What is the main approach for solving the Count Number of Rectangles Containing Each Point problem?

The primary approach involves scanning the array of rectangles and using hash lookups or binary search techniques to efficiently count how many rectangles contain each point.

How can I optimize the brute-force solution to Count Number of Rectangles Containing Each Point?

You can optimize the solution by leveraging binary search on sorted rectangles or points, and by using pre-computation for faster lookups.

What is the maximum time complexity for this problem?

The maximum time complexity occurs when using a brute-force approach, which is O(N*M), where N is the number of rectangles and M is the number of points.

What key patterns should I focus on for this problem?

Key patterns include array scanning plus hash lookup, binary search optimization, and pre-computation based on given constraints.

How can GhostInterview help me with the Count Number of Rectangles Containing Each Point problem?

GhostInterview offers tailored guidance, helping you optimize your approach and providing instant feedback to ensure efficiency and correctness.

#2250

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

统计包含每个点的矩形数目

给你一个二维整数数组 rectangles ，其中 rectangles[i] = [l i , h i ] 表示第 i 个矩形长为 l i 高为 h i 。给你一个二维整数数组 points ，其中 points[j] = [x j , y j ] 是坐标为 (x j , y j ) 的一个点。 …

数组哈希表二分查找树状数组排序

题目描述

给你一个二维整数数组 rectangles ，其中 rectangles[i] = [l_i, h_i] 表示第 i 个矩形长为 l_i 高为 h_i 。给你一个二维整数数组 points ，其中 points[j] = [x_j, y_j] 是坐标为 (x_j, y_j) 的一个点。

第 i 个矩形的 左下角 在 (0, 0) 处，右上角 在 (l_i, h_i) 。

请你返回一个整数数组 count ，长度为 points.length，其中 count[j]是包含第 j 个点的矩形数目。

如果 0 <= x_j <= l_i 且 0 <= y_j <= h_i ，那么我们说第 i 个矩形包含第 j 个点。如果一个点刚好在矩形的边上，这个点也被视为被矩形包含。

示例 1：

输入：rectangles = [[1,2],[2,3],[2,5]], points = [[2,1],[1,4]]
输出：[2,1]
解释：
第一个矩形不包含任何点。
第二个矩形只包含一个点 (2, 1) 。
第三个矩形包含点 (2, 1) 和 (1, 4) 。
包含点 (2, 1) 的矩形数目为 2 。
包含点 (1, 4) 的矩形数目为 1 。
所以，我们返回 [2, 1] 。

示例 2：

输入：rectangles = [[1,1],[2,2],[3,3]], points = [[1,3],[1,1]]
输出：[1,3]
解释：
第一个矩形只包含点 (1, 1) 。
第二个矩形只包含点 (1, 1) 。
第三个矩形包含点 (1, 3) 和 (1, 1) 。
包含点 (1, 3) 的矩形数目为 1 。
包含点 (1, 1) 的矩形数目为 3 。
所以，我们返回 [1, 3] 。

提示：

1 <= rectangles.length, points.length <= 5 * 10⁴
rectangles[i].length == points[j].length == 2
1 <= l_i, x_j <= 10⁹
1 <= h_i, y_j <= 100
所有 rectangles 互不相同 。
所有 points 互不相同 。

lightbulb

解题思路

方法一：排序 + 二分

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class Solution:
    def countRectangles(
        self, rectangles: List[List[int]], points: List[List[int]]
    ) -> List[int]:
        d = defaultdict(list)
        for x, y in rectangles:
            d[y].append(x)
        for y in d.keys():
            d[y].sort()
        ans = []
        for x, y in points:
            cnt = 0
            for h in range(y, 101):
                xs = d[h]
                cnt += len(xs) - bisect_left(xs, x)
            ans.append(cnt)
        return ans

speed

复杂度分析

指标	值
时间	complexity depends on the approach chosen. The brute force solution involves checking every point against all rectangles, leading to O(N*M) complexity, where N is the number of rectangles and M is the number of points. Optimizations such as binary search or pre-computation can reduce this. Space complexity also depends on the approach, with storage for rectangles and points taking O(N + M) space in the worst case.
空间	Depends on the final approach

指标

值

时间

complexity depends on the approach chosen. The brute force solution involves checking every point against all rectangles, leading to O(N*M) complexity, where N is the number of rectangles and M is the number of points. Optimizations such as binary search or pre-computation can reduce this. Space complexity also depends on the approach, with storage for rectangles and points taking O(N + M) space in the worst case.

空间

Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate's ability to optimize brute force methods for large inputs.
question_mark
Awareness of sorting and binary search optimization techniques.
question_mark
Understanding of hash table usage for efficient lookup and pre-computation.

warning

常见陷阱

外企场景

error
Incorrectly assuming all rectangles are relevant for each point, leading to excessive checks.
error
Failure to optimize searching for rectangles by overlooking sorting or binary search options.
error
Not considering space and time trade-offs when scaling to the upper constraint limits.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if rectangles or points had larger or smaller values?
arrow_right_alt
How would the problem change if rectangles were not guaranteed to have unique dimensions?
arrow_right_alt
How can we modify this approach to handle a dynamically changing set of rectangles or points?

help

常见问题

外企场景

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