LeetCode #3623 — MEDIUM

Count Number of Trapezoids I

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 2D integer array points, where points[i] = [x_i, y_i] represents the coordinates of the i^th point on the Cartesian plane.

A horizontal trapezoid is a convex quadrilateral with at least one pair of horizontal sides (i.e. parallel to the x-axis). Two lines are parallel if and only if they have the same slope.

Return the number of unique horizontal trapezoids that can be formed by choosing any four distinct points from points.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: points = [[1,0],[2,0],[3,0],[2,2],[3,2]]

Output: 3

Explanation:

There are three distinct ways to pick four points that form a horizontal trapezoid:

Using points [1,0], [2,0], [3,2], and [2,2].
Using points [2,0], [3,0], [3,2], and [2,2].
Using points [1,0], [3,0], [3,2], and [2,2].

Example 2:

Input: points = [[0,0],[1,0],[0,1],[2,1]]

Output: 1

Explanation:

There is only one horizontal trapezoid that can be formed.

Constraints:

4 <= points.length <= 10⁵
–10⁸ <= x_i, y_i <= 10⁸
All points are pairwise distinct.

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array points, where points[i] = [xi, yi] represents the coordinates of the ith point on the Cartesian plane. A horizontal trapezoid is a convex quadrilateral with at least one pair of horizontal sides (i.e. parallel to the x-axis). Two lines are parallel if and only if they have the same slope. Return the number of unique horizontal trapezoids that can be formed by choosing any four distinct points from points. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[[1,0],[2,0],[3,0],[2,2],[3,2]]

Example 2

[[0,0],[1,0],[0,1],[2,1]]

Step 02

Core Insight

What unlocks the optimal approach

For a line parallel to the x‑axis, all its points must share the same y‑coordinate.
Group the points by their y‑coordinate.
Choose two distinct groups (two horizontal lines), and from each group select two points to form a trapezoid.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3623: Count Number of Trapezoids I
class Solution {
    public int countTrapezoids(int[][] points) {
        final int mod = (int) 1e9 + 7;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (var p : points) {
            cnt.merge(p[1], 1, Integer::sum);
        }
        long ans = 0, s = 0;
        for (int v : cnt.values()) {
            long t = 1L * v * (v - 1) / 2;
            ans = (ans + s * t) % mod;
            s += t;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #3623: Count Number of Trapezoids I
func countTrapezoids(points [][]int) int {
	const mod = 1_000_000_007
	cnt := make(map[int]int)
	for _, p := range points {
		cnt[p[1]]++
	}

	var ans, s int64
	for _, v := range cnt {
		t := int64(v) * int64(v-1) / 2
		ans = (ans + s*t) % mod
		s += t
	}
	return int(ans)
}

# Accepted solution for LeetCode #3623: Count Number of Trapezoids I
class Solution:
    def countTrapezoids(self, points: List[List[int]]) -> int:
        mod = 10**9 + 7
        cnt = Counter(p[1] for p in points)
        ans = s = 0
        for v in cnt.values():
            t = v * (v - 1) // 2
            ans = (ans + s * t) % mod
            s += t
        return ans

// Accepted solution for LeetCode #3623: Count Number of Trapezoids I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3623: Count Number of Trapezoids I
// class Solution {
//     public int countTrapezoids(int[][] points) {
//         final int mod = (int) 1e9 + 7;
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (var p : points) {
//             cnt.merge(p[1], 1, Integer::sum);
//         }
//         long ans = 0, s = 0;
//         for (int v : cnt.values()) {
//             long t = 1L * v * (v - 1) / 2;
//             ans = (ans + s * t) % mod;
//             s += t;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #3623: Count Number of Trapezoids I
function countTrapezoids(points: number[][]): number {
    const mod = 1_000_000_007;
    const cnt = new Map<number, number>();

    for (const p of points) {
        cnt.set(p[1], (cnt.get(p[1]) ?? 0) + 1);
    }

    let ans = 0;
    let s = 0;
    for (const v of cnt.values()) {
        const t = (v * (v - 1)) / 2;
        const mul = BigInt(s) * BigInt(t);
        ans = Number((BigInt(ans) + mul) % BigInt(mod));
        s += t;
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.