Problem Description

The problem presents us with a set of rectangles and a set of points. Each rectangle is defined by its length and height, with the bottom-left corner fixed at the origin (0, 0) and the top-right corner at coordinates (length, height). The points are given with their x and y coordinates.

Your task is to determine for each point how many rectangles out of the given set contain it. A point is considered contained within a rectangle if it lies inside the rectangle or on its edges, meaning the point's x-coordinate is less than or equal to the rectangle's length and its y-coordinate is less than or equal to the rectangle's height.

To sum it up: for every point given, count how many rectangles have both their length and height greater than or equal to the point's x and y coordinates, respectively.

Intuition

The intuition behind the solution involves preprocessing the rectangles to make the query for each point efficient. Since we only care about whether a point is contained within a rectangle, we can organize our rectangles by their heights. For each possible height, we keep a list of lengths of rectangles that have at least that height.

Once we have this structure, we can quickly determine how many rectangles contain a point with a given y-value. For a point with coordinates (x, y), we need to count the number of rectangles that have a height >= y and a width >= x.

This is where binary search comes into play. Because each list of lengths for a given height is sorted, we can use binary search to efficiently find the index of the smallest length that is still >= x. All lengths to the right of this index will also be >= x, so the number of such lengths is the total length of the list minus the index.

By summing up the counts for each height from y to the maximum possible height (which in this case is set to 100, the assumed maximum height), we get the total number of rectangles that contain the point. We repeat this process for each point and return the accumulated counts as our answer.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The implementation of the solution can be divided into the following steps:

1. First, we initialize a dictionary (defaultdict) named d that will hold lists of rectangle lengths indexed by their heights.

2. We then iterate through each rectangle in the rectangles list, add the length of each rectangle to the corresponding list in d indexed by its height.

3. After storing all the rectangles, we sort each list of lengths in d. This allows us to apply binary search on these lists.

4. We then initialize an empty list ans which will hold our final counts for each point.

5. Now we iterate through each given point in points. For the current point, we start a counter at 0. Then we check every possible rectangle height starting from the point's y-coordinate up to the maximum height (which is set to 101 to include the maximum height of 100).

6. For each height h that is equal to or greater than the y-coordinate of the point, we get the list of rectangle lengths xs from the dictionary d.

7. We perform a binary search using the bisect_left function from Python’s bisect module to find the index where the point's x-coordinate could be inserted into the xs list to maintain sorted order. This index also represents the number of lengths that are less than the point's x-coordinate.

8. To find the number of lengths that are greater than or equal to the point's x-coordinate, we subtract the found index from the total number of lengths (len(xs) - bisect_left(xs, x)).

9. We add this number to our counter cnt and, after checking all the possible heights, append cnt to our ans list.

10. Finally, after calculating the counts for all the points, we return the list ans as our output.

Let's go through an example to better understand the algorithm using its data structures and patterns:

• Suppose we have rectangles = [[1,2], [2,3], [2,5]] and points = [[1,1], [1,4]].
• Our dictionary d after processing rectangles will look like: {2: [1], 3: [2], 5: [2]}
• After sorting, it will be {2: [1], 3: [2], 5: [2]} (no change in this case because there's only one length for each height).
• Now for point [1,1], we look at d[1] upwards: d[1] doesn't exist, but d[2] does, and since 1 can be inserted at index 0 in [1], there is 1 - 0 = 1 rectangle that contains it; looking at d[3] and d[5], both have lists [2] which would include the point since the bisect_left([2], 1) would give 0 resulting again in one rectangle each. Hence cnt = 1+1+1.
• The process is repeated for the second point [1,4], where now only d[5] can include the point, resulting in cnt = 1.
• Finally, we have the answer [3,1].

This solution utilizes a combination of hashmap to collate per-height rectangle lengths, sorting to prepare for binary search, and binary search to effectively answer queries for each point on how many rectangles contain it.

