2943. Maximize Area of Square Hole in Grid
Problem Description
In this LeetCode problem, we're tasked with finding the largest possible square-shaped hole that can be created in a grid by removing certain bars. The grid consists of n + 2
horizontal bars and m + 2
vertical bars, forming 1 x 1
unit cells. The bars are 1-indexed for reference.
To create a hole, we have the option of removing horizontal bars at positions listed in hBars
, and vertical bars at positions listed in vBars
. Each array contains distinct positions, and they fall within certain rangesāhBars
values are within [2, n + 1]
, and vBars
within [2, m + 1]
.
Our goal is to return the maximum area of the square hole. To visualize this, imagine removing consecutive bars to create the largest possible square void in the grid's structure.
Intuition
Firstly, we need to understand that the maximum square hole we can make in the grid is limited by the number of consecutive bars we can remove either horizontally or vertically. Therefore, the key to this problem is figuring out the longest sequence of consecutive bar numbers in the hBars
and vBars
that weāre permitted to take out.
Let's consider any sequence of removable bars. The longest consecutive run of bar numbers in either direction will ultimately determine the maximum size of the square-shaped hole we can achieve. For instance, if we can remove three consecutive horizontal bars and two consecutive vertical bars, the largest square hole we can create is 2 x 2
, since the square cannot exceed the smaller side.
The solution approach here is quite clever. We sort each bar array to make it easier to find consecutive sequences. Then, using a helper function f(nums)
, we traverse through the sorted array and count the length of the longest increasing consecutive subsequence. This function returns the longest length plus one (to account for the extra space at the end of the last bar, as a square hole need one more unit space to be formed).
After calculating this for both hBars
and vBars
, the side length of the largest possible square hole is the minimum of these two lengths. Since square area is equal to the side length squared, we finally return the side length squared (calculated minimum) as the answer.
Learn more about Sorting patterns.
Solution Approach
The implementation relies on a straightforward but effective sorting and counting approach.
Here's a step-by-step breakdown:
-
Sort both arrays,
hBars
andvBars
. Sorting brings all consecutive sequences into adjacent positions in the array, which makes it easy to count them in the next step. -
Define a helper function
f(nums)
which takes a sorted list of bar positions. Initialize two variables:ans
andcnt
to1
. Here,ans
will store the length of the longest consecutive sequence we find, andcnt
is a counter for the current consecutive sequence. -
Loop through the
nums
list starting from the second element (indexed at1
), and compare each element with its predecessor.-
If the current element is exactly one more than the previous (
nums[i] == nums[i - 1] + 1
), it means that we've found consecutive bars. Incrementcnt
to extend the current consecutive sequence. Updateans
to hold the maximum value betweenans
and the currentcnt
. -
If the current element is not consecutive, reset
cnt
to1
because we've hit a break in the sequence, and we'll need to start counting a new sequence from this point.
-
-
When the loop finishes, add
1
to the finalans
. This is because the square hole size is actually one unit larger than the number of bars removed (the last bar removed implies there is one extra unit of open space that completes the square). -
Apply the helper function to both
hBars
andvBars
to get the size of the longest consecutive sequence for both horizontal and vertical bars. -
The side length of the largest square hole we can create is the minimum of these two sizes. This is because the shape has to be a square, therefore, it is limited by the shorter side.
-
Calculate the area of the largest square hole by squaring the side length found in the previous step. The square of the minimum consecutive sequence length provides us with the maximum square hole area after removing the bars, which is the required output.
Using the steps from the approach above, the code neatly encapsulates the logic within one main function maximizeSquareHoleArea
and one helper function f
. The use of sorting and a single pass counting mechanism ensures an efficient solution, with the overall time complexity being dominated by the sorting step, which is O(n log n)
for each array.
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Start EvaluatorExample Walkthrough
To illustrate the solution approach, let's walk through a small example. Suppose we have the following input:
n + 2 = 5
horizontal bars (indexed from 1 to 5)m + 2 = 5
vertical bars (indexed from 1 to 5)hBars
= [3, 4]vBars
= [2, 4]
Step 1: Sort both arrays, hBars
= [3, 4], vBars
= [2, 4].
The arrays are already sorted in this example, so we can proceed directly to the next step.
Step 2: We will use a helper function f(nums)
to find the length of the longest consecutive sequence.
