LeetCode #3625 — HARD

Count Number of Trapezoids II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

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.

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

A trapezoid is a convex quadrilateral with at least one pair of parallel sides. Two lines are parallel if and only if they have the same slope.

Example 1:

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

Output: 2

Explanation:

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

The points [-3,2], [2,3], [3,2], [2,-3] form one trapezoid.
The points [2,3], [3,2], [3,0], [2,-3] form another trapezoid.

Example 2:

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

Output: 1

Explanation:

There is only one trapezoid which can be formed.

Constraints:

4 <= points.length <= 500
–1000 <= x_i, y_i <= 1000
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. Return the number of unique trapezoids that can be formed by choosing any four distinct points from points. A trapezoid is a convex quadrilateral with at least one pair of parallel sides. Two lines are parallel if and only if they have the same slope.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[[-3,2],[3,0],[2,3],[3,2],[2,-3]]

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Hash every point-pair by its reduced slope <code>(dy,dx)</code> (normalize with GCD and fix signs).
In each slope-bucket of size <code>k</code>, there are <code>C(k,2)</code> ways to pick two segments as the trapezoid's parallel bases.
Skip any base-pair that shares an endpoint since it would not form a quadrilateral.
Subtract one count for each parallelogram. Each parallelogram was counted once for each of its two parallel-side pairs, so after subtracting once, every quadrilateral with at least one pair of parallel sides, including parallelograms, contributes exactly one to the final total.
Final answer = total valid base-pairs minus parallelogram overcounts.

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

Largest constraint values

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 #3625: Count Number of Trapezoids II
class Solution {
    public int countTrapezoids(int[][] points) {
        int n = points.length;
        Map<Double, Map<Double, Integer>> cnt1 = new HashMap<>(n * n);
        Map<Integer, Map<Double, Integer>> cnt2 = new HashMap<>(n * n);

        for (int i = 0; i < n; ++i) {
            int x1 = points[i][0], y1 = points[i][1];
            for (int j = 0; j < i; ++j) {
                int x2 = points[j][0], y2 = points[j][1];
                int dx = x2 - x1, dy = y2 - y1;
                double k = dx == 0 ? Double.MAX_VALUE : 1.0 * dy / dx;
                double b = dx == 0 ? x1 : 1.0 * (y1 * dx - x1 * dy) / dx;
                if (k == -0.0) {
                    k = 0.0;
                }
                if (b == -0.0) {
                    b = 0.0;
                }
                cnt1.computeIfAbsent(k, _ -> new HashMap<>()).merge(b, 1, Integer::sum);
                int p = (x1 + x2 + 2000) * 4000 + (y1 + y2 + 2000);
                cnt2.computeIfAbsent(p, _ -> new HashMap<>()).merge(k, 1, Integer::sum);
            }
        }

        int ans = 0;
        for (var e : cnt1.values()) {
            int s = 0;
            for (int t : e.values()) {
                ans += s * t;
                s += t;
            }
        }
        for (var e : cnt2.values()) {
            int s = 0;
            for (int t : e.values()) {
                ans -= s * t;
                s += t;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3625: Count Number of Trapezoids II
func countTrapezoids(points [][]int) int {
	n := len(points)
	cnt1 := make(map[float64]map[float64]int, n*n)
	cnt2 := make(map[int]map[float64]int, n*n)

	for i := 0; i < n; i++ {
		x1, y1 := points[i][0], points[i][1]
		for j := 0; j < i; j++ {
			x2, y2 := points[j][0], points[j][1]
			dx, dy := x2-x1, y2-y1

			var k, b float64
			if dx == 0 {
				k = 1e9
				b = float64(x1)
			} else {
				k = float64(dy) / float64(dx)
				b = float64(int64(y1)*int64(dx)-int64(x1)*int64(dy)) / float64(dx)
			}

			if cnt1[k] == nil {
				cnt1[k] = make(map[float64]int)
			}
			cnt1[k][b]++

			p := (x1+x2+2000)*4000 + (y1 + y2 + 2000)
			if cnt2[p] == nil {
				cnt2[p] = make(map[float64]int)
			}
			cnt2[p][k]++
		}
	}

	ans := 0
	for _, e := range cnt1 {
		s := 0
		for _, t := range e {
			ans += s * t
			s += t
		}
	}
	for _, e := range cnt2 {
		s := 0
		for _, t := range e {
			ans -= s * t
			s += t
		}
	}
	return ans
}

