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AshleyCoder3/anithingButFive.py

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Python brain teasers
Raw
anithingButFive.py
# Given a start integer and an ending integer (both inclusive), write a function that returns the count (not the sum) of all integers in the range (except integers that contain the digit 5).
# Examples:
# csAnythingButFive(1, 5) -> 1, 2, 3, 4, -> 4 (there are 4 integers in the range that do not contain the digit 5)
# csAnythingButFive(1, 9) -> 1, 2, 3, 4, 6, 7, 8, 9 -> 8
# csAnythingButFive(4, 17) -> 4,6,7,8,9,10,11,12,13,14,16,17 -> 12
# Notes:
# The output can contain the digit 5.
# The start number will always be less than the end number (both numbers can also be negative).
# [execution time limit] 4 seconds (py3)
# [input] integer start
# [input] integer end
# [output] integer
def csAnythingButFive(start, end):
count = 0
new_list = []
for num in range(start, end + 1):
str_num = str(num)
if '5' in str_num :
continue
count += 1
return count
print(csAnythingButFive(1, 5))
print(csAnythingButFive(1, 9))
print(csAnythingButFive(4, 17))
print(csAnythingButFive(-4, 17))
print(csAnythingButFive(0, 7))
print(csAnythingButFive(-14, -5))
#uper
# input 2 integers
# output 1 integer = a count of numbers in range(start, end)
# start < end
# the 2 nums cand be negative
# PLAN
# for loop in range(start, end + 1)
# if num == 5 skip
# else count(num)
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binaryToASCII.py
# Given a binary string (ASCII encoded), write a function that returns the equivalent decoded text.
# Every eight bits in the binary string represents one character on the ASCII table.
# Examples:
# csBinaryToASCII("011011000110000101101101011000100110010001100001") -> "lambda"
# 01101100 -> 108 -> "l"
# 01100001 -> 97 -> "a"
# 01101101 -> 109 -> "m"
# 01100010 -> 98 -> "b"
# 01100100 -> 100 -> "d"
# 01100001 -> 97 -> "a"
# csBinaryToASCII("") -> ""
# Notes:
# The input string will always be a valid binary string.
# Characters can be in the range from "00000000" to "11111111" (inclusive).
# In the case of an empty input string, your function should return an empty string.
# [execution time limit] 4 seconds (py3)
# [input] string binary
# [output] string
def csBinaryToASCII(binary):
binary_letters = []
letters = ''
if binary == "":
return ""
for index in range(0, len(binary), 8):
binary_letters.append(binary[index:index + 8])
print(binary_letters)
for string in binary_letters:
binary_int = v=chr(int(string, 2))
print(binary_int)
letters += binary_int
return letters
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findAddedLetter.py
# You are given two strings, str_1 and str_2, where str_2 is generated by randomly shuffling str_1 and then adding one letter at a random position.
# Write a function that returns the letter that was added to str_2.
# Examples:
# csFindAddedLetter(str_1 = "bcde", str_2 = "bcdef") -> "f"
# csFindAddedLetter(str_1 = "", str_2 = "z") -> "z"
# csFindAddedLetter(str_1 = "b", str_2 = "bb") -> "b"
# csFindAddedLetter(str_1 = "bf", str_2 = "bfb") -> "b"
# Notes:
# str_1 and str_2 both consist of only lowercase alpha characters.
# [execution time limit] 4 seconds (py3)
# [input] string str_1
# [input] string str_2
# [output] string
def csFindAddedLetter(str_1, str_2):
m1 = {}
for i in str_2:
if i in m1:
m1[i] += 1
else:
m1[i] =1
for i in str_1:
m1[i] -= 1
for h1 in m1:
if m1[h1] == 1:
return h1
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findTheSingleNumber.py
You are given a non-empty array of integers.
One element appears exactly once, with every other element appearing at least twice, perhaps more.
