Title refers to my runtime rankings against other users not how long it took me to complete these. Ran into a couple of stubborn bugs, one of which was failing to read the question properly because I jumped ahead - doh.
Problem: Return the uncommon words from two sentences
Note: a word is uncommon if it appears no more than once in either string.
This solution works regardless of the amount of sentences given. Thinking about finding uncommon words in just two sentences seems like a
from collections import Counterclass Solution(object):def uncommonFromSentences(self, A, B):""":type A: str:type B: str:rtype: List[str]"""word_counter = Counter()for word in A.split(' ') + B.split(' '):word_counter[word] += 1uncommon_words = for key, val in word_counter.items():if val == 1:uncommon_words.append(key)return uncommon_words
Add all words to a Counter Dictionary, return the words with a value of
Problem: Search a Binary Tree.
My solution is a recursive search.
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution(object):def searchBST(self, root, val):""":type root: TreeNode:type val: int:rtype: TreeNode"""if not root:return Noneif root.val == val:return rootelif val < root.val:return self.searchBST(root.left, val)return self.searchBST(root.right, val)
Nothing complicated to see here.
Problem: Implement the infamous FizzBuzz algorithm.
This was my first time 'solving' this in Python. I almost forgot about having to shift the iteration range (1-100 instead of 0-99).
class Solution:def fizzBuzz(self, n):""":type n: int:rtype: List[str]"""str_reps = for i in range(1, n+1):if i % 3 == 0 and i % 5 == 0:str_reps.append('FizzBuzz')elif i % 3 == 0:str_reps.append('Fizz')elif i % 5 == 0:str_reps.append('Buzz')else:str_reps.append(str(i))return str_reps
Day 5 complete, 16 problems solved in 2019.