I’ve been going through the problems roughly sorted by their acceptance rate: higher
-> lower. Today, the difficulty is slowly creeping up. I’m looking forward to tackling some Mediums soon.
Problem: Given a list of
words, return only those that can be typed using one keyboard row
This was a neat problem to solve. At one point, I was
lower()ing the words to account for capitalization but simply adding capital letters to the Sets decreased the runtime by ~13%.
class Solution(object): def findWords(self, words): """ :type words: List[str] :rtype: List[str] """ rows = [set('qwertyuiopQWERTYUIOP'), set('asdfghjklASDFGHJKL'), set('zxcvbnmZXCVBNM')] one_row_words =  for word in words: for row in rows: if word in row: if all(char in row for char in word[1:]): one_row_words.append(word) break return one_row_words
I made sure to break once the word’s row had been found to save on loop cycles over ‘dead’ words.
Problem: Given an array of
logs with alphanumeric identifiers at the start, sort them so that letter-only logs come before digit-only logs.
I’ve added in some comments where there are specific rules but I recommend checking their description as this one’s problem statement is harder to understand than any solution.
This was a fun one to solve.
class Solution(object): def reorderLogFiles(self, logs): """ :type logs: List[str] :rtype: List[str] """ letter_logs =  digit_logs =  for log in logs: log = log.split() # check ascii value to determine digit or letter log if ord(log) < 66: # the digit-logs should be put in their original order digit_logs.append(log) else: letter_logs.append(log) # the letter-logs are ordered lexicographically ignoring identifier, # with the identifier used in case of ties letter_logs.sort(key=lambda x: x[1:] + x[0:1]) return [' '.join(log) for log in letter_logs + digit_logs]
I’m getting a little better with Python generators. I really like how they feel to use, and how they affect the structure of these programs.
Runtime complexity: Separate logs
O(n), sort letter logs
O(n log n), join logs
Problem: Given an array
N of integers, numbers all appear twice except for one. Return that integer.
I knew this could be XORed but I couldn’t remember so implemented it with a Set and made a note to check after.
class Solution(object): def singleNumber(self, nums): """ :type nums: List[int] :rtype: int """ nums_set = set() for i in nums: if i in nums_set: nums_set.remove(i) else: nums_set.add(i) for i in nums_set: return i
And now without that extra space, using XOR.
class Solution(object): def singleNumber(self, nums): """ :type nums: List[int] :rtype: int """ result = 0 for i in nums: result ^= i return result
See you next time.