Inspired by this article
https://mail.google.com/mail/u/0/#inbox/FMfcgxwLswDnDZZvQsSGqCpXgqNGTQbZ
Given an integer, return the integer with reversed digits.
Note: The integer could be either positive or negative.
-231=>-132345=>543
Solution
def solution(n):
return f'-{str(n)[1:][::-1]}' if n<0 else str(n)[::-1]For a given sentence, return the average word length.
Note: Remember to remove punctuation first.
sentence1 = "Hi all, my name is Tom...I am originally from Australia."
sentence2 = "I need to work very hard to learn more about algorithms in Python!"
assert solution(sentence1)==3.82
assert solution(sentence2)==4.08Solution
import re
def solution(sentence):
def avg(iterable):
ret = 0
n = 0.0 # if using python 2 force to float
for x in iterable:
ret+=x
n+=1.0 # counting like this because `iterable` is a generator and generators has no `len` method
return ret/n # if using python 2 remember to cast to float
return round(avg(map(len, re.findall(r'\w+', sentence))), 2)Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
Solution
from functools import reduce # only needed for python 3
def solution(num1, num2):
n1 = reduce(lambda a, b: a*10+b-48, map(ord, num1), 0)
n2 = reduce(lambda a, b: a*10+b-48, map(ord, num2), 0)
return n1+n2Given a string, find the first non-repeating character in it and return its index.
If it doesn't exist, return -1.
Note: all the input strings are already lowercase.
Solution
def solution(s):
seen = [None]*256 # let's assume only ascii characters 0..255
for i, x in enumerate(s):
n = ord(x)
if seen[n] is None:
seen[n] = i
elif seen[n] is not False:
seen[n] = False
try:
return min(x for x in seen if x not in (False, None))
except ValueError: # empty generator above
return -1Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
The string will only contain lowercase characters a-z.
Solution
def solution(s):
if s[:len(s)//2+1] == s[:len(s)//2-1:-1]:
return True # already a palindrome
for i in range(len(s)):
w = s[:i] + s[i+1:]
if w[:len(w)//2+1] == w[:len(w)//2-1:-1]:
return True
return FalseGiven an array of integers, determine whether the array is monotonic or not.
A = [6,5,4,4]
B = [1,1,1,3,3,4,3,2,4,2]
C = [1,1,2,3,7]
assert solution(A)
assert not solution(B)
assert solution(C)Solution
def solution(nums):
return all(a>=b for a, b in zip(nums, nums[1:])) or \
all(a<=b for a, b in zip(nums, nums[1:]))Given an array nums, write a function to move all zeroes to the end of it while maintaining the relative order of the non-zero elements.
array1 = [0,1,0,3,12]
array2 = [1,7,0,0,8,0,10,12,0,4]
assert solution(array1) == [1, 3, 12, 0, 0]
assert solution(array2) == [1, 7, 8, 10, 12, 4, 0, 0, 0, 0]Solution
def solution(nums): # NOT changing the input
ret = [x for x in nums if x!=0]
return ret + [0] * (len(nums)-len(ret))
from collections import deque
def solution_2(nums): # changing the input (inplace)
zeros = deque()
for i in range(len(nums)):
x = nums[i]
if x==0:
zeros.append(i)
elif len(zeros)>0:
idx = zeros.popleft()
nums[idx] = x
nums[i] = 0
zeros.append(i)
return numsGiven an array containing None values fill in the None values with most recent non-None value in the array
array1 = [1, None, 2, 3, None, None, 5, None]
assert solution(array1) == [1, 1, 2, 3, 3, 3, 5, 5]Solution
def solution(arr):
ret = []
for x in arr:
valid = valid if x is None else x
ret.append(valid)
return retAnother solution
def first_true(iter):
for x in iter:
if x is not None:
return x
raise ValueError("All the values are None")
def solution(a):
return [first_true(x) for x in zip(*(([None] * i + a[:len(a)-i]) for i in range(len(a))))]That can become even more obscure
def solution(a):
return [
next(y for y in x if y is not None)
for x in zip(*(([None] * i + a[:len(a)-i]) for i in range(len(a))))
]Basically I am creating a matrix and rotating it with zip. Then for each column (that now is a row), taking the first not None element.
The first solution is the more readable and the fastest one.
>>> timeit.timeit('solution([1, None, 2, 3, None, None, 5, None])', '''
def solution(arr):
ret = []
for x in arr:
valid = valid if x is None else x
ret.append(valid)
return ret
''')
0.6686493580054957
>>> timeit.timeit('solution([1, None, 2, 3, None, None, 5, None])', '''
def first_true(iter):
for x in iter:
if x is not None:
return x
raise ValueError("All the values are None")
def solution(a):
return [first_true(x) for x in zip(*(([None] * i + a[:len(a)-i]) for i in range(len(a))))]
''')
4.560916772999917
>>> timeit.timeit('solution([1, None, 2, 3, None, None, 5, None])', '''
def solution(a):
return [
next(y for y in x if y is not None)
for x in zip(*(([None] * i + a[:len(a)-i]) for i in range(len(a))))
]
''')
5.971725533992867Given two sentences, return an array that has the words that appear in one sentence and not the other and an array with the words in common.
sentence1 = 'We are really pleased to meet you in our city'
sentence2 = 'The city was hit by a really heavy storm'
assert solution(sentence1, sentence2) == ({'We', 'to', 'heavy', 'The', 'storm', 'meet', 'hit', \
'pleased', 'are', 'by', 'a', 'in', 'was', 'you', 'our'}, {'really', 'city'})Solution
def solution(sentence1, sentence2):
set1 = set(sentence1.split())
set2 = set(sentence2.split())
return set1^set2, set1&set2Given k numbers which are less than n, return the set of prime number among them
Note: The task is to write a program to print all Prime numbers in an Interval.
Definition: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
assert solution(35) == [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]Solution
def solution(n):
primes = []
for num in range(2, n):
for p in primes:
if num % p == 0:
break
else:
primes.append(num)
return primes