list [1, 2, 3, 5].
   • Determine all pairwise distances: [(2-1), (3-1), (5-1), (3-2), (5-2), (5-3)], which simplifies to [1, 2, 4, 1, 3, 2].
   • The set hs of unique distances becomes {1, 2, 3, 4}.

2. Calculate Vertical Distances:

   • Apply the helper function f to vFences with k = 6.
   • Add the boundary fences, so the modified fence list is [1, 2, 4, 5, 6].
   • After sorting, we have the sorted list [1, 2, 4, 5, 6].
   • Determine all pairwise distances: [(2-1), (4-1), (5-1), (6-1), (4-2), (5-2), (6-2), (5-4), (6-4), (6-5)], which simplifies to [1, 3, 4, 5, 2, 3, 4, 1, 2, 1].
   • The set vs of unique distances becomes {1, 2, 3, 4, 5}.

3. Find Intersecting Distances:

   • Find the intersection of hs and vs: {1, 2, 3, 4} ∩ {1, 2, 3, 4, 5} which is {1, 2, 3, 4}.

4. Determine the Largest Square:

   • The maximum value in the intersection is 4, so the side length of the largest possible square is 4.
   • Therefore, the largest area is 4^2, which equals 16.
   • Since 16 is smaller than 10^9 + 7, we do not need to take the modulus.

The solution, therefore, is that the largest square that can be formed given hFences = [2, 3] and vFences = [2, 4, 5] on a field of size (5, 6) has an area of 16. If no common distances were identified, we would return -1.

Solution Implementation

Time and Space Complexity

Time Complexity

The primary contributors to time complexity are the sorting operations and the set comprehensions which include the generation of combinations of fence positions.

1. nums.sort() is called twice, once for horizontal fences (hFences) and once for vertical fences (vFences). The time complexity for sorting is O(k*log(k)) where k is the number of elements in the list being sorted. Since this is done for both hFences and vFences, it will be O(h*log(h)) + O(v*log(v)) where h is the number of horizontal fences and v is the number of vertical fences.

2. The combinations(nums, 2) function generates all possible pairs of fence positions, which is then used to calculate the gaps between fences. Since combinations are taken 2 at a time from a list with h+2 or v+2 (after adding the borders), the complexity of generating combinations is O((h+2)C2) for horizontal and O((v+2)C2) for vertical, respectively, which simplifies to O(h^2) and O(v^2).

3. The intersection operation hs & vs, while generally faster than the worst-case linear time in the size of the sets, is dependent on the number of elements generated in the previous step. Therefore, it will be bound by O(min(h^2, v^2)) in the worst case.

4. The final part involves a comparison and a modulo operation which are constant time operations.

Combining these together, the overall time complexity is dominated by the combination's computations, leading to O(h^2 + v^2).

Space Complexity

The space complexity is analyzed based on the additional space needed for computation, besides the input:

1. Additional lists created by extending hFences and vFences with the border positions do not significantly increase the space, effectively remaining O(h) and O(v).

2. The set comprehensions in the function f generate all pairs of fence positions. The number of gaps generated will thus be O(h^2) for hs and O(v^2) for vs.

3. The intersection hs & vs can have at most min(h^2, v^2) elements stored in a set.

Combining these space requirements, the overall space complexity is O(h^2 + v^2), which agrees with the reference answer.

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