Determine overlapping rectangles

https://github.com/Premiumlab/Python-for-Algorithms--Data-Structures--and-Interviews/blob/master/Mock%20Interviews/Large%20E-Commerce%20Company/E-Commerce%20Company%20-%20Interview%20Problems%20-%20SOLUTIONS/On-Site%20Question%203%20-%20SOLUTION.ipynb

On-Site Question 3 - SOLUTION

Problem

Given two rectangles, determine if they overlap. The rectangles are defined as a Dictionary, for example:

In [2]:

r1 = {

         # x and y coordinates of the bottom-left corner of the rectangle
         'x': 2 , 'y': 4,

         # Width and Height of rectangle
         'w':5,'h':12}

 

If the rectangles do overlap, return the dictionary which describes the overlapping section

 

Requirements

Make sure the dictionary you output is in the same form as the input.

Feel free to use an IDE for the code, but make sure you use paper/pencil or whiteboard to draw out your plan and logic

 

Solution

This is a problem where it helps a lot to draw out your thinking. There are a few things we will need to think about:

• How can we determine an intersection?
• What if a rectangle is fully inside another rectangle?
• What if there is no intersection, but the rectangles share an edge?

The key to solving this problem is to break it up in to sub-problems. We can split up the problem into an x-axis problem and a y-axis problem.

We will create a function that can detect overlap in 1 dimension. Then we will split the rectangles into x and width, and y and height components. We can then determine that if there is overlap on both dimensions, then the rectangles themselves intersect!

In order to understand the calc_overlap function, draw out two flat lines and follow along with the function and notice how it detects an overlap!

Let's begin by creating a general function to detect overlap in a single dimension:

def calc_overlap(coor1,dim1,coor2,dim2):
    """
    Takes in 2 coordinates and their length in that dimension
    """

    # Find greater of the two coordinates
    # (this is either the point to the most right
    #  or the higher point, depending on the dimension)

    # The greater point would be the start of the overlap
    greater = max(coor1,coor2)

    # The lower point is the end of the overlap
    lower = min(coor1+dim1,coor2+dim2)

    # Return a tuple of Nones if there is no overlap

    if greater >= lower:
        return (None,None)

    # Otherwise, get the overlap length
    overlap = lower-greater

    return (greater,overlap)

 

Now let's use this function to detect if the rectangles overlap!

 

 

def calc_rect_overlap(r1,r2):

    x_overlap, w_overlap = calc_overlap(r1['x'],r1['w'],r2['x'],r2['w'])

    y_overlap, h_overlap = calc_overlap(r1['y'],r1['h'],r2['y'],r2['h'])

    # If either returned None tuples, then there is no overlap!
    if not w_overlap or not h_overlap:
        print 'There was no overlap!'
        return None

    # Otherwise return the dictionary format of the overlapping rectangle
    return { 'x':x_overlap,'y': y_overlap,'w':w_overlap,'h':h_overlap}

 

 

Our solution is O(1) for both time and space! Let's see it in action:

In [5]:

r1 = {'x': 2 , 'y': 4,'w':5,'h':12}
r2 = {'x': 1 , 'y': 5,'w':7,'h':14}
calc_rect_overlap(r1,r2)

Out[5]:

{'h': 11, 'w': 5, 'x': 2, 'y': 5}

 

Make sure to review the answer and practice writing it out by hand!

Good Job!