LeetCode #2001 — MEDIUM

Number of Pairs of Interchangeable Rectangles

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

The Problem

Problem Statement

You are given n rectangles represented by a 0-indexed 2D integer array rectangles, where rectangles[i] = [width_i, height_i] denotes the width and height of the i^th rectangle.

Two rectangles i and j (i < j) are considered interchangeable if they have the same width-to-height ratio. More formally, two rectangles are interchangeable if width_i/height_i == width_j/height_j (using decimal division, not integer division).

Return the number of pairs of interchangeable rectangles in rectangles.

Example 1:

Input: rectangles = [[4,8],[3,6],[10,20],[15,30]]
Output: 6
Explanation: The following are the interchangeable pairs of rectangles by index (0-indexed):
- Rectangle 0 with rectangle 1: 4/8 == 3/6.
- Rectangle 0 with rectangle 2: 4/8 == 10/20.
- Rectangle 0 with rectangle 3: 4/8 == 15/30.
- Rectangle 1 with rectangle 2: 3/6 == 10/20.
- Rectangle 1 with rectangle 3: 3/6 == 15/30.
- Rectangle 2 with rectangle 3: 10/20 == 15/30.

Example 2:

Input: rectangles = [[4,5],[7,8]]
Output: 0
Explanation: There are no interchangeable pairs of rectangles.

Constraints:

n == rectangles.length
1 <= n <= 10⁵
rectangles[i].length == 2
1 <= width_i, height_i <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given n rectangles represented by a 0-indexed 2D integer array rectangles, where rectangles[i] = [widthi, heighti] denotes the width and height of the ith rectangle. Two rectangles i and j (i < j) are considered interchangeable if they have the same width-to-height ratio. More formally, two rectangles are interchangeable if widthi/heighti == widthj/heightj (using decimal division, not integer division). Return the number of pairs of interchangeable rectangles in rectangles.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[[4,8],[3,6],[10,20],[15,30]]

Example 2

[[4,5],[7,8]]

Core Insight

What unlocks the optimal approach

Store the rectangle height and width ratio in a hashmap.
Traverse the ratios, and for each ratio, use the frequency of the ratio to add to the total pair count.

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 #2001: Number of Pairs of Interchangeable Rectangles
class Solution {
    public long interchangeableRectangles(int[][] rectangles) {
        long ans = 0;
        int n = rectangles.length + 1;
        Map<Long, Integer> cnt = new HashMap<>();
        for (var e : rectangles) {
            int w = e[0], h = e[1];
            int g = gcd(w, h);
            w /= g;
            h /= g;
            long x = (long) w * n + h;
            ans += cnt.getOrDefault(x, 0);
            cnt.merge(x, 1, Integer::sum);
        }
        return ans;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #2001: Number of Pairs of Interchangeable Rectangles
func interchangeableRectangles(rectangles [][]int) int64 {
	ans := 0
	n := len(rectangles)
	cnt := map[int]int{}
	for _, e := range rectangles {
		w, h := e[0], e[1]
		g := gcd(w, h)
		w, h = w/g, h/g
		x := w*(n+1) + h
		ans += cnt[x]
		cnt[x]++
	}
	return int64(ans)
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

# Accepted solution for LeetCode #2001: Number of Pairs of Interchangeable Rectangles
class Solution:
    def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
        ans = 0
        cnt = Counter()
        for w, h in rectangles:
            g = gcd(w, h)
            w, h = w // g, h // g
            ans += cnt[(w, h)]
            cnt[(w, h)] += 1
        return ans

// Accepted solution for LeetCode #2001: Number of Pairs of Interchangeable Rectangles
use std::collections::HashMap;
impl Solution {
    pub fn compute_gcd(mut a: i32, mut b: i32) -> i32 {
        while a > 0 && b > 0 {
            if a > b {
                a %= b;
            } else {
                b %= a;
            }
        }
        if a == 0 {
            b
        } else {
            a
        }
    }
    pub fn interchangeable_rectangles(rectangles: Vec<Vec<i32>>) -> i64 {
        rectangles
            .into_iter()
            .map(|v| {
                let gcd = Solution::compute_gcd(v[0], v[1]);
                (v[0] / gcd, v[1] / gcd)
            })
            .fold(HashMap::new(), |mut dict, t| {
                dict.entry(t).and_modify(|cnt| *cnt += 1).or_insert(1);
                dict
            })
            .into_iter()
            .filter(|(_, v)| *v > 1)
            .map(|(_, v)| v)
            .fold(0, |mut cnt, v| {
                cnt += v * (v - 1) / 2;
                cnt
            })
    }
}

// Accepted solution for LeetCode #2001: Number of Pairs of Interchangeable Rectangles
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2001: Number of Pairs of Interchangeable Rectangles
// class Solution {
//     public long interchangeableRectangles(int[][] rectangles) {
//         long ans = 0;
//         int n = rectangles.length + 1;
//         Map<Long, Integer> cnt = new HashMap<>();
//         for (var e : rectangles) {
//             int w = e[0], h = e[1];
//             int g = gcd(w, h);
//             w /= g;
//             h /= g;
//             long x = (long) w * n + h;
//             ans += cnt.getOrDefault(x, 0);
//             cnt.merge(x, 1, Integer::sum);
//         }
//         return ans;
//     }
// 
//     private int gcd(int a, int b) {
//         return b == 0 ? a : gcd(b, a % b);
//     }
// }

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 × log M)

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.