699. Falling Squares
Description
There are several squares being dropped onto the X-axis of a 2D plane.
You are given a 2D integer array positions
where positions[i] = [lefti, sideLengthi]
represents the ith
square with a side length of sideLengthi
that is dropped with its left edge aligned with X-coordinate lefti
.
Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands on the top side of another square or on the X-axis. A square brushing the left/right side of another square does not count as landing on it. Once it lands, it freezes in place and cannot be moved.
After each square is dropped, you must record the height of the current tallest stack of squares.
Return an integer array ans
where ans[i]
represents the height described above after dropping the ith
square.
Example 1:
Input: positions = [[1,2],[2,3],[6,1]] Output: [2,5,5] Explanation: After the first drop, the tallest stack is square 1 with a height of 2. After the second drop, the tallest stack is squares 1 and 2 with a height of 5. After the third drop, the tallest stack is still squares 1 and 2 with a height of 5. Thus, we return an answer of [2, 5, 5].
Example 2:
Input: positions = [[100,100],[200,100]] Output: [100,100] Explanation: After the first drop, the tallest stack is square 1 with a height of 100. After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100. Thus, we return an answer of [100, 100]. Note that square 2 only brushes the right side of square 1, which does not count as landing on it.
Constraints:
1 <= positions.length <= 1000
1 <= lefti <= 108
1 <= sideLengthi <= 106
Solutions
Solution 1: Segment Tree
According to the problem description, we need to maintain a set of intervals that support modification and query operations. In this case, we can use a segment tree to solve the problem.
A segment tree divides the entire interval into multiple non-contiguous sub-intervals, with the number of sub-intervals not exceeding $\log(width)$, where $width$ is the length of the interval. To update the value of an element, we only need to update $\log(width)$ intervals, and these intervals are all contained within a larger interval that includes the element. When modifying intervals, we need to use lazy propagation to ensure efficiency.
- Each node of the segment tree represents an interval;
- The segment tree has a unique root node representing the entire statistical range, such as $[1, n]$;
- Each leaf node of the segment tree represents a primitive interval of length 1, $[x, x]$;
- For each internal node $[l, r]$, its left child is $[l, \textit{mid}]$, and its right child is $[\textit{mid} + 1, r]$, where $\textit{mid} = \frac{l + r}{2}$;
For this problem, the information maintained by the segment tree nodes includes:
- The maximum height $v$ of the blocks in the interval
- Lazy propagation marker $add$
Additionally, since the range of the number line is very large, up to $10^8$, we use dynamic node creation.
In terms of time complexity, each query and modification has a time complexity of $O(\log n)$, and the total time complexity is $O(n \log n)$. The space complexity is $O(n)$.
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