0614MediumMathGeometry

Total Rug Coverage

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Trains the technique from

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.

Two rugs lie flat on a floor, each with its sides parallel to the floor's grid lines. A survey marker sits at the origin, so a coordinate may be negative.

The first rug spans from the corner (ax1, ay1) to the opposite corner (ax2, ay2), with ax1 the smaller horizontal coordinate and ay1 the smaller vertical one. The second rug spans from (bx1, by1) to (bx2, by2) under the same rule. A rug may be folded flat, in which case one of its sides has length zero and it covers nothing.

Return the area of floor that has at least one rug over it. Floor that both rugs cover counts once. Two rugs whose edges only meet along a line, or at a single point, do not share any area.

Examples

Example 1

Input: ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = 0, by1 = 0, bx2 = 4, by2 = 3
Output: 24

The first rug covers 16 units of floor and the second covers 12. The floor they share runs horizontally from 0 to 2 and vertically from 0 to 2, so 4 units are under both and are counted once: 16 + 12 - 4 = 24.

Example 2

Input: ax1 = 0, ay1 = 0, ax2 = 2, ay2 = 2, bx1 = 2, by1 = 0, bx2 = 4, by2 = 2
Output: 8

The rugs meet along the vertical line at 2 and share no area, so the total is 4 + 4 = 8.

Example 3

Input: ax1 = -5, ay1 = -5, ax2 = 5, ay2 = 5, bx1 = -1, by1 = -1, bx2 = 1, by2 = 1
Output: 100

The smaller rug lies entirely on top of the larger one, so the covered floor is just the larger rug's 100 units.

Constraints

-10^4 <= ax1 <= ax2 <= 10^4
-10^4 <= ay1 <= ay2 <= 10^4
-10^4 <= bx1 <= bx2 <= 10^4
-10^4 <= by1 <= by2 <= 10^4

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 compute_area(ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:

Java

public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2)

The intended approach

Time O(1) — a fixed number of arithmetic operations.Space O(1)

Each rug's own area is the product of its two side lengths. Adding the two areas counts the floor under both rugs twice, so that part is subtracted once. The part under both is a rectangle bounded horizontally by the larger of the two left edges and the smaller of the two right edges, and vertically the same way; when either of those two spans is zero or negative the rugs overlap in nothing, so zero is subtracted. Everything is arithmetic on the eight inputs.

Pythonentry point: compute_area

def compute_area(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2):
    # O(1) time, O(1) space. Add both rug areas, which double counts exactly
    # the part lying under both, then subtract that part once. The shared part
    # is itself a rectangle whose horizontal extent runs from the larger left
    # edge to the smaller right edge, and likewise vertically; if either extent
    # comes out zero or negative the rugs share no area at all.
    area_a = (ax2 - ax1) * (ay2 - ay1)
    area_b = (bx2 - bx1) * (by2 - by1)
    width = min(ax2, bx2) - max(ax1, bx1)
    height = min(ay2, by2) - max(ay1, by1)
    shared = width * height if width > 0 and height > 0 else 0
    return area_a + area_b - shared

The obvious version

Cut the plane along every one of the four given horizontal coordinates and every one of the four given vertical coordinates. That splits the region into at most nine boxes, each of which is entirely inside or entirely outside each rug. Add the area of every box that falls inside at least one rug.

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_compute_area(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2):
    xs = sorted({ax1, ax2, bx1, bx2})
    ys = sorted({ay1, ay2, by1, by2})
    total = 0
    for i in range(len(xs) - 1):
        for j in range(len(ys) - 1):
            x0 = xs[i]
            x1 = xs[i + 1]
            y0 = ys[j]
            y1 = ys[j + 1]
            under_a = ax1 <= x0 and x1 <= ax2 and ay1 <= y0 and y1 <= ay2
            under_b = bx1 <= x0 and x1 <= bx2 and by1 <= y0 and y1 <= by2
            if under_a or under_b:
                total += (x1 - x0) * (y1 - y0)
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 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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