2975. Maximum Square Area by Removing Fences From a Field

2975. Maximum Square Area by Removing Fences From a Field

Difficulty: Medium

Topics: Array, Hash Table, Enumeration, Weekly Contest 377

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.

Hint:

1. Put 1 and m into hFences. The differences of any two values in the new hFences can be a horizontal edge of a rectangle.
2. Similarly, put 1 and n into vFences. The differences of any two values in the new vFences can be a vertical edge of a rectangle.
3. Our goal is to find the maximum common value in both parts.

Solution:

We are given a large field with boundaries at x=1, x=m, y=1, y=n. Inside, there are horizontal fences (parallel to x-axis) at given hFences (x-coordinates) and vertical fences (parallel to y-axis) at given vFences (y-coordinates). We can remove some fences to create a larger open area. We want the maximum area of a square that can be formed by removing some fences.

Approach:

1. Expand fence arrays: Add the boundary fences (1 and m to hFences, 1 and n to vFences) since these cannot be removed but define the available segments.
2. Find all possible horizontal segments: Compute all possible distances (differences) between any two horizontal fences by iterating through all pairs in the expanded hFences array.
3. Find all possible vertical segments: Similarly compute all possible distances between any two vertical fences in the expanded vFences array.
4. Find maximum common distance: Look for the largest distance that appears in both the horizontal and vertical segment sets. This represents the maximum possible square side length.
5. Calculate area: If a common distance exists, return (side × side) modulo 10⁹+7. Otherwise return -1.

Let's implement this solution in PHP: 2975. Maximum Square Area by Removing Fences From a Field

<?php
/**
 * @param Integer $m
 * @param Integer $n
 * @param Integer[] $hFences
 * @param Integer[] $vFences
 * @return Integer
 */
function maximizeSquareArea(int $m, int $n, array $hFences, array $vFences): int
{
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo maximizeSquareArea(4, 3, [2,3], [2]) . "\n";       // Output: 4
echo maximizeSquareArea(6, 7, [2]. [4]) . "\n";         // Output: -1
?>

Explanation:

• The key insight is that any two horizontal fences define a possible vertical side length for a rectangle, and any two vertical fences define a possible horizontal side length.
• To form a square, we need the same length to be available in both horizontal and vertical directions.
• By adding the boundaries (1, m) and (1, n), we ensure all possible segments are considered, including those using the outer boundaries.
• The algorithm efficiently finds all possible segment lengths using O(k²) comparisons where k is the number of fences (≤ 600), which is manageable.
• The solution works within the constraints by avoiding the massive m,n values (up to 10⁹) directly and only working with the fence positions.

Complexity:

• Time: O(H² + V²) where H and V are the numbers of horizontal and vertical fences after adding boundaries
• Space: O(H²) for storing the horizontal differences

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