Intersection over Union (IoU)

mediumFREEObject DetectionDetectionCV: Recognition

You are given two axis-aligned bounding boxes and need to calculate their Intersection over Union (IoU), the standard metric for evaluating object detection accuracy.

Bounding boxes are represented as [x1, y1, x2, y2] where (x1, y1) is the top-left corner and (x2, y2) is the bottom-right corner.

$IoU = \frac{Area_{intersection}}{Area_{union}} = \frac{Area_{intersection}}{Area_A + Area_B - Area_{intersection}}$

The algorithm:

Find the intersection rectangle (if any):
- x1_inter = max(x1_A, x1_B)
- y1_inter = max(y1_A, y1_B)
- x2_inter = min(x2_A, x2_B)
- y2_inter = min(y2_A, y2_B)
Compute intersection area (0 if boxes don't overlap)
Compute union = Area_A + Area_B - intersection
IoU = intersection / union

IoU ranges from 0 (no overlap) to 1 (perfect overlap).

Example

Input

box1 = [0, 0, 10, 10]
box2 = [5, 5, 15, 15]

Output

0.1429

Reasoning

Inputs: box1 and box2 are corner lists $[x_1, y_1, x_2, y_2]$ . box1 covers $x \in [0, 10]$ , $y \in [0, 10]$ ; box2 covers $x \in [5, 15]$ , $y \in [5, 15]$ . Each is a $10 \times 10$ square, so $Area_A = Area_B = 100$ .
Overlap along x: the two intervals $[0, 10]$ and $[5, 15]$ share only $[5, 10]$ , a width of $5$ . The y axis is identical, giving a height of $5$ .
Intersection: the shared region is the square from $(5, 5)$ to $(10, 10)$ , so $Area_{intersection} = 5 \times 5 = 25$ .
Union: $100 + 100 - 25 = 175$ . The overlap is subtracted once because adding both areas counts that $5 \times 5$ patch twice.
IoU: $\frac{25}{175} = \frac{1}{7} \approx 0.142857$ , rounded to 4 decimal places gives 0.1429.
Notice: the boxes share a quarter of each area, yet IoU is only about $0.14$ , because the union also counts both non-overlapping parts. Sliding box2 until its intervals no longer meet box1's leaves zero width, hence an IoU of $0$ .

Constraints

Boxes are in [x1, y1, x2, y2] format (top-left and bottom-right corners)
x2 > x1 and y2 > y1 for valid boxes
Coordinates can be any real numbers
Return IoU rounded to 4 decimal places

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Concept primer

You want to compute how much two axis-aligned rectangles overlap, expressed as a ratio between 0 and 1 called Intersection over Union (IoU). This is a pure geometry / implementation exercise with constant-time operations.

1. Background Knowledge

In object detection, a model predicts bounding boxes for objects in an image, and these predictions are compared against ground-truth boxes. The most common accuracy metric is Intersection over Union (IoU), which measures how much the predicted box overlaps with the ground-truth box relative to their total combined area.

Each bounding box is typically represented as $[x_1, y_1, x_2, y_2]$ , where:

$(x_1, y_1)$ is the top-left corner,
$(x_2, y_2)$ is the bottom-right corner,
and the box is axis-aligned (its edges are parallel to the image axes).

The IoU between two boxes $A$ and $B$ is:

IoU = \frac{Area_{intersection}}{Area_{union}} = \frac{Area_{intersection}}{Area_A + Area_B - Area_{intersection}}

If boxes do not overlap, $Area_{intersection} = 0$ and so $IoU = 0$ . If they match exactly, intersection = union and $IoU = 1$ .

2. Algorithm / Approach

The key pattern is:

Treat overlap in x and y dimensions independently.
Compute a 1D overlap length along x and along y using max/min.
Multiply these to get the intersection area (or 0 if no overlap).
Compute each box’s area.
Compute union using the inclusion–exclusion formula.
Return intersection / union.

This is a standard geometry / interval-overlap problem in 2D, framed as two 1D overlaps (x and y).

3. Step-by-Step Strategy

Unpack coordinates For box A: $(x1_A, y1_A, x2_A, y2_A)$ For box B: $(x1_B, y1_B, x2_B, y2_B)$
Compute intersection rectangle coordinates

$x1_{inter} = \max(x1_A, x1_B)$
$y1_{inter} = \max(y1_A, y1_B)$
$x2_{inter} = \min(x2_A, x2_B)$
$y2_{inter} = \min(y2_A, y2_B)$

Compute intersection width and height

inter_w = x2_inter - x1_inter
inter_h = y2_inter - y1_inter

If inter_w <= 0 or inter_h <= 0, then no overlap: intersection area = 0.

Intersection area

inter_area = max(inter_w, 0) * max(inter_h, 0)

Individual box areas

area_A = max(x2_A - x1_A, 0) * max(y2_A - y1_A, 0)
area_B = max(x2_B - x1_B, 0) * max(y2_B - y1_B, 0)

Union area

union_area = area_A + area_B - inter_area

Compute IoU

If union_area == 0 (degenerate boxes), define IoU as 0 (or handle specially).

iou = inter_area / union_area if union_area > 0 else 0.0

4. Common Pitfalls

Not checking for non-overlap: Directly multiplying negative inter_w or inter_h gives a negative “area.” Always clamp with max(..., 0) or explicitly check with if inter_w <= 0 or inter_h <= 0.
Coordinate convention confusion: The problem states that $(x_1, y_1)$ is top-left and $(x_2, y_2)$ is bottom-right. Make sure you do not assume center-width-height form or bottom-left origin.
Degenerate boxes: When $x_1 = x_2$ or $y_1 = y_2$ , area is zero. Decide what IoU should be if both boxes have zero area and overlap at a point; for coding problems, usually return 0 to avoid division by zero.
Integer division: In some languages, dividing integers yields integer results. Cast to float to avoid IoU being truncated to 0 or 1.
Wrong formula for union: Union is not simply area_A + area_B. You must subtract the intersection once: union = area_A + area_B - intersection.

5. Time & Space Complexity

Time complexity: All operations are constant-time arithmetic; no loops or recursion. O(1).
Space complexity: Only a fixed number of variables are used. O(1) extra space.

This is essentially a small, careful implementation of interval intersection plus an area calculation.

solution.py

Python 3.13

Test results