Step 3: Let's apply function f(nums)
on hBars
:
- We start with
ans = 1
andcnt = 1
. - Loop through
hBars
, starting from the second element:- Compare
hBars[1]
tohBars[0]
. Since4 == 3 + 1
, they are consecutive, so increasecnt
to 2. - Since we are at the end of the array, we finish the loop with
ans
being the value ofcnt
, which is2
.
- Compare
Step 4: Add 1
to the final ans
, yielding 2 + 1 = 3
. This means horizontally, we can fit a square hole of side length 3.
Step 5: Repeat steps 3 and 4 for vBars
:
cnt
starts at1
, andans
starts at1
.- Comparing
vBars[1]
tovBars[0]
, we see that4 != 2 + 1
, so these are not consecutive. - The loop ends with
ans
still equal to1
.
Step 6: Add 1
to the final ans
, yielding 1 + 1 = 2
. This means vertically, the largest square hole we can form has a side length of 2.
Step 7: The comparison of the two sizes (3 and 2)
shows that the maximum square hole we can create is limited by the vertical bars, which allow for a side length of 2
.
Step 8: The area of the square is the side length squared, so in this example, the largest square hole we can create has an area of 2 x 2 = 4
.
Following this approach, the maximizeSquareHoleArea
function would return the value 4
as the answer for this example.
Solution Implementation
1class Solution:
2 def maximizeSquareHoleArea(self, n: int, m: int, hBars: List[int], vBars: List[int]) -> int:
3 # Define a helper function to find the largest square hole along one direction (either horizontally or vertically).
4 def find_largest_consecutive_sequence(nums: List[int]) -> int:
5 nums.sort() # Sort the array to find consecutive numbers.
6 max_sequence_length = current_sequence_length = 1 # Initialize counters for sequences.
7 for i in range(1, len(nums)):
8 if nums[i] == nums[i - 1] + 1:
9 current_sequence_length += 1 # Increase the sequence length if consecutive.
10 max_sequence_length = max(max_sequence_length, current_sequence_length)
11 else:
12 current_sequence_length = 1 # Reset the sequence length if not consecutive.
13 return max_sequence_length + 1 # Add 1 to include the space on both ends of the sequence.
14
15 # Calculate the maximum square hole size for both horizontal and vertical bars
16 # by finding the smallest of the two maximum consecutive sequences.
17 max_hole_size = min(find_largest_consecutive_sequence(hBars), find_largest_consecutive_sequence(vBars))
18
19 # Since the problem is about finding the area of the largest square hole,
20 # we square the maximum hole size to get the answer.
21 return max_hole_size ** 2
22
1class Solution {
2 public int maximizeSquareHoleArea(int n, int m, int[] horizontalBars, int[] verticalBars) {
3 // Find the maximum consecutive bars for both the horizontal and vertical arrays
4 int maxConsecutiveBars = Math.min(findMaxConsecutiveBars(horizontalBars), findMaxConsecutiveBars(verticalBars));
5 // The area of the largest square hole is the square of the number of maximum consecutive bars
6 return maxConsecutiveBars * maxConsecutiveBars;
7 }
8
9 // Helper method to find the length of the maximum set of consecutive bars
10 private int findMaxConsecutiveBars(int[] bars) {
11 // Sort the array to easily find consecutive numbers
12 Arrays.sort(bars);
13 // Initialize variables to store the current count of consecutive numbers and the maximum found so far
14 int maxConsecutive = 1;
15 int currentConsecutiveCount = 1;
16
17 // Iterate through the sorted array to find the maximum set of consecutive numbers
18 for (int i = 1; i < bars.length; ++i) {
19 // If the current bar is consecutive to the previous one, increase the count
20 if (bars[i] == bars[i - 1] + 1) {
21 currentConsecutiveCount++;
22 // Update maxConsecutive with the maximum value found so far
23 maxConsecutive = Math.max(maxConsecutive, currentConsecutiveCount);
24 } else {
25 // Reset the count if the current bar is not consecutive
26 currentConsecutiveCount = 1;
27 }
28 }
29 // Return the maximum consecutive count plus one (to account for the space between bars)
30 return maxConsecutive + 1;
31 }
32}
33
1#include <vector>
2#include <algorithm>
3using namespace std;
4
5class Solution {
6public:
7 // Method to maximize the square hole area given horizontal and vertical bars
8 int maximizeSquareHoleArea(int n, int m, vector<int>& horizontalBars, vector<int>& verticalBars) {
9 // Lambda function to calculate the largest square hole from a sequence of bars
10 auto findLargestSquare = [](vector<int>& bars) {
11 int largestSquareSide = 1; // Initial largest square side length
