Leetcode #699: Falling Squares
In this guide, we solve Leetcode #699 Falling Squares in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Segment Tree, Array, Ordered Set
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
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].
Python Solution
class Node:
def __init__(self, l, r):
self.left = None
self.right = None
self.l = l
self.r = r
self.mid = (l + r) >> 1
self.v = 0
self.add = 0
class SegmentTree:
def __init__(self):
self.root = Node(1, int(1e9))
def modify(self, l, r, v, node=None):
if l > r:
return
if node is None:
node = self.root
if node.l >= l and node.r <= r:
node.v = v
node.add = v
return
self.pushdown(node)
if l <= node.mid:
self.modify(l, r, v, node.left)
if r > node.mid:
self.modify(l, r, v, node.right)
self.pushup(node)
def query(self, l, r, node=None):
if l > r:
return 0
if node is None:
node = self.root
if node.l >= l and node.r <= r:
return node.v
self.pushdown(node)
v = 0
if l <= node.mid:
v = max(v, self.query(l, r, node.left))
if r > node.mid:
v = max(v, self.query(l, r, node.right))
return v
def pushup(self, node):
node.v = max(node.left.v, node.right.v)
def pushdown(self, node):
if node.left is None:
node.left = Node(node.l, node.mid)
if node.right is None:
node.right = Node(node.mid + 1, node.r)
if node.add:
node.left.v = node.add
node.right.v = node.add
node.left.add = node.add
node.right.add = node.add
node.add = 0
class Solution:
def fallingSquares(self, positions: List[List[int]]) -> List[int]:
ans = []
mx = 0
tree = SegmentTree()
for l, w in positions:
r = l + w - 1
h = tree.query(l, r) + w
mx = max(mx, h)
ans.append(mx)
tree.modify(l, r, h)
return ans
Complexity
The time complexity is O(n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.