2975. Maximum Square Area by Removing Fences From a Field LeetCode Solution

Last updated: March 5, 2025

In this guide, you will get 2975. Maximum Square Area by Removing Fences From a Field LeetCode Solution with the best time and space complexity. The solution to Maximum Square Area by Removing Fences From a Field problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Maximum Square Area by Removing Fences From a Field solution in C++
Maximum Square Area by Removing Fences From a Field solution in Java
Maximum Square Area by Removing Fences From a Field solution in Python
Additional Resources

2975. Maximum Square Area by Removing Fences From a Field LeetCode Solution image

Problem Statement of Maximum Square Area by Removing Fences From a Field

There is a large (m – 1) x (n – 1) rectangular field with corners at (1, 1) and (m, n) containing some horizontal and vertical fences given in arrays hFences and vFences respectively.
Horizontal fences are from the coordinates (hFences[i], 1) to (hFences[i], n) and vertical fences are from the coordinates (1, vFences[i]) to (m, vFences[i]).
Return the maximum area of a square field that can be formed by removing some fences (possibly none) or -1 if it is impossible to make a square field.
Since the answer may be large, return it modulo 109 + 7.
Note: The field is surrounded by two horizontal fences from the coordinates (1, 1) to (1, n) and (m, 1) to (m, n) and two vertical fences from the coordinates (1, 1) to (m, 1) and (1, n) to (m, n). These fences cannot be removed.

Example 1:

Input: m = 4, n = 3, hFences = [2,3], vFences = [2]
Output: 4
Explanation: Removing the horizontal fence at 2 and the vertical fence at 2 will give a square field of area 4.

Example 2:

Input: m = 6, n = 7, hFences = [2], vFences = [4]
Output: -1
Explanation: It can be proved that there is no way to create a square field by removing fences.

Constraints:

3 <= m, n <= 109
1 <= hFences.length, vFences.length <= 600
1 < hFences[i] < m
1 < vFences[i] < n
hFences and vFences are unique.

Complexity Analysis

Time Complexity: O(|\texttt{hFences}|^2 + |\texttt{vFences}|^2)
Space Complexity: O(|\texttt{hFences}|^2 + |\texttt{vFences}|^2)

2975. Maximum Square Area by Removing Fences From a Field LeetCode Solution in C++

class Solution {
 public:
  int maximizeSquareArea(int m, int n, vector<int>& hFences,
                         vector<int>& vFences) {
    constexpr int kMod = 1'000'000'007;

    hFences.push_back(1);
    hFences.push_back(m);
    vFences.push_back(1);
    vFences.push_back(n);

    ranges::sort(hFences);
    ranges::sort(vFences);

    const unordered_set<int> hGaps = getGaps(hFences);
    const unordered_set<int> vGaps = getGaps(vFences);
    int maxGap = -1;

    for (const int hGap : hGaps)
      if (vGaps.contains(hGap))
        maxGap = max(maxGap, hGap);

    return maxGap == -1 ? -1 : static_cast<long>(maxGap) * maxGap % kMod;
  }

 private:
  unordered_set<int> getGaps(const vector<int>& fences) {
    unordered_set<int> gaps;
    for (int i = 0; i < fences.size(); ++i)
      for (int j = 0; j < i; ++j)
        gaps.insert(fences[i] - fences[j]);
    return gaps;
  }
};
/* code provided by PROGIEZ */

2975. Maximum Square Area by Removing Fences From a Field LeetCode Solution in Java

class Solution {
  public int maximizeSquareArea(int m, int n, int[] hFences, int[] vFences) {
    final int kMod = 1_000_000_007;

    hFences = Arrays.copyOf(hFences, hFences.length + 2);
    vFences = Arrays.copyOf(vFences, vFences.length + 2);

    hFences[hFences.length - 2] = 1;
    hFences[hFences.length - 1] = m;
    vFences[vFences.length - 2] = 1;
    vFences[vFences.length - 1] = n;

    Arrays.sort(hFences);
    Arrays.sort(vFences);

    Set<Integer> hGaps = getGaps(hFences);
    Set<Integer> vGaps = getGaps(vFences);
    int maxGap = -1;

    for (final int hGap : hGaps)
      if (vGaps.contains(hGap))
        maxGap = Math.max(maxGap, hGap);

    return maxGap == -1 ? -1 : (int) ((long) maxGap * maxGap % kMod);
  }

  private Set<Integer> getGaps(int[] fences) {
    Set<Integer> gaps = new HashSet<>();
    for (int i = 0; i < fences.length; ++i)
      for (int j = 0; j < i; ++j)
        gaps.add(fences[i] - fences[j]);
    return gaps;
  }
}
// code provided by PROGIEZ

2975. Maximum Square Area by Removing Fences From a Field LeetCode Solution in Python

class Solution:
  def maximizeSquareArea(
      self,
      m: int,
      n: int,
      hFences: list[int],
      vFences: list[int],
  ) -> int:
    hFences = sorted(hFences + [1, m])
    vFences = sorted(vFences + [1, n])
    hGaps = {hFences[i] - hFences[j]
             for i in range(len(hFences))
             for j in range(i)}
    vGaps = {vFences[i] - vFences[j]
             for i in range(len(vFences))
             for j in range(i)}
    maxGap = next((hGap
                  for hGap in sorted(hGaps, reverse=True)
                  if hGap in vGaps), -1)
    return -1 if maxGap == -1 else maxGap**2 % (10**9 + 7)
 # code by PROGIEZ

Additional Resources

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