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.
- Majority Element
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 = 1current = numsfor i in nums[1:]:if i != current:counter -= 1if counter < 0:current = icounter = 1else:counter += 1return current
Its all about tracking who is in front of the pack and swapping when a challenger comes along.
- Detect Capital
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) < 97if is_cap(word) and len(word) > 1 and is_cap(word):for c in word:if not is_cap(c):return Falseelif is_cap(word):for c in word[1:]:if is_cap(c):return Falseelse:for c in word[1:]:if is_cap(c):return Falsereturn 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.
- Not Boring Movies
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 belowSELECT * FROM cinema WHERE id % 2 != 0 AND description != 'boring' ORDER BY rating DESC;