How do you solve Find the Number of Ways to Place People I on LeetCode?

Find the Number of Ways to Place People I (LeetCode #3025) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps to systematically check pairs.

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

The optimal solution for Find the Number of Ways to Place People I runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n), and maintaining the set takes O(n) in the worst case.

What is the brute force solution for Find the Number of Ways to Place People I?

The brute force approach for Find the Number of Ways to Place People I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every pair of points to see if one is in the upper left of the other and if the rectangle they form is empty. This is simple but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find the Number of Ways to Place People I?

Find the Number of Ways to Place People I uses the following concepts: Array, Math, Geometry, Sorting, Enumeration. The recommended patterns to study are: Sorting, Set, Two Pointers.

What are the interview tips for Find the Number of Ways to Place People I?

Clarify the problem and constraints before coding. Think about sorting and data structures that can help reduce complexity. Practice dry-running your logic with small examples.

What are common mistakes when solving Find the Number of Ways to Place People I?

Not considering edge cases where points are on the same line. Forgetting to check the rectangle's boundaries.

#3025

Find the Number of Ways to Place People I

Medium

ArrayMathGeometrySortingEnumerationSortingSetTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

By sorting the points and using a data structure to track the points we've seen, we can efficiently count valid pairs without checking every rectangle.

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Algorithm

3 steps

1Step 1: Sort the points based on x-coordinates, and then by y-coordinates.
2Step 2: Use a data structure (like a set) to keep track of y-coordinates of points we've seen.
3Step 3: For each point, count how many points in the set have a y-coordinate greater than the current point's y-coordinate.

solution.py8 lines

1def countPairs(points):
2    points.sort()  # Sort by x, then by y
3    count = 0
4    seen = set()
5    for x, y in points:
6        count += sum(1 for s in seen if s > y)
7        seen.add(y)
8    return count

ℹ

Complexity note: The sorting step takes O(n log n), and maintaining the set takes O(n) in the worst case.

1Sorting helps to systematically check pairs.
2Using a set allows for efficient counting of valid y-coordinates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.