While the "Dynamic programming and recursion" section hosts this problem, the
optimal solution does not use recursion. Many cite the Fibonacci problem as a
quintessential dynamic programming question. I assume these people expect an
answer like:
```python
def fib(n):
cache = {0: 0, 1: 1}
def do_fib(n):
if n in cache:
return cache[n]
else:
cache[n - 1] = do_fib(n - 1)
cache[n - 2] = do_fib(n - 2)
return cache[n - 1] + cache[n - 2]
return do_fib(n)
```
The cache turns the runtime of the classic Fibonacci solution...
```python
def fib(n):
if n in {0, 1}:
return n
return fib(n - 1) + fib(n - 2)
```
... from O(2^n) to a O(n). But both the cache itself and the additional stacks
that the runtime allocates for each recursive call create an O(n) space
complexity.
InterviewCake wants the answer to be solved in O(n) time with O(1)
space. To achieve this, instead of solving fib(n) from the top-down, we solve it
from the bottom-up.
I found this problem to be satisfying to solve.
Write a function to find a duplicate item in a list of numbers. The values are
in the range [1, n]; the length of the list is n + 1. The solution should run in
linear time and consume constant space.
The solution is to construct a graph from the list. Each graph will have a cycle
where the last element in the cycle is a duplicate value.
See the solution for specific techniques on how to compute the length the cycle
without infinitely looping.
Write a function that returns the shortest path between nodes A and B in an
unweighted graph.
I know two algorithms for finding the shortest path in a *weighted* graph:
- Use a heap as a priority queue instead of the regular queue that you would use
when doing a BFT. This is called Dijkstra's algorithm. You can also use
Dijkstra's algorithm in an unweight graph by imaginging that all of the
weights on the edges are the same value (e.g. 1).
- Map the weighted graph into an unweighted graph by inserting N nodes between
each node, X and Y, where N is equal to the weight of the edge between X and
Y. After you map the weighted graph into an unweighted graph, perform a BFT
from A to B. A BFT will always find the shortest path between nodes A and B in
an unweighted graph.
I had forgotten that a BFT in an unweighted graph will always return the
shortest path between two nodes. I learned two things from InterviewCake.com's
solution:
1. I remembered that a BFT in an unweighted graph will return the shortest
path (if one exists).
2. I learned to use a dictionary to store the edge information and then
back-tracking to reconstruct the shortest path.
Return a function that returns the second largest item in a binary search
tree (i.e. BST).
A BST is a tree where each node has no more than two children (i.e. one left
child and one right child). All of the values in a BST's left subtree must be
less than the value of the root node; all of the values in a BST's right subtree
must be greater than the value of the root node; both left and right subtrees
must also be BSTs themselves.
I solved this problem thrice -- improving the performance profile each time. The
final solution has a runtime complexity of O(n) and a spacetime complexity of
O(1).
Write a function that returns true if a given binary tree is a valid binary
search tree (i.e. if all of root's left nodes are less than root.value, all of
root's right nodes are greater than root.value, and both left and right subtrees
are also valid binary search trees).
Write a predicate for determining if a binary tree is "super balanced", which
means that the depths of all of the tree's leaves are equal or differ by at most
one.
I wrongfully assumed that the relationship between a question and a question
category was one-to-one; it is actually one-to-many. This explains why I
completed the "Cafe Order Checker" and "Top Scores" questions twice.
I'm marking the questions that I've completed as DONE because I would prefer to
do every question once and then prioritize repeating the questions with which I
experienced difficulty.
Write a function to sort a list of scores for a game in linear time. While I had
previously solved this in python, I hadn't marked the todo.org file, so I ended
up doing this again.
"Perfect practice makes perfect."
Write a function that accepts a rotated cycle of alphabetically sorted strings
and returns the index what should be the first element if the elements were not
rotated.
This problem challenged me: without using division, write a function that maps a
list of integers into a list of the product of every integer in the list except
for the integer at that index.
This was another greedy algorithm. The take-away is to first solve the problem
using brute force; this yields an algorithm with O(n*(n-1)) time
complexity. Instead of a quadratic time complexity, a linear time complexity can
be achieved my iterating over the list of integers twice:
1. Compute the products of every number to the left of the current number.
2. Compute the products of every number to the right of the current number.
Finally, iterate over each of these and compute lhs * rhs. Even though I've
solved this problem before, I used InterviewCake's hints because I was stuck
without them.
I should revisit this problem in a few weeks.
Write a function that returns the highest product of three integers within a
list of integers. This solution uses a greedy algorithm that solves for the
answer in linear time. The space complexity is constant.
Write a function that returns the maximum profit that a trader could have made
in a day. I solved this using a greedy algorithm which constantly sets the
maximum profit by tracking the lowest price we've encountered.
Wrote a function to reverse the words in a list of characters. A word is a
space-delimited strings of characters.
The trick here is to first reverse the entire string and then reverse each word
individually.
Write a function to merge meeting times. Added an in-place solution, which the
"Bonus" section suggested attempting to solve.
- Added some simple benchmarks to test the performance differences between the
in-place and not-in-place variants. To my surprise, the in-place solution was
consistently slower than the not-in-place solution.
I had a spare fifteen minutes and decided that I should tidy up my
monorepo. The work of tidying up is not finished; this is a small step in the
right direction.
TL;DR
- Created a tools directory
- Created a scratch directory (see README.md for more information)
- Added README.md to third_party
- Renamed delete_dotfile_symlinks -> symlinkManager
- Packaged symlinkManager as an executable symlink-mgr using buildGo
2020-02-12 16:58:29 +00:00
Renamed from deepmind/part_two/todo.org (Browse further)
