How do you solve Falling Squares on LeetCode?

Falling Squares (LeetCode #699) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a Segment Tree allows efficient height tracking.

What is the brute force solution for Falling Squares?

The brute force approach for Falling Squares is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each square against all previously dropped squares to determine where it lands. This is straightforward but inefficient, as it requires multiple comparisons for each square. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Falling Squares?

Falling Squares uses the following concepts: Array, Segment Tree, Ordered Set. The recommended patterns to study are: Segment Tree, Dynamic Programming.

What are the interview tips for Falling Squares?

Explain your thought process clearly. Consider edge cases, such as overlapping squares. Practice implementing data structures like Segment Trees.

What are common mistakes when solving Falling Squares?

Not considering the height of previously dropped squares. Failing to update the data structure correctly.

#699

Falling Squares

Hard

ArraySegment TreeOrdered SetSegment TreeDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using a data structure like a Segment Tree allows us to efficiently track the heights of the squares and quickly find the maximum height in the overlapping regions. This significantly reduces the time complexity compared to the brute force method.

⚙️

Algorithm

3 steps

1Step 1: Use a Segment Tree to maintain the heights of the squares.
2Step 2: For each square, calculate its landing height by querying the Segment Tree for the maximum height in the range it occupies.
3Step 3: Update the Segment Tree with the new height of the square after it lands.

solution.py41 lines

1class SegmentTree:
2    def __init__(self, size):
3        self.size = size
4        self.tree = [0] * (4 * size)
5
6    def update(self, index, value, node=1, start=0, end=None):
7        if end is None:
8            end = self.size - 1
9        if start == end:
10            self.tree[node] = max(self.tree[node], value)
11        else:
12            mid = (start + end) // 2
13            if index <= mid:
14                self.update(index, value, node * 2, start, mid)
15            else:
16                self.update(index, value, node * 2 + 1, mid + 1, end)
17            self.tree[node] = max(self.tree[node * 2], self.tree[node * 2 + 1])
18
19    def query(self, left, right, node=1, start=0, end=None):
20        if end is None:
21            end = self.size - 1
22        if left > end or right < start:
23            return 0
24        if left <= start and end <= right:
25            return self.tree[node]
26        mid = (start + end) // 2
27        return max(self.query(left, right, node * 2, start, mid), self.query(left, right, node * 2 + 1, mid + 1, end))
28
29
30def fallingSquares(positions):
31    max_position = 0
32    for left, side in positions:
33        max_position = max(max_position, left + side)
34    seg_tree = SegmentTree(max_position)
35    result = []
36    for left, side in positions:
37        right = left + side - 1
38        height = seg_tree.query(left, right)
39        seg_tree.update(left, height + side)
40        result.append(max(seg_tree.tree))
41    return result

ℹ

Complexity note: The time complexity is O(n log n) because each update and query operation in the Segment Tree takes O(log n) time, and we perform these operations for each square. The space complexity is O(n) due to the storage used for the Segment Tree.

1Using a Segment Tree allows efficient height tracking.
2Understanding the overlap of squares is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.