Physical Address
304 North Cardinal St.
Dorchester Center, MA 02124
Interviewing for a software engineering position at Amazon? Start with our practice programming interview questions. We’ve helped several engineers get the job at Amazon already.
A crack team of love scientists from OkEros (a hot new dating site) have devised a way to represent dating profiles as rectangles on a two-dimensional plane.
They need help writing an algorithm to find the intersection of two users’ love rectangles. They suspect finding that intersection is the key to a matching algorithm so powerful it will cause an immediate acquisition by Google or Facebook or Obama or something.
Write a function to find the rectangular intersection of two given love rectangles.
As with the example above, love rectangles are always “straight” and never “diagonal.” More rigorously: each side is parallel with either the x-axis or the y-axis.
They are defined as dictionaries ↴ like this:
my_rectangle = {
# Coordinates of bottom-left corner
'left_x' : 1,
'bottom_y' : 1,
# Width and height
'width' : 6,
'height' : 3,
}
Your output rectangle should use this format as well.
What if there is no intersection? Does your function do something reasonable in that case?
What if one rectangle is entirely contained in the other? Does your function do something reasonable in that case?
What if the rectangles don’t really intersect but share an edge? Does your function do something reasonable in that case?
Do some parts of your function seem very similar? Can they be refactored so you repeat yourself less?
Let’s break this problem into subproblems. How can we divide this problem into smaller parts?
We could look at the two rectangles’ “horizontal overlap” or “x overlap” separately from their “vertical overlap” or “y overlap.”
Lets start with a helper function find_x_overlap().
Need help finding the x overlap?
Since we’re only working with the x dimension, we can treat the two rectangles’ widths as ranges on a 1-dimensional number line.
What are the possible cases for how these ranges might overlap or not overlap? Draw out some examples!
There are four relevant cases:
1) The ranges partially overlap:
2) One range is completely contained in the other:
3) The ranges don’t overlap:
4) The ranges “touch” at a single point:
Let’s start with the first 2 cases. How do we compute the overlapping range?
One of our ranges starts “further to the right” than the other. We don’t know ahead of time which one it is, but we can check the starting points of each range to see which one has the highest_start_point. That highest_start_point is always the left-hand side of the overlap, if there is one.
Not convinced? Draw some examples!
Similarly, the right-hand side of our overlap is always the lowest_end_point. That may or may not be the end point of the same input range that had the highest_start_point—compare cases (1) and (2).
This gives us our x overlap! So we can handle cases (1) and (2). How do we know when there is no overlap?
If highest_start_point > lowest_end_point, the two rectangles do not overlap.
But be careful—is it just greater than or is it greater than or equal to?
It depends how we want to handle case (4) above.
If we use greater than, we treat case (4) as an overlap. This means we could end up returning a rectangle with zero width, which … may or may not be what we’re looking for. You could make an argument either way.
Let’s say a rectangle with zero width (or zero height) isn’t a rectangle at all, so we should treat that case as “no intersection.”
Can you finish find_x_overlap()?
Here’s one way to do it:
def find_x_overlap(x1, width1, x2, width2): # Find the highest ("rightmost") start point and
# lowest ("leftmost") end point
highest_start_point = max(x1, x2)
lowest_end_point = min(x1 + width1, x2 + width2)
# Return null overlap if there is no overlap
if highest_start_point >= lowest_end_point:
return (None, None)
# Compute the overlap width
overlap_width = lowest_end_point - highest_start_point
return (highest_start_point, overlap_width)
How can we adapt this for the rectangles’ ys and heights?
Can we just make one find_range_overlap() function that can handle x overlap and y overlap?
Yes! We simply use more general parameter names:
def find_range_overlap(point1, length1, point2, length2): # Find the highest start point and lowest end point.
# The highest ("rightmost" or "upmost") start point is
# the start point of the overlap.
# The lowest end point is the end point of the overlap.
highest_start_point = max(point1, point2)
lowest_end_point = min(point1 + length1, point2 + length2)
# Return null overlap if there is no overlap
if highest_start_point >= lowest_end_point:
return (None, None)
# Compute the overlap length
overlap_length = lowest_end_point - highest_start_point
return (highest_start_point, overlap_length)
We’ve solved our subproblem of finding the x and y overlaps! Now we just need to put the results together.
We divide the problem into two halves:
Both problems are basically the same as finding the intersection of two “ranges” on a 1-dimensional number line.
So we write a helper function find_range_overlap() that can be used to find both the x overlap and the y overlap, and we use it to build the rectangular overlap:
def find_range_overlap(point1, length1, point2, length2): # Find the highest start point and lowest end point.
# The highest ("rightmost" or "upmost") start point is
# the start point of the overlap.
# The lowest end point is the end point of the overlap.
highest_start_point = max(point1, point2)
lowest_end_point = min(point1 + length1, point2 + length2)
# Return null overlap if there is no overlap
if highest_start_point >= lowest_end_point:
return (None, None)
# Compute the overlap length
overlap_length = lowest_end_point - highest_start_point
return (highest_start_point, overlap_length)
def find_rectangular_overlap(rect1, rect2):
# Get the x and y overlap points and lengths
x_overlap_point, overlap_width = find_range_overlap(rect1['left_x'],
rect1['width'],
rect2['left_x'],
rect2['width'])
y_overlap_point, overlap_height = find_range_overlap(rect1['bottom_y'],
rect1['height'],
rect2['bottom_y'],
rect2['height'])
# Return null rectangle if there is no overlap
if not overlap_width or not overlap_height:
return {
'left_x' : None,
'bottom_y' : None,
'width' : None,
'height' : None,
}
return {
'left_x' : x_overlap_point,
'bottom_y' : y_overlap_point,
'width' : overlap_width,
'height' : overlap_height,
}
O(1)O(1) time and O(1)O(1) space.
What if we had a list of rectangles and wanted to find all the rectangular overlaps between all possible pairs of two rectangles within the list? Note that we’d be returning a list of rectangles.
What if we had a list of rectangles and wanted to find the overlap between all of them, if there was one? Note that we’d be returning a single rectangle.
This is an interesting one because the hard part isn’t the time or space optimization—it’s getting something that works and is readable.
For problems like this, I often see candidates who can describe the strategy at a high level but trip over themselves when they get into the details.
Don’t let it happen to you. To keep your thoughts clear and avoid bugs, take time to:
In our experience, we suggest you solve this Practice Questions for the Amazon Interview and gain some new skills from Professionals completely free and we assure you will be worth it.
If you are stuck anywhere between any coding problem, just visit Queslers to get the Practice Questions for the Amazon Interview
I hope this Practice Questions for the Amazon Interview would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.
This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.
Keep Learning!
More Coding Solutions >>