0890MediumArrayHash TableMathCountingNumber Theory

Panes Cut to the Same Shape

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Trains the technique from

LeetCode 2001Number of Pairs of Interchangeable Rectangles

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A glazier's stock is given as rectangles, where rectangles[i] is [width, height].

Two panes are the same shape when their width divided by their height comes to the same value.

Return how many pairs of panes in the stock are the same shape.

Examples

Example 1

Input: rectangles = [[24, 36], [14, 21], [90000, 60000], [15, 10], [7, 11]]
Output: 2

The first two both reduce to 2 by 3, and the next two both reduce to 3 by 2, giving one pair from each group. The last pane is on its own.

Example 2

Input: rectangles = [[1, 7], [2, 14], [3, 21], [4, 28], [5, 35], [6, 42]]
Output: 15

All six panes reduce to 1 by 7, so every one of the fifteen pairs matches.

Example 3

Input: rectangles = [[2, 3], [3, 2]]
Output: 0

Turning a pane on its side gives a different proportion, so these two are not the same shape.

Constraints

1 <= rectangles.length <= 10^5
rectangles[i].length == 2
1 <= rectangles[i][0] <= 10^5
1 <= rectangles[i][1] <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def interchangeable_rectangles(rectangles: list[list[int]]) -> int:

Java

public long interchangeableRectangles(int[][] rectangles)

The intended approach

Time O(n log V) for n panes and side lengths up to V, the logarithm being the divisor step.Space O(n) for the tally of shapes.

Comparing every pane against every other is too slow, and comparing the divisions themselves invites rounding trouble, so give each pane an exact name for its shape: divide its width and height by their greatest common divisor and use the reduced pair. Two panes share a name exactly when their shapes match. Tally the names in one pass, and a name held by `c` panes contributes `c` choose two pairs, which is `c` times `c - 1` halved. Adding those up over the tally gives the answer.

Pythonentry point: interchangeable_rectangles

def interchangeable_rectangles(rectangles):
    from math import gcd
    tally = {}
    for width, height in rectangles:
        g = gcd(width, height)
        key = (width // g, height // g)
        tally[key] = tally.get(key, 0) + 1
    total = 0
    for count in tally.values():
        total += count * (count - 1) // 2
    return total

The obvious version

Compare every pair of panes directly, cross-multiplying the two width and height pairs so the test stays in whole numbers. Quadratic in the pane count and it never names a shape at all.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_interchangeable_rectangles(rectangles):
    n = len(rectangles)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            if rectangles[i][0] * rectangles[j][1] == rectangles[j][0] * rectangles[i][1]:
                total += 1
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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