I’m really learning a lot by reading the discussion boards after submitting each of my near-optimal solutions — especially, optimization. I’m also noticing holes in my toolset. E.g., bitwise — as we’ll see in just a second.
This Leetcode series started after I felt like I wasn’t getting that much practical knowledge going over algorithm lectures and books. I knew enough of the basics to start putting solving and basic analysis into practise .. so I did. Solving three problems everyday increase my producivity as well which is a huge bonus.
The most beneficial path to me at this point is to use both types of resources.
Since I have a fixed number that I aim for each day, I am finding myself shying away to the more simple problems but this isn’t nesscarily a bad thing. The problems, the research, and the post-solve reading serve as building blocks for taking on the harder ones.
Problem: Given an array of integers, find the only element that doesn’t appear three times.
I knew that the most optimal solution to this problem was going to be by bitwise methods but it wasn’t coming to me. My bitwise capabilities feel below par.
I attempted to solve it as best as I could and ended up with a neat linear-time program.
from collections import Counter class Solution: def singleNumber(self, nums): """ :type nums: List[int] :rtype: int """ ints = Counter() maybes = set() for i in nums: maybes.add(i) ints[i] += 1 if ints[i] == 3: maybes.remove(i) for i in maybes: return i
For every element in the array, there are a few more Dictionary operations than one might like but they are all constant time, and the code is clean and precise.
n find the
nth Fibonacci number.
It’s a shame that this problem doesn’t support memoization because that would make it a little more interesting.
As it is, the only mistep you can make is going for a heavy-handed recursive solution. Mine just iterates and adds.
class Solution: def fib(self, N): """ :type N: int :rtype: int """ if N == 0: return 0 last = 1 curr = 1 i = 2 while i < N: temp = curr curr += last last = temp i += 1 return curr
A fairly neat solution. Only beaten if you use math which brings a constant runtime. E.g., using a formula for the n^th Fibonacci sequence in terms of the golden ratio.
Problem: Given a ransom note string and a magazine string work out whether the ransom note can be constructed from the magazine.
I first attempted to solve this with a pair of Dictionaries, then a single Dictionary, then I remembered that when dealing with single letters it’s usual viable to use an array instead.
import string class Solution(object): def canConstruct(self, ransomNote, magazine): """ :type ransomNote: str :type magazine: str :rtype: bool """ letters =  * 26 for i in magazine: letters[string.lowercase.index(i)] += 1 for i in ransomNote: letters[string.lowercase.index(i)] -= 1 if letters[string.lowercase.index(i)] < 0: return False return True
Mostly pretty simple solutions today. I worked through some tree problems but ran into some edge cases. Presumably at some point I will post up a whole swathe of tree and graph problems!
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