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:
-
Helper Function
f
:- This function takes a list of fence positions (
nums
) and the maximum extent of the field (k
), which would bem
for horizontal andn
for vertical fences. - It first adds the boundary positions,
1
andk
, 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.
- This function takes a list of fence positions (
-
Finding Common Distances:
- The
f
function is called twice, once forhFences
withm
and once forvFences
withn
, returning setshs
andvs
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.
- The
-
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 squaringans
and then take the modulus with10**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
.
- From the set intersection, we find the maximum value (
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.
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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.
-
Calculate Horizontal Distances:
- Apply the helper function
f
tohFences
withk = 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}
.
- Apply the helper function
-
Calculate Vertical Distances:
- Apply the helper function
f
tovFences
withk = 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}
.
- Apply the helper function
-
Find Intersecting Distances:
- Find the intersection of
hs
andvs
:{1, 2, 3, 4}
∩{1, 2, 3, 4, 5}
which is{1, 2, 3, 4}
.
- Find the intersection of
-
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 equals16
. - Since
16
is smaller than10^9 + 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
1from typing import List, Set
2from itertools import combinations
3
4class Solution:
5 def maximizeSquareArea(self, m: int, n: int, hFences: List[int], vFences: List[int]) -> int:
6 # Helper function to find the differences between consecutive elements
7 # in a sorted version of nums extended with 1 and k.
8 def find_intervals(nums: List[int], k: int) -> Set[int]:
9 # Add borders as possible fence positions
10 nums.extend([1, k])
11 nums.sort()
12 # Create a set of all possible intervals between fences
13 return {b - a for a, b in combinations(nums, 2)}
14
15 # Modulus for the final answer to prevent large integer issues
16 MOD = 10**9 + 7
17
18 # Calculate horizontal and vertical intervals
19 horizontal_intervals = find_intervals(hFences, m)
20 vertical_intervals = find_intervals(vFences, n)
21
22 # Find the largest common interval (square side length) between horizontal and vertical
23 max_square_side = max(horizontal_intervals & vertical_intervals, default=0)
24
25 # Compute the area of the square with side length max_square_side
26 # If max_square_side is 0, return -1; otherwise, return the area modulo MOD.
27 return max_square_side ** 2 % MOD if max_square_side else -1
28
1import java.util.Set;
2import java.util.HashSet;
3import java.util.Arrays;
4
5class Solution {
6 public int maximizeSquareArea(int m, int n, int[] horizontalFences, int[] verticalFences) {
7 // Get the set of all possible fence lengths for horizontal and vertical fences
8 Set<Integer> horizontalSet = calculateFenceLengths(horizontalFences, m);
9 Set<Integer> verticalSet = calculateFenceLengths(verticalFences, n);
10
11 // Retain only the common lengths in both sets to ensure it can be a square
12 horizontalSet.retainAll(verticalSet);
13
14 // Initialize the answer as -1 to signify no square found
15 int maxSquareArea = -1;
16
17 // Define the modulo due to large numbers
18 final int mod = (int) 1e9 + 7;
19
20 // Iterate through the common fence lengths to find the maximum square area
21 for (int length : horizontalSet) {
22 maxSquareArea = Math.max(maxSquareArea, length);
23 }
24
25 // If a valid square is found, calculate its area else return -1
26 return maxSquareArea > 0 ? (int) (1L * maxSquareArea * maxSquareArea % mod) : -1;
27 }
28
29 // Helper method to calculate possible fence lengths given the fences and the total length
30 private Set<Integer> calculateFenceLengths(int[] fences, int totalLength) {
31 int n = fences.length;
32 // Extend the array to include the first and last fence
33 fences = Arrays.copyOf(fences, n + 2);
34 fences[n] = 1;
35 fences[n + 1] = totalLength;
36 // Sort the fence array to evaluate the fence lengths in order
37 Arrays.sort(fences);
38
39 // Use a Set to avoid duplicates and store all possible lengths
40 Set<Integer> lengthsSet = new HashSet<>();
41
42 // Calculate all the differences between pairs of fences which represent all possible lengths
43 for (int i = 0; i < fences.length; ++i) {
44 for (int j = 0; j < i; ++j) {
45 lengthsSet.add(fences[i] - fences[j]);
46 }
47 }
48
49 return lengthsSet;
50 }
51}
52
1#include <vector>
2#include <unordered_set>
3#include <algorithm>
4
5class Solution {
6public:
7 // Function to find potential square side lengths
8 // given a set of fence positions and a dimension limit.
