We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Maximum Square Area by Removing Fences From a Field

Number: 3250

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

Given a rectangular field defined by its boundary fences at (1, 1) and (m, n) along with internal removable horizontal fences at positions specified in hFences and vertical fences specified in vFences, the task is to remove some of the internal fences (possibly none) such that the remaining fences form the boundary of a square field. Note that the external boundary fences cannot be removed. Return the maximum square area possible (i.e. side^2) modulo 10^9+7 if a square can be formed; otherwise return -1.

Key Insights

  • The final square field must have its four boundaries chosen from the set of fences that remain. Since the boundaries are only the fences that originally exist, you can only use fence positions from {1} ∪ hFences ∪ {m} for horizontal boundaries and {1} ∪ vFences ∪ {n} for vertical.
  • You can remove any internal fence to “merge” intervals. Thus, you can effectively choose any two fence positions from the available set to form the boundary as long as they are not removed.
  • The question reduces to finding a value s such that there exists a pair of horizontal fences with difference s and also a pair of vertical fences with the same difference. You then return s^2 (modulo 10^9+7) with s as large as possible.
  • If no common gap exists between the horizontal and vertical candidates, it is impossible to form a square field; hence, the answer is -1.

Space and Time Complexity

Time Complexity: O(k^2) for each of the horizontal and vertical arrays (k ≤ 602), which is acceptable. Space Complexity: O(k^2) in the worst case to store differences, which is negligible given the constraints.

Solution

First, construct the complete sets of fence positions for horizontal and vertical directions by including the boundaries (1 and m for horizontal, 1 and n for vertical) along with the removable fences. Sort the lists. Next, compute the differences between every pair of positions in the horizontal list; do the same for the vertical list. Since you are allowed to remove fences arbitrarily, any two positions can serve as boundaries as long as they come from the available set. Then, find the intersection of the differences from horizontal and vertical directions. The maximum value in this intersection (if any exists) will be the side length of the largest square field that can be formed. Compute and return its square modulo 10^9+7; if no common side can be formed, return -1.

Code Solutions

# Python solution with detailed comments

MOD = 10**9 + 7

def maxSquareArea(m, n, hFences, vFences):
    # Create complete sorted lists of fence positions for horizontal and vertical sides
    horizontal = sorted(set([1, m] + hFences))
    vertical = sorted(set([1, n] + vFences))
    
    # Function to calculate all possible differences from a sorted list of fence positions
    def get_differences(positions):
        differences = set()
        length = len(positions)
        for i in range(length):
            for j in range(i+1, length):
                diff = positions[j] - positions[i]
                differences.add(diff)
        return differences
    
    # Compute all differences in horizontal and vertical directions
    hDiff = get_differences(horizontal)
    vDiff = get_differences(vertical)
    
    # Find the common differences that appear in both horizontal and vertical directions
    common = hDiff.intersection(vDiff)
    
    # If there is no common gap, return -1
    if not common:
        return -1
    
    # Maximum possible square side is the maximum common difference
    max_side = max(common)
    # Return the area modulo MOD
    return (max_side * max_side) % MOD

# Example usage:
#print(maxSquareArea(4, 3, [2, 3], [2]))  # Expected output: 4 for Example 1
#print(maxSquareArea(6, 7, [2], [4]))      # Expected output: -1 for Example 2
← Back to All Questions