But it doesn't have to end here! Sign up for the 7-day coding interview crash course and you'll get a free Interview Cake problem every week.
You're in!
Find a duplicate, Space Edition™ BEAST MODE
In Find a duplicate, Space Edition™, we were given a vector of integers where:
These properties mean the vector must have at least 1 duplicate. Our challenge was to find a duplicate number without modifying the input and optimizing for space. We used a divide and conquer approach, iteratively cutting the vector in half to find a duplicate integer in time and space (sort of a modified binary search).
But we can actually do better. We can find a duplicate integer in time while keeping our space cost at .
This is a tricky one to derive (unless you have a strong background in graph theory), so we'll get you started:
Imagine each item in the vector as a node in a linked list. In any linked list, each node has a value and a "next" pointer. In this case:
Here’s a full example:
Notice we're using "positions" and not "indices." For this problem, we'll use the word "position" to mean something like "index," but different: indices start at 0, while positions start at 1. More rigorously: position = index + 1.
Using this, find a duplicate integer in time while keeping our space cost at . Just like before, don't modify the input.
Drawing pictures will help a lot with this one. Grab some paper and pencil (or a whiteboard, if you have one).
We don’t need any new data structures. Your final space cost must be .
We can do this without destroying the input.
Log in or sign up with one click to get immediate access to free mock interview questions
We'll never post on your wall or message your friends.
Actually, we don't support password-based login. Never have. Just the OAuth methods above. Why?
Log in or sign up with one click to get immediate access to free mock interview questions
We'll never post on your wall or message your friends.
Actually, we don't support password-based login. Never have. Just the OAuth methods above. Why?
time and space.
Our space cost is because all of our additional variables are integers, which each take constant space.
For our runtime, we iterate over the array a constant number of times, and each iteration takes time in its worst case. So we traverse the linked list more than once, but it's still a constant number of times—about 3.
There another approach using randomized algorithms that is time and space. Can you come up with that one? (Hint: You'll want to focus on the median.)
Log in or sign up with one click to get immediate access to free mock interview questions
We'll never post on your wall or message your friends.
Actually, we don't support password-based login. Never have. Just the OAuth methods above. Why?
Reset editor
Powered by qualified.io