The three-a-day streak nearly ended today.
I came close to completing a couple of Medium problems but I didn’t want to brute force the answers. I have a set of unwritten rules about what I’m allowed to look up in a book or google. All solutions need to be within the range of ‘optimal’ as well.
The goal is to get better at solving problems and algorithm analysis — not to get the green text that reads Accepted. I ended up searching for some Easys to complete so I could get to bed on time.
Problem: Given an array of size
n find the majority element that appears more than
I wrote about the algoritm that solves this problem for the first post on this blog. The Boyer–Moore majority vote algorithm.
Let’s take a look at its linear runtime and constant space goodness.
class Solution: def majorityElement(self, nums): """ :type nums: List[int] :rtype: int """ counter = 1 current = nums for i in nums[1:]: if i != current: counter -= 1 if counter < 0: current = i counter = 1 else: counter += 1 return current
Its all about tracking who is in front of the pack and swapping when a challenger comes along.
Problem: Judge whether a word uses correct capitals.
The three types of correctness:
I wanted to test the runtime between using Python’s built-ins and a custom solution for Leetcode’s test cases.
So, the complex way:
class Solution: def detectCapitalUse(self, word): """ :type word: str :rtype: bool """ def is_cap(c): return ord(c) < 97 if is_cap(word) and len(word) > 1 and is_cap(word): for c in word: if not is_cap(c): return False elif is_cap(word): for c in word[1:]: if is_cap(c): return False else: for c in word[1:]: if is_cap(c): return False return True
This had a runtime of 76ms.
Instead, we could have just written:
class Solution: def detectCapitalUse(self, word): """ :type word: str :rtype: bool """ return word.isupper() or word.islower() or word.istitle()
This much shorter version had a runtime of .. 76ms. The exact same result!
I’m fairly sure that as
n tends to infinity, the more complex solution will surely win out.
Problem: Given a SQL table
cinema find movies with an odd numbered ID and a description that is not (literally) ‘boring’.
Not much to comment on this one. I read through the discussion board and some people found that using
(not .. ) for their test expressions ran slower but I suppose that will be down to implementation details.
# Write your MySQL query statement below SELECT * FROM cinema WHERE id % 2 != 0 AND description != 'boring' ORDER BY rating DESC;