9 std::unordered_set<int> findPotentialSides(std::vector<int>& positions, int limit) {
10 // Adding boundary limits to the vector
11 positions.push_back(limit);
12 positions.push_back(1);
13
14 // Sort the positions to allow for positional comparisons
15 std::sort(positions.begin(), positions.end());
16
17 // Create a set to store unique differences (potential side lengths)
18 std::unordered_set<int> sideLengths;
19
20 // Calculate all possible gaps between fences (side lengths for squares)
21 for (size_t i = 0; i < positions.size(); ++i) {
22 for (size_t j = 0; j < i; ++j) {
23 // Insert the difference in positions as a potential side length
24 sideLengths.insert(positions[i] - positions[j]);
25 }
26 }
27
28 // Return set containing all potential side lengths
29 return sideLengths;
30 }
31
32 // Main function to maximize the area of a square within given horizontal and vertical fence limits
33 int maximizeSquareArea(int m, int n, std::vector<int>& hFences, std::vector<int>& vFences) {
34 // Find horizontal and vertical potential side lengths
35 std::unordered_set<int> horizontalSides = findPotentialSides(hFences, m);
36 std::unordered_set<int> verticalSides = findPotentialSides(vFences, n);
37
38 int maxSideLength = 0; // Initialize the maximum side length to 0
39
40 // Look for the largest common side length in the horizontal and vertical sets
41 for (const int& side : horizontalSides) {
42 if (verticalSides.count(side)) {
43 maxSideLength = std::max(maxSideLength, side);
44 }
45 }
46
47 const int mod = 1e9 + 7; // Define modulus value for result
48
49 // Return the square of the maximal side length modulo 1e9 + 7,
50 // or -1 if no square can be formed
51 return maxSideLength > 0 ? static_cast<long long>(maxSideLength) * maxSideLength % mod : -1;
52 }
53};
54
1// Calculates the maximum area of a square that can fit within a grid with the
2// removal of horizontal and vertical fences.
3function maximizeSquareArea(rows: number, cols: number, horizontalFences: number[], verticalFences: number[]): number {
4 // Helper function to calculate the set of all possible distances along
5 // a dimension, given the fences removed and the length of that dimension.
6 const calculateDistances = (fences: number[], dimensionLength: number): Set<number> => {
7 // Add the boundaries to the fences array.
8 fences.push(1, dimensionLength);
9 // Sort the array to establish ordered fence positions.
10 fences.sort((a, b) => a - b);
11 const distances: Set<number> = new Set();
12
13 // Calculate all possible distances between fences and add to the set.
14 for (let i = 0; i < fences.length; ++i) {
15 for (let j = 0; j < i; ++j) {
16 distances.add(fences[i] - fences[j]);
17 }
18 }
19 return distances;
20 };
21
22 // Calculate all possible distances for both horizontal and vertical dimensions.
23 const horizontalDistances = calculateDistances(horizontalFences, rows);
24 const verticalDistances = calculateDistances(verticalFences, cols);
25
26 // Initialize the maximum square side length to 0.
27 let maxSquareSide = 0;
28
29 // Iterate over the horizontal distances to find the largest square's side length,
30 // which fits both horizontally and vertically.
31 for (const distance of horizontalDistances) {
32 if (verticalDistances.has(distance)) {
33 maxSquareSide = Math.max(maxSquareSide, distance);
34 }
35 }
36
37 // Calculate the area of the square with the found side length, and apply
38 // the modulo operation as specified.
39 // If no valid square is found, return -1.
40 return maxSquareSide ? Number(BigInt(maxSquareSide) ** 2n % 1000000007n) : -1;
41}
42
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.
-
nums.sort()
is called twice, once for horizontal fences (hFences
) and once for vertical fences (vFences
). The time complexity for sorting isO(k*log(k))
wherek
is the number of elements in the list being sorted. Since this is done for bothhFences
andvFences
, it will beO(h*log(h)) + O(v*log(v))
whereh
is the number of horizontal fences andv
is the number of vertical fences. -
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 withh+2
orv+2
(after adding the borders), the complexity of generating combinations isO((h+2)C2)
for horizontal andO((v+2)C2)
for vertical, respectively, which simplifies toO(h^2)
andO(v^2)
. -
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 byO(min(h^2, v^2))
in the worst case. -
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:
-
Additional lists created by extending
hFences
andvFences
with the border positions do not significantly increase the space, effectively remainingO(h)
andO(v)
. -
The set comprehensions in the function
f
generate all pairs of fence positions. The number of gaps generated will thus beO(h^2)
forhs
andO(v^2)
forvs
. -
The intersection
hs & vs
can have at mostmin(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.
How would you design a stack which has a function min
that returns the minimum element in the stack, in addition to push
and pop
? All push
, pop
, min
should have running time O(1)
.
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