Is Find the Number of Ways to Place People I easy or hard?

The problem is rated Medium on LeetCode. The brute force logic is straightforward, but recognizing that sorting and a sweep-line constraint can remove the expensive third loop requires stronger algorithmic intuition.

Find the Number of Ways to Place People I Python/Java solution

Most implementations follow the same structure: sort the list of coordinate pairs, then run nested loops to sweep potential rectangles while maintaining a boundary value. The logic translates directly across Python, Java, C++, and C# because it relies only on sorting and iteration.

How to solve Find the Number of Ways to Place People I in O(n^2)?

Sort all coordinates by x ascending and by y descending for ties. For each point as the potential top-left corner, sweep through points to its right. Track the highest y encountered that is still below the starting point's y. Count a pair only when the new point's y is below the start but above the blocking boundary, ensuring no interior point exists.

What is the best approach for Find the Number of Ways to Place People I?

The most effective approach sorts the points by x ascending and y descending, then uses a sweep to evaluate valid top-left and bottom-right pairs. While scanning to the right, track the highest y that blocks rectangles. This avoids checking every interior point and reduces the complexity to O(n^2) after sorting.

Is Find the Number of Ways to Place People I asked at Google/Amazon/Meta?

Problems combining geometry, sorting, and pair enumeration are common in interviews at large tech companies. Variants of this pattern appear in Google and Amazon interview prep because they test spatial reasoning and the ability to optimize brute force checks using ordering.

What data structure is used in Find the Number of Ways to Place People I?

The solution mainly relies on arrays to store coordinates and sorting to order them. The optimized algorithm uses a sweep-line style traversal with a few scalar variables to track boundaries instead of additional data structures.

What is the time complexity of Find the Number of Ways to Place People I?

The brute force solution runs in O(n^3) because every pair of points requires scanning all remaining points to check if the rectangle is empty. The optimized solution sorts the points and performs a quadratic sweep, resulting in O(n^2) time and O(1) extra space.

Description Hints 2 Solution Video 42:42

3025. Find the Number of Ways to Place People I

Medium64.0% acceptance4 companies

You are given a 2D array points of size n x 2 representing integer coordinates of some points on a 2D plane, where points[i] = [x_i, y_i].

Count the number of pairs of points (A, B), where

A is on the upper left side of B, and
there are no other points in the rectangle (or line) they make (including the border).

Return the count.

Example 1:

Input: points = [[1,1],[2,2],[3,3]]

Output: 0

Explanation:

There is no way to choose A and B so A is on the upper left side of B.

Example 2:

Input: points = [[6,2],[4,4],[2,6]]

Output: 2

Explanation:

The left one is the pair (points[1], points[0]), where points[1] is on the upper left side of points[0] and the rectangle is empty.
The middle one is the pair (points[2], points[1]), same as the left one it is a valid pair.
The right one is the pair (points[2], points[0]), where points[2] is on the upper left side of points[0], but points[1] is inside the rectangle so it's not a valid pair.

Example 3:

Input: points = [[3,1],[1,3],[1,1]]

Output: 2

Explanation:

The left one is the pair (points[2], points[0]), where points[2] is on the upper left side of points[0] and there are no other points on the line they form. Note that it is a valid state when the two points form a line.
The middle one is the pair (points[1], points[2]), it is a valid pair same as the left one.
The right one is the pair (points[1], points[0]), it is not a valid pair as points[2] is on the border of the rectangle.

Constraints:

2 <= n <= 50
points[i].length == 2
0 <= points[i][0], points[i][1] <= 50
All points[i] are distinct.

Asked at 4 companies

Google Amazon Meta Bloomberg

Topics

Hints open one at a time

Hint 1

We can enumerate all the upper-left and lower-right corners.

Hint 2

If the upper-left corner is (x1, y1) and lower-right corner is (x2, y2), check that there is no point (x, y) such that x1 <= x <= x2 and y2 <= y <= y1.

42:42

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3025. Find the Number of Ways to Place People I

Asked at 4 companies

Topics

Hints open one at a time

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