12 int consecutive = 1; // Count of consecutive bars
13 sort(bars.begin(), bars.end()); // Sort the bars to find consecutive numbers
14
15 // Iterate through the sorted bars to find the max count of consecutive numbers
16 for (int i = 1; i < bars.size(); ++i) {
17 if (bars[i] == bars[i - 1] + 1) {
18 // If consecutive, increment count
19 largestSquareSide = max(largestSquareSide, ++consecutive);
20 } else {
21 // Reset count if not consecutive
22 consecutive = 1;
23 }
24 }
25 return largestSquareSide + 1; // Add 1 to get the side length of the hole
26 };
27
28 // Call the lambda function for both horizontal and vertical bars
29 int maxHorizontal = findLargestSquare(horizontalBars);
30 int maxVertical = findLargestSquare(verticalBars);
31
32 // Find the minimum of max horizontal and vertical squares to form a square hole
33 int minSideLength = min(maxHorizontal, maxVertical);
34
35 // Calculate and return the area of the largest square hole
36 return minSideLength * minSideLength;
37 }
38};
39
1// Function to find the maximum size for a square hole that can be formed
2// by removing certain horizontal and vertical bars.
3// n: The total number of horizontal bars
4// m: The total number of vertical bars
5// hBars: List of positions of removable horizontal bars
6// vBars: List of positions of removable vertical bars
7function maximizeSquareHoleArea(n: number, m: number, hBars: number[], vBars: number[]): number {
8 // Helper function to calculate the maximum number of consecutive bars
9 const getMaxConsecutiveBars = (bars: number[]): number => {
10 // Sort the bars array to count consecutive numbers
11 bars.sort((a, b) => a - b);
12 let maxSize = 1; // Initialize maximum size to 1
13 let consecutiveCount = 1; // Count of consecutive bars, start with 1 as we count the first bar
14
15 // Iterate over the bars to find the maximum consecutive sequence
16 for (let i = 1; i < bars.length; ++i) {
17 // If the current bar is consecutive to the previous one
18 if (bars[i] === bars[i - 1] + 1) {
19 consecutiveCount++; // Increase count
20 maxSize = Math.max(maxSize, consecutiveCount); // Update the max size found
21 } else {
22 // Reset count when no longer consecutive
23 consecutiveCount = 1;
24 }
25 }
26
27 // Return the size including the space without bars
28 return maxSize + 1;
29 };
30
31 // Calculate the maximum size for both horizontal and vertical bars
32 const horizontalMaxSize = getMaxConsecutiveBars(hBars);
33 const verticalMaxSize = getMaxConsecutiveBars(vBars);
34
35 // The size of the square hole will be the minimum of the two sizes squared
36 // because the hole must be square, and its sides depend on the minimum of horizontal and vertical spacing
37 return Math.min(horizontalMaxSize, verticalMaxSize) ** 2;
38}
39
Time and Space Complexity
The function maximizeSquareHoleArea
includes two calls to the helper function f
, once for hBars
and once for vBars
. The complexity analysis will be the same for both calls since the operations performed by f
are identical for both lists.
Time Complexity
The time complexity is dominated by the sorting operation inside the function f
. Since Python's sort method typically uses Timsort, which has an average and worst-case time complexity of O(n log n)
where n
is the length of the list being sorted. The subsequent loop runs in linear time, O(n)
, since it iterates through the sorted list once.
Therefore, the time complexity of the function f
is O(n log n) + O(n)
which simplifies to O(n log n)
because the n log n
term dominates.
Since we call the function f
twice, once for hBars
and once for vBars
, and assuming n
refers to the longer of the two lists for worst-case analysis, then the overall time complexity remains O(n log n)
since these two calls are sequential, not nested.
Space Complexity
The space complexity of the function f
is O(n)
because of the space required to sort the list. The sort operation may require additional space to hold elements temporarily during the sort. The variables ans
and cnt
use constant space and do not scale with input size.
Overall, since we consider the larger of the two lists hBars
and vBars
, the total space complexity of the maximizeSquareHoleArea
function is O(n)
, where n
is the length of the larger input list.
Learn more about how to find time and space complexity quickly using problem constraints.
Which of the following array represent a max heap?
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