How do you solve Count Number of Trapezoids I on LeetCode?

Count Number of Trapezoids I (LeetCode #3623) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: We can check all combinations of four points to see if they form a trapezoid. This is straightforward but inefficient for larger datasets.

What is the time complexity of Count Number of Trapezoids I?

The optimal solution for Count Number of Trapezoids I runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Count Number of Trapezoids I?

Count Number of Trapezoids I uses the following concepts: Array, Hash Table, Math, Geometry. The recommended patterns to study are: .

#3623

Count Number of Trapezoids I

Medium

ArrayHash TableMathGeometry

LeetCode ↗

Approaches

💡

Intuition

Time Space

We can check all combinations of four points to see if they form a trapezoid. This is straightforward but inefficient for larger datasets.

⚙️

Algorithm

3 steps

1Step 1: Iterate through all combinations of four distinct points.
2Step 2: Check if two pairs of points have the same y-coordinate (horizontal sides).
3Step 3: Count valid trapezoids.

solution.py8 lines

1from itertools import combinations
2
3def count_trapezoids(points):
4    count = 0
5    for p1, p2, p3, p4 in combinations(points, 4):
6        if (p1[1] == p2[1] and p3[1] == p4[1]) or (p1[1] == p3[1] and p2[1] == p4[1]) or (p1[1] == p4[1] and p2[1] == p3[1]):
7            count += 1
8    return count % (10**9 + 7)

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