Leetcode #850: Rectangle Area II
In this guide, we solve Leetcode #850 Rectangle Area II 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
You are given a 2D array of axis-aligned rectangles. Each rectangle[i] = [xi1, yi1, xi2, yi2] denotes the ith rectangle where (xi1, yi1) are the coordinates of the bottom-left corner, and (xi2, yi2) are the coordinates of the top-right corner.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Segment Tree, Array, Ordered Set, Line Sweep
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: rectangles = [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
Output: 6
Explanation: A total area of 6 is covered by all three rectangles, as illustrated in the picture.
From (1,1) to (2,2), the green and red rectangles overlap.
From (1,0) to (2,3), all three rectangles overlap.
Python Solution
class Node:
def __init__(self):
self.l = self.r = 0
self.cnt = self.length = 0
class SegmentTree:
def __init__(self, nums):
n = len(nums) - 1
self.nums = nums
self.tr = [Node() for _ in range(n << 2)]
self.build(1, 0, n - 1)
def build(self, u, l, r):
self.tr[u].l, self.tr[u].r = l, r
if l != r:
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
def modify(self, u, l, r, k):
if self.tr[u].l >= l and self.tr[u].r <= r:
self.tr[u].cnt += k
else:
mid = (self.tr[u].l + self.tr[u].r) >> 1
if l <= mid:
self.modify(u << 1, l, r, k)
if r > mid:
self.modify(u << 1 | 1, l, r, k)
self.pushup(u)
def pushup(self, u):
if self.tr[u].cnt:
self.tr[u].length = self.nums[self.tr[u].r + 1] - self.nums[self.tr[u].l]
elif self.tr[u].l == self.tr[u].r:
self.tr[u].length = 0
else:
self.tr[u].length = self.tr[u << 1].length + self.tr[u << 1 | 1].length
def length(self):
return self.tr[1].length
class Solution:
def rectangleArea(self, rectangles: List[List[int]]) -> int:
segs = []
alls = set()
for x1, y1, x2, y2 in rectangles:
segs.append((x1, y1, y2, 1))
segs.append((x2, y1, y2, -1))
alls.update([y1, y2])
segs.sort()
alls = sorted(alls)
tree = SegmentTree(alls)
m = {v: i for i, v in enumerate(alls)}
ans = 0
for i, (x, y1, y2, k) in enumerate(segs):
if i:
ans += tree.length * (x - segs[i - 1][0])
tree.modify(1, m[y1], m[y2] - 1, k)
ans %= int(1e9 + 7)
return ans
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
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.