2943. Maximize Area of Square Hole in Grid

2943. Maximize Area of Square Hole in Grid

Difficulty: Medium

Topics: Array, Sorting, Biweekly Contest 118

You are given the two integers, n and m and two integer arrays, hBars and vBars. The grid has n + 2 horizontal and m + 2 vertical bars, creating 1 x 1 unit cells. The bars are indexed starting from 1.

You can remove some of the bars in hBars from horizontal bars and some of the bars in vBars from vertical bars. Note that other bars are fixed and cannot be removed.

Return an integer denoting the maximum area of a square-shaped hole in the grid, after removing some bars (possibly none).

Example 1:

• Input: n = 2, m = 1, hBars = [2,3], vBars = [2]
• Output: 4
• Explanation:
  • The left image shows the initial grid formed by the bars. The horizontal bars are [1,2,3,4], and the vertical bars are [1,2,3].
  • One way to get the maximum square-shaped hole is by removing horizontal bar 2 and vertical bar 2.

Example 2:

• Input: n = 1, m = 1, hBars = [2], vBars = [2]
• Output: 4
• Explanation: To get the maximum square-shaped hole, we remove horizontal bar 2 and vertical bar 2.

Example 3:

• Input: n = 2, m = 3, hBars = [2,3], vBars = [2,4]
• Output: 4
• Explanation: One way to get the maximum square-shaped hole is by removing horizontal bar 3, and vertical bar 4.

Constraints:

• 1 <= n <= 10⁹
• 1 <= m <= 10⁹
• 1 <= hBars.length <= 100
• 2 <= hBars[i] <= n + 1
• 1 <= vBars.length <= 100
• 2 <= vBars[i] <= m + 1
• All values in hBars are distinct.
• All values in vBars are distinct.

Hint:

1. Sort hBars and vBars and consider them separately.
2. Compute the longest sequence of consecutive integer values in each array, denoted as [hx, hy] and [vx, vy], respectively.
3. The maximum square length we can get is min(hy - hx + 2, vy - vx + 2).
4. Square the maximum square length to get the area.

Solution:

We are given the problem of maximizing the area of a square hole in a grid by removing some bars.
The grid has horizontal bars from 1 to n+2 and vertical bars from 1 to m+2.
We are given two arrays: hBars (horizontal bars that we can remove) and vBars (vertical bars that we can remove).

Approach:

1. Sort both arrays: Sort hBars and vBars in ascending order to easily identify consecutive sequences.
2. Find longest consecutive sequences: For each sorted array, find the length of the longest consecutive sequence of integers.
3. Calculate gap lengths: For a sequence of k consecutive removable bars, you can create a gap of length (k + 1) units (since the fixed bars at both ends remain).
4. Determine square side: The maximum square hole side length is the minimum of the maximum horizontal gap and maximum vertical gap.
5. Return area: Square the side length to get the area.

Let's implement this solution in PHP: 2943. Maximize Area of Square Hole in Grid

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

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

Explanation:

• Consecutive bar removal: When you remove consecutive horizontal bars (like bars 2, 3, 4), you effectively merge multiple adjacent 1×1 cells into a larger rectangular hole. The same applies to vertical bars.
• Gap calculation: If you remove bars [x, x+1, x+2, ..., x+k-1] (k consecutive bars), the gap between the fixed bar at (x-1) and (x+k) becomes (k + 1) units.
• Square constraint: A square hole requires equal horizontal and vertical dimensions. The maximum square size is limited by the smaller of the maximum horizontal and vertical gaps you can create.
• Fixed bars: Note that bars 1 and (n+2) (for horizontal) and bars 1 and (m+2) (for vertical) are always fixed and cannot be removed. The arrays hBars and vBars contain indices of removable bars between these fixed boundaries.

Key Insight: The problem reduces to finding the longest consecutive sequence in each array, then using min(horizontal_gap, vertical_gap) as the square side length.

Complexity

• Time Complexity: O(h log h + v log v) where h = length of hBars, v = length of vBars (due to sorting).
• Space Complexity: O(1) extra space (excluding input storage).

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