Example Walkthrough

Let's walk through a small example to illustrate the approach provided in the solution. We have a set of rectangles and a set of points defined as follows:

• rectangles = [[1,2], [2,3], [2,5]]
• points = [[1,1], [1,4]]

Step 1: The dictionary d is initialized. After processing rectangles, d is filled as follows:

• For the rectangle [1,2], we add 1 to the list at d[2].
• For the rectangle [2,3], we add 2 to the list at d[3].
• For the rectangle [2,5], we add 2 to the list at d[5].

The dictionary now looks like this:

d = {
  2: [1],
  3: [2],
  5: [2]
}

Step 2: As each list only contains one element, the lists are already sorted. If there were multiple elements, we would sort them to efficiently apply binary search later on.

Step 3: We initialize ans as an empty list to hold the count for each point.

Step 4-9: We iterate through the points [1,1] and [1,4] to count how many rectangles contain these points:

• For point [1,1], we start with a counter cnt set to 0. We look at possible heights starting from y=1:

  • At height h=2 (since d[1] doesn't exist), we find the list [1]. Since 1 can be inserted at index 0 to maintain order, it means there is 1 - 0 = 1 rectangle of height at least 2 that contains the point.
  • At height h=3, bisect_left([2], 1) is 0, so there is 1 - 0 = 1 rectangle of height at least 3 that contains the point.
  • At height h=5, bisect_left([2], 1) is again 0, so there is another rectangle here. The accumulated count cnt becomes 1 + 1 + 1 = 3. We append this count to ans.

• For point [1,4], cnt is again set to 0. Since we only have heights up to 5, we look at:

  • At height h=5, bisect_left([2], 1) gives 0, so there is 1 - 0 = 1 rectangle of height at least 5 that contains the point. No other heights are considered since they are less than the y-coordinate of the point. The final count for this point is 1, which we append to ans.

Step 10: The final output [3,1] is obtained after processing both points, which indicates that the first point [1,1] is contained within 3 rectangles and the second point [1,4] is contained within 1 rectangle.

In this example, we see the practical use of organizing rectangle dimensions by height to efficiently query with binary search the count of rectangles that include each point.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code consists of multiple parts:

1. Building the dictionary d: We iterate over each rectangle in rectangles, resulting in O(n) operations, where n is the number of rectangles. At most, we are storing 100 keys (because height h could be from 1 to 100 according to the problem constraints).

2. Sorting the lists in the dictionary: For each key in the dictionary, which represents a unique height h, we sort the list of associated x coordinates. The sort operation has a time complexity of O(k log k) for each list of length k. Since there could be up to n x values per height, and there are at most 100 heights, the worst-case complexity of this part is O(100 * n log n).

3. Processing the points: We iterate over every point in points, resulting in O(m) operations, where m is the number of points. For each point (x, y), we have an inner loop where we iterate potentially from y to 100, let's denote this range as R = 100 - y + 1. Inside this loop, we perform a binary search using bisect_left which has a complexity of O(log k) for each height h. The total time complexity for processing points translates to O(m*R*log n).

Combining these complexities, the overall time complexity of the function is O(n + 100 * n log n + m*R*log n). Since R and the number 100 are constants, and in the context of big O notation we drop constants and lower-order terms, the dominant term is O(n log n + m*log n). When n (the number of rectangles) is large compared to m (the number of points), the time complexity approximates to O(n log n); otherwise, it will be closer to O((n + m) log n).

Space Complexity

The space complexity is determined by:

1. The dictionary d storage: The dictionary itself will take O(h) space, with h being the unique heights, which is at most 100. Each list in the dictionary can collectively hold up to O(n) coordinates. Therefore, the space complexity for the dictionary is O(h + n). Since h is a constant (at most 100), we can simply say O(n).

2. The output list ans: It stores the result for each point, so it takes O(m) space.

Thus, the final space complexity of the function is O(n + m), for storing all the rectangles and output results. In the contexts where n is much larger than m, it simplifies to O(n).

Learn more about how to find time and space complexity quickly using problem constraints.