# Write a function that can find and return the element that appears exactly once.
# Example 1:
# Input: [1,1,2,1]
# Output: 2
# Example 2:
# Input: [1,2,1,2,1,2,80]
# Output: 80
# Note: You should be able to develop a solution that has O(n) time complexity.
# [execution time limit] 4 seconds (py3)
# [input] array.integer nums
# [output] integer
from collections import Counter
def csFindTheSingleNumber(nums):
frequency = Counter(nums)
for i in frequency:
if frequency[i] == 1:
return i
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makeItJazzy.py
#Create a function that concatenates the number 7 to the end of every chord in a list. If a chord already ends with a 7, ignore that chord.
def csMakeItJazzy(chords):
for index in range(len(chords)):
if chords[index].__contains__("7"):
continue
elif chords == []:
return []
else:
chords[index] = chords[index] + "7"
return chords
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maxNumberOfLambdas.py
# Given a string text, you need to use the characters of text to form as many instances of the word "lambda" as possible.
# You can use each character in text at most once.
# Write a function that returns the maximum number of instances of "lambda" that can be formed.
# Input: text = "mbxcdatlas"
# Output: 1
# Example 2:
# Input: text = "lalaaxcmbdtsumbdav"
# Output: 2
# Example 3:
# Input: text = "sctlamb"
# Output: 0
# Notes:
# text consists of lowercase English characters only
# [execution time limit] 4 seconds (py3)
# [input] string text
# [output] integer
def csMaxNumberOfLambdas(text):
sub_string = "lambda"
lambda_count = {'l': 0, 'a': 0, 'm': 0, 'b': 0, 'd': 0, 'a': 0}
counts = []
for letter in text:
if letter in lambda_count:
lambda_count[letter] += 1
for key, value in lambda_count.items():
counts.append(value)
return min(counts)
Raw
python_challenges.py
"""
Challenge #6:
Return the number (count) of vowels in the given string.
We will consider `a, e, i, o, u as vowels for this challenge (but not y).
The input string will only consist of lower case letters and/or spaces.
"""
def get_count(input_str):
# Your code here
vowel_counts = {} #this is a dictionary to store the vowels
for vowel in "aeiou":
count = input_str.count(vowel)
vowel_counts[vowel] = count
print(vowel_counts)
counts = vowel_counts.values()
return sum(counts)
print(get_count("adela are mere"))
"""
Challenge #5:
Create a function that returns the data type of a given argument. There are
seven data types this challenge will be testing for:
- List
- Dictionary
- String
- Integer
- Float
- Boolean
- Date
Examples:
- data_type([1, 2, 3, 4]) ➞ "list"
- data_type({'key': "value"}) ➞ "dictionary"
- data_type("This is an example string.") ➞ "string"
- data_type(datetime.date(2018,1,1)) ➞ "date"
Notes:
- Return the name of the data type as a lowercase string.
"""
import datetime
def data_type(value):
# Your code here
date = datetime.date
if type(value) is int:
return "integer"
elif type(value) is str:
return "string"
elif type(value) is list:
return "list"
elif type(value) is dict:
return "dictionary"
elif type(value) is float:
return "float"
elif type(value) is bool:
return "boolean"
elif type(value) == date:
return "date"
print(data_type([1, 2, 3, 4]))
print(data_type({'key': "value"}))
print(data_type("This is an example string."))
print(data_type(datetime.date(2018,1,1)))
"""
Challenge #4:
Create a function that changes specific words into emoticons. Given a sentence
as a string, replace the words `smile`, `grin`, `sad`, and `mad` with their
corresponding emoticons.
word -> emoticon
---
smile -> :D
grin -> :)
sad -> :(
mad -> :P
Examples:
- emotify("Make me smile") ➞ "Make me :D"
- emotify("Make me grin") ➞ "Make me :)"
- emotify("Make me sad") ➞ "Make me :("
Notes:
- The sentence always starts with "Make me".
- Try to solve this without using conditional statements like if/else.
