Leetcode #497: Random Point in Non-overlapping Rectangles
In this guide, we solve Leetcode #497 Random Point in Non-overlapping Rectangles 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 an array of non-overlapping axis-aligned rectangles rects where rects[i] = [ai, bi, xi, yi] indicates that (ai, bi) is the bottom-left corner point of the ith rectangle and (xi, yi) is the top-right corner point of the ith rectangle. Design an algorithm to pick a random integer point inside the space covered by one of the given rectangles.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Reservoir Sampling, Array, Math, Binary Search, Ordered Set, Prefix Sum, Randomized
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input
["Solution", "pick", "pick", "pick", "pick", "pick"]
[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]
Output
[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]
Explanation
Solution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);
solution.pick(); // return [1, -2]
solution.pick(); // return [1, -1]
solution.pick(); // return [-1, -2]
solution.pick(); // return [-2, -2]
solution.pick(); // return [0, 0]
Python Solution
class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects
self.s = [0] * len(rects)
for i, (x1, y1, x2, y2) in enumerate(rects):
self.s[i] = self.s[i - 1] + (x2 - x1 + 1) * (y2 - y1 + 1)
def pick(self) -> List[int]:
v = random.randint(1, self.s[-1])
idx = bisect_left(self.s, v)
x1, y1, x2, y2 = self.rects[idx]
return [random.randint(x1, x2), random.randint(y1, y2)]
# Your Solution object will be instantiated and called as such:
# obj = Solution(rects)
# param_1 = obj.pick()
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
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.