# Accepted solution for LeetCode #3625: Count Number of Trapezoids II
class Solution:
    def countTrapezoids(self, points: List[List[int]]) -> int:
        n = len(points)

        # cnt1: k -> (b -> count)
        cnt1: dict[float, dict[float, int]] = defaultdict(lambda: defaultdict(int))
        # cnt2: p -> (k -> count)
        cnt2: dict[int, dict[float, int]] = defaultdict(lambda: defaultdict(int))

        for i in range(n):
            x1, y1 = points[i]
            for j in range(i):
                x2, y2 = points[j]
                dx, dy = x2 - x1, y2 - y1

                if dx == 0:
                    k = 1e9
                    b = x1
                else:
                    k = dy / dx
                    b = (y1 * dx - x1 * dy) / dx

                cnt1[k][b] += 1

                p = (x1 + x2 + 2000) * 4000 + (y1 + y2 + 2000)
                cnt2[p][k] += 1

        ans = 0

        for e in cnt1.values():
            s = 0
            for t in e.values():
                ans += s * t
                s += t

        for e in cnt2.values():
            s = 0
            for t in e.values():
                ans -= s * t
                s += t

        return ans

// Accepted solution for LeetCode #3625: Count Number of Trapezoids II
// 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 #3625: Count Number of Trapezoids II
// class Solution {
//     public int countTrapezoids(int[][] points) {
//         int n = points.length;
//         Map<Double, Map<Double, Integer>> cnt1 = new HashMap<>(n * n);
//         Map<Integer, Map<Double, Integer>> cnt2 = new HashMap<>(n * n);
// 
//         for (int i = 0; i < n; ++i) {
//             int x1 = points[i][0], y1 = points[i][1];
//             for (int j = 0; j < i; ++j) {
//                 int x2 = points[j][0], y2 = points[j][1];
//                 int dx = x2 - x1, dy = y2 - y1;
//                 double k = dx == 0 ? Double.MAX_VALUE : 1.0 * dy / dx;
//                 double b = dx == 0 ? x1 : 1.0 * (y1 * dx - x1 * dy) / dx;
//                 if (k == -0.0) {
//                     k = 0.0;
//                 }
//                 if (b == -0.0) {
//                     b = 0.0;
//                 }
//                 cnt1.computeIfAbsent(k, _ -> new HashMap<>()).merge(b, 1, Integer::sum);
//                 int p = (x1 + x2 + 2000) * 4000 + (y1 + y2 + 2000);
//                 cnt2.computeIfAbsent(p, _ -> new HashMap<>()).merge(k, 1, Integer::sum);
//             }
//         }
// 
//         int ans = 0;
//         for (var e : cnt1.values()) {
//             int s = 0;
//             for (int t : e.values()) {
//                 ans += s * t;
//                 s += t;
//             }
//         }
//         for (var e : cnt2.values()) {
//             int s = 0;
//             for (int t : e.values()) {
//                 ans -= s * t;
//                 s += t;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3625: Count Number of Trapezoids II
function countTrapezoids(points: number[][]): number {
    const n = points.length;

    const cnt1: Map<number, Map<number, number>> = new Map();
    const cnt2: Map<number, Map<number, number>> = new Map();

    for (let i = 0; i < n; i++) {
        const [x1, y1] = points[i];
        for (let j = 0; j < i; j++) {
            const [x2, y2] = points[j];
            const [dx, dy] = [x2 - x1, y2 - y1];

            const k = dx === 0 ? 1e9 : dy / dx;
            const b = dx === 0 ? x1 : (y1 * dx - x1 * dy) / dx;

            if (!cnt1.has(k)) {
                cnt1.set(k, new Map());
            }
            const mapB = cnt1.get(k)!;
            mapB.set(b, (mapB.get(b) || 0) + 1);

            const p = (x1 + x2 + 2000) * 4000 + (y1 + y2 + 2000);

            if (!cnt2.has(p)) {
                cnt2.set(p, new Map());
            }
            const mapK = cnt2.get(p)!;
            mapK.set(k, (mapK.get(k) || 0) + 1);
        }
    }

    let ans = 0;
    for (const e of cnt1.values()) {
        let s = 0;
        for (const t of e.values()) {
            ans += s * t;
            s += t;
        }
    }
    for (const e of cnt2.values()) {
        let s = 0;
        for (const t of e.values()) {
            ans -= s * t;
            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^2)

Space

O(n^2)

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.