"""
def emotify(txt):
# Your code here
#````another option```
# new = txt.split(' ')
# print(new)
#````````````
new_list = list(txt)
#print(new_list)
sliced_list = new_list[8:]
emotion = "".join(sliced_list)
print(emotion)
if emotion == "smile":
return "Make me :D"
elif emotion == "grin":
return "Make me :)"
else:
return "Make me :("
print(emotify("Make me smile"))
print(emotify("Make me grin"))
print(emotify("Make me sad"))
import math
"""
Challenge #3:
Given a string of numbers separated by a comma and space, return the product of the numbers.
Examples:
- multiply_nums("2, 3") ➞ 6
- multiply_nums("1, 2, 3, 4") ➞ 24
- multiply_nums("54, 75, 453, 0") ➞ 0
- multiply_nums("10, -2") ➞ -20
Notes:
- Bonus: Try to complete this challenge in one line!
"""
def multiply_nums(nums):
# Your code here
#nums = "2, 4, 6"
integer_list = []
new_list = list(nums)
sliced_list = new_list[0::3]
print(sliced_list)
for elem in sliced_list:
int_elem = int(elem)
integer_list.append(int_elem)
print(type(int_elem))
return math.prod(integer_list)
print(multiply_nums("2, 3"))
"""
Challenge #2:
Given a list of numbers, create a function that returns the list but with each
element's index in the list added to itself. You should add 0 to the number at
index 0, add 1 to the number at index 1, etc.
Examples:
- add_indexes([0, 0, 0, 0, 0]) ➞ [0, 1, 2, 3, 4]
- add_indexes([1, 2, 3, 4, 5]) ➞ [1, 3, 5, 7, 9]
- add_indexes([5, 4, 3, 2, 1]) ➞ [5, 5, 5, 5, 5]
Notes:
- The input list will only contain integers.
"""
def add_indexes(numbers):
# Your code here
new_list = []
for index, value in enumerate(numbers):
sum = index + value
new_list.append(sum)
return new_list
enumerate(my_list) # tuples of the index and the value of the list
list(enumerate(a_list)) #and makes it into an array
for item in enumerate(a_list):
print(item)
for index, value in enumerate(a_list):
print(index) or print(value)
"""
Challenge #1:
Write a function that retrieves the last n elements from a list.
Examples:
- last([1, 2, 3, 4, 5], 1) ➞ [5]
- last([4, 3, 9, 9, 7, 6], 3) ➞ [9, 7, 6]
- last([1, 2, 3, 4, 5], 7) ➞ "invalid"
- last([1, 2, 3, 4, 5], 0) ➞ []
Notes:
- Return "invalid" if n exceeds the length of the list.
- Return an empty list if n == 0.
"""
#list[-1] = last element
#list[-2] = last 2 elements of the list
def last(a, n):
# Your code here
if n > len(a):
return "invalid"
elif n == 0 :
return []
else:
return a[-n:] #the last 3/n elements
print(last([1, 2, 3, 4, 5], 1))
"""
Challenge #9:
Given a string, write a function that returns the "middle" character of the
word.
If the word has an odd length, return the single middle character. If the word
has an even length, return the middle two characters.
Examples:
- get_middle("test") -> "es"
- get_middle("testing") -> "t"
- get_middle("middle") -> "dd"
- get_middle("A") -> "A"
"""
def get_middle(input_str):
# Your code here
return input_str[(len(input_str) -1) // 2: (len(input_str) + 2) // 2]
print(get_middle("test"))
print(get_middle("testing"))
print(get_middle("middle"))
print(get_middle("A"))
print(get_middle("beyoudre"))
print(get_middle("you"))
#Given a string of words, return the length of the shortest word(s).
#The input string will never be empty and you do not need to validate for different data types.
def csShortestWord(input_str):
word = map(len, input_str.split())
return min(word)
# Given an array of integers, return the sum of all the positive integers in the array.
# Examples:
# csSumOfPositive([1, 2, 3, -4, 5]) -> 1 + 2 + 3 + 5 = 11
# csSumOfPositive([-3, -2, -1, 0, 1]) -> 1
# csSumOfPositive([-3, -2]) -> 0
# Notes:
# If the input_arr does not contain any positive integers, the default sum should be 0.
import math
def csSumOfPositive(input_arr):
new_array = []
for num in input_arr:
if num > 0:
new_array.append(num)
else:
continue
return math.fsum(new_array)
"""
Challenge #10:
Given a string of space separated integers, write a function that returns the
maximum and minimum integers.
Example:
- max_and_min("1 2 3 4 5") -> "5 1"
- max_and_min("1 2 -3 4 5") -> "5 -3"
- max_and_min("1 9 3 4 -5") -> "9 -5"
Notes:
- All inputs are valid integers.
- There will always be at least one number in the input string.
- The return string must be two numbers separated by a single space, and
the maximum number is first.
"""
import functools
import operator
def max_and_min(input_str):
# Your code here
new_list = list(map(int, input_str.split()))
new_string = str(max(new_list)) + " " + str(min(new_list))
print(type(new_string))
return new_string
print(max_and_min("1 2 3 4 5"))
print(max_and_min("1 2 -3 4 5"))
print(max_and_min("1 9 3 4 -5"))
"""
Challenge #10:
Create a function that applies a discount d to every number in the list.
Examples:
- get_discounts([2, 4, 6, 11], "50%") ➞ [1, 2, 3, 5.5]
- get_discounts([10, 20, 40, 80], "75%") ➞ [7.5, 15, 30, 60]
- get_discounts([100], "45%") ➞ [45]
Notes:
- The discount is the percentage of the original price (i.e the discount of
"75%" to 12 would be 9 as opposed to taking off 75% (making 3)).
- There won't be any awkward decimal numbers, only 0.5 to deal with.
"""
import math
def get_discounts(nums, percentage):
# Your code here
new_percentage = list(percentage)
discount_str = int(new_percentage[0] + new_percentage[1])
discount = int(discount_str) / 100
print(discount)
d = []
for num in nums:
num = discount * num
s = str(num)
new_num = int(s.rstrip(".0")) if ".0" in s else float(s)
d.append(new_num)
return d
print(get_discounts([2, 4, 6, 11], "50%"))
print(get_discounts([10, 20, 40, 80], "75%"))
print(get_discounts([100], "45%"))
Raw
python_exercises.py
import re
text = "Adela, hi!"
# def csOppositeReverse(txt):
# for i in range(len(txt)):
# if re.match("^[a-z]+$", txt[i]):
# txt[i] = txt[i].upper()
# return txt[::-1]
# elif re.match("^[A-Z]+$", txt[i]):
# txt[i] = txt[i].lower()
# return txt[::-1]
def csOppositeReverse(txt):
return txt.swapcase()[::-1]
print(csOppositeReverse(text))
#puterea a doua a fiecarui digit intr un numar
def csSquareAllDigits(n):
return int(''.join(str(int(i)**2) for i in str(n)))
#take out the vowels of a string
import re
def csRemoveTheVowels(txt):
# vowels = ["a", "e", "i", "o", "u"]
# return "".join([l for l in txt if l not in vowels])
return re.sub(r'[AEIOU]', "", txt, flags=re.IGNORECASE)
print(csRemoveTheVowels(text))
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rainDrops.py
# Given a number, write a function that converts that number into a string that contains "raindrop sounds" corresponding to certain potential factors. A factor is a number that evenly divides into another number, leaving no remainder. The simplest way to test if one number is a factor of another is to use the modulo operator.
# Here are the rules for csRaindrop. If the input number:
# has 3 as a factor, add "Pling" to the result.
# has 5 as a factor, add "Plang" to the result.
# has 7 as a factor, add "Plong" to the result.
# does not have any of 3, 5, or 7 as a factor, the result should be the digits of the input number.
# Examples:
# csRaindrops(28) -> "Plong"
# 28 has 7 as a factor, but not 3 or 5.
# csRaindrops(30) -> "PlingPlang"
# 30 has both 3 and 5 as factors, but not 7.
# csRaindrops(34) -> "34"
# 34 is not factored by 3, 5, or 7.
# [execution time limit] 4 seconds (py3)
# [input] integer number
# [output] string
def csRaindrops(number):
output_string = ""
has_3_factor = output_string + "Pling"
if number % 3 == 0:
output_string += "Pling"
elif number % 5 == 0:
output_string += "Plang"
elif number % 7 == 0:
output_string += "Plong"
elif output_string is "":
output_string += str(number)
return output_string
Raw
repeat_it.py
"""
Challenge #7:
Given a string of lowercase and uppercase alpha characters, write a function
that returns a string where each character repeats in an increasing pattern,
starting at 1. Each character repetition starts with a capital letter and the
rest of the repeated characters are lowercase. Each repetition segment is
separated by a `-` character.
Examples:
- repeat_it("abcd") -> "A-Bb-Ccc-Dddd"
- repeat_it("RqaEzty") -> "R-Qq-Aaa-Eeee-Zzzzz-Tttttt-Yyyyyyy"
- repeat_it("cwAt") -> "C-Ww-Aaa-Tttt"
"""
def repeat_it(input_str):
# Your code here
new_list = list(input_str)
empty = []
for index, letter in enumerate(new_list):
letter = "-" + letter.upper() + (index * letter).lower()
empty.append(letter)
new_str = "".join(empty)
transformed = list(new_str)
good_string = transformed[1:]
return ''.join(good_string)
print(repeat_it("cwAt"))
print(repeat_it("RqaEzty"))
print(repeat_it("abcd"))
Raw
reverseIntegerBits.py
# Given an integer, write a function that reverses the bits (in binary) and returns the integer result.
# Examples:
# csReverseIntegerBits(417) -> 267
# 417 in binary is 110100001. Reversing the binary is 100001011, which is 267 in decimal.
# csReverseIntegerBits(267) -> 417
# csReverseIntegerBits(0) -> 0
# Notes:
# The input integer will not be negative.
# [execution time limit] 4 seconds (py3)
# [input] integer n
# [output] integer
def csReverseIntegerBits(n):
reversed_num = 0
while(n > 0):
reversed_num = reversed_num << 1
if ( n & 1 == 1):
reversed_num = reversed_num ^ 1
n = n >> 1
return reversed_num
Raw
schoolGroups.py
# Write a function that returns the groups in the school by year (as a string), separated with a comma and space in the form of "1a, 1b, 1c, 1d, 2a, 2b (....) 6d, 7a, 7b, 7c, 7d".
# Examples:
# csSchoolYearsAndGroups(years = 7, groups = 4) ➞ "1a, 1b, 1c, 1d, 2a, 2b, 2c, 2d, 3a, 3b, 3c, 3d, 4a, 4b, 4c, 4d, 5a, 5b, 5c, 5d, 6a, 6b, 6c, 6d, 7a, 7b, 7c, 7d"
# Notes:
# 1 <= years <= 10
# 1 <= groups <=26
# [execution time limit] 4 seconds (py3)
# [input] integer years
# [input] integer groups
# [output] string
def schoolGroups(years, groups):
alphabet = list(map(chr, range(97, 123)))
emp_list = []
for year in range(1, years + 1):
for group in range(0, groups):
emp_list.append(f'{year}{alphabet[group]}')
#print(emp_list)
return ', '.join(emp_list)
print(schoolGroups(1, 4))
print(schoolGroups(6, 7))
Raw
validParenthasesSequence.py
# You are given a parentheses sequence, check if it's regular.
# Example
# For s = "()()(())", the output should be
# validParenthesesSequence(s) = true;
# For s = "()()())", the output should be
# validParenthesesSequence(s) = false.
# Input/Output
# [execution time limit] 4 seconds (py3)
# [input] string s
# A string, consisting only of '(' and ')'.
# Guaranteed constraints:
# 0 ≤ s.length ≤ 105.
# [output] boolean
# true is the sequence is regular and false otherwise.
def validParenthesesSequence(s):
stack = []
# Traversing the Expression
for char in s:
if char in ["(", "{", "["]:
stack.append(char)
else:
if not stack:
return False
current_char = stack.pop()
if current_char == '(':
if char != ")":
return False
if current_char == '{':
if char != "}":
return False
if current_char == '[':
if char != "]":
return False
if stack:
return False
return True
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