CodingbatPythonQuestionsandAnswersSection1

Codingbat Python Questions and Answers

Section 1

This document is prepared and can be used only for educational purposes. All questions are taken from which contains great questions about Python and Java. Please go to original website and solve

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The answers are done by me in my spare times, Codingbat says all answers are true. I used some of questions

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use/modify/distribute answers of me or this document it under the terms of the GNU General Public License

as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later

version.

Document is prepared with Google Docs and syntax highlighter is GeSHi, which is a great open source you have any questions, please email me*******************Have fun with Python!Samet Atdag

This document contains 44 questions in these sections:Warmup-1

●●String-1●List-1●Logic-1Warmup-1

_inThe parameter weekday is True if it is a weekday, and the parameter vacation is

True if we are on vacation. We sleep in if it is not a weekday or we're on vacation.

Return True if we sleep in.

sleep_in(False, False) → Truesleep_in(True, False) → Falsesleep_in(False, True) → True

My solution:def sleep_in(weekday, vacation): if(not weekday or vacation): return True else: return False

_troubleWe have two monkeys, a and b, and the parameters a_smile and b_smile indicate

if each is smiling. We are in trouble if they are both smiling or if neither of them is

smiling. Return True if we are in trouble.

monkey_trouble(True, True) → Truemonkey_trouble(False, False) → Truemonkey_trouble(True, False) → False

My solution:def monkey_trouble(a_smile, b_smile): if((a_smile and b_smile) or ((not a_smile) and (not b_smile))): return True else: return False

_doubleGiven two int values, return their sum. Unless the two values are the same, then

return double their sum.

sum_double(1, 2) → 3sum_double(3, 2) → 5sum_double(2, 2) → 8

My solution:def sum_double(a, b): result = a+b if(a == b): result = 2*result return result

21Given an int n, return the absolute difference between n and 21, except return

double the absolute difference if n is over 21.

diff21(19) → 2diff21(10) → 11diff21(21) → 0

My solution:def diff21(n): if n>21: result = 2*(n-21) else: result = 21-n return result

_troubleWe have a loud talking parrot. The "hour" parameter is the current hour time in the

range 0..23. We are in trouble if the parrot is talking and the hour is before 7 or

after 20. Return True if we are in trouble.

parrot_trouble(True, 6) → Trueparrot_trouble(True, 7) → Falseparrot_trouble(False, 6) → False

My solution:def parrot_trouble(talking, hour): if talking: if hour<7 or hour>20: return True else: return False else: return False

10Given 2 ints, a and b, return True if one if them is 10 or if their sum is 10.

makes10(9, 10) → Truemakes10(9, 9) → Falsemakes10(1, 9) → True

My solution:def makes10(a, b): if a==10 or b==10 or a+b==10: return True else: return False

_hundredGiven an int n, return True if it is within 10 of 100 or 200. Note: abs(num)

computes the absolute value of a number.

near_hundred(93) → Truenear_hundred(90) → Truenear_hundred(89) → False

My solution:def near_hundred(n): if abs(100-n)<=10 or abs(200-n)<=10: return True else: return False

_negGiven 2 int values, return True if one is negative and one is positive. Unless the

parameter "negative" is True, then they both must be negative.

pos_neg(1, -1, False) → Truepos_neg(-1, 1, False) → Truepos_neg(1, 1, False) → False

My solution:def pos_neg(a, b, negative): if negative: return (a < 0 and b < 0) else: return ((a < 0 and b > 0) or (a > 0 and b < 0))

_stringGiven a string, return a new string where "not " has been added to the front.

However, if the string already begins with "not", return the string unchanged.

not_string('candy') → 'not candy'not_string('x') → 'not x'not_string('not bad') → 'not bad'

My solution:def not_string(str): a = ('not') if len(a) > 1 and a[0]=="": return str else: return "not " + str

g_charGiven a non-empty string and an int n, return a new string where the char at index

n has been removed. The value of n will be a valid index of a char in the original

string (i.e. n will be in the range 0..len(str)-1 inclusive).

missing_char('kitten', 1) → 'ktten'missing_char('kitten', 0) → 'itten'missing_char('kitten', 4) → 'kittn'

My solution:def missing_char(str, n): return str[0:n] + str[n+1:]

_backGiven a string, return a new string where the first and last chars have been

exchanged.

front_back('code') → 'eodc'front_back('a') → 'a'front_back('ab') → 'ba'

My solution:def front_back(str): if len(str) == 1: return str elif len(str) == 2: return str[1] + str[0] else: return str[-1:] + str[1:-1] + str[:1]

3Given a string, we'll say that the front is the first 3 chars of the string. If the string

length is less than 3, the front is whatever is there. Return a new string which is 3

copies of the front.

front3('Java') → 'JavJavJav'front3('Chocolate') → 'ChoChoCho'front3('abc') → 'abcabcabc'

My solution:def front3(str): if len(str) < 3: return str+str+str else: return str[:3] + str[:3] + str[:3]

_nameGiven a string name, e.g. "Bob", return a greeting of the form "Hello Bob!".

hello_name('Bob') → 'Hello Bob!'hello_name('Alice') → 'Hello Alice!'hello_name('X') → 'Hello X!'

My solution:def hello_name(name): return "Hello " + name + "!"

_abbaGiven two strings, a and b, return the result of putting them together in the order

abba, e.g. "Hi" and "Bye" returns "HiByeByeHi".

make_abba('Hi', 'Bye') → 'HiByeByeHi'make_abba('Yo', 'Alice') → 'YoAliceAliceYo'make_abba('x', 'y') → 'xyyx'

My solution:def make_abba(a, b): return a + b + b + a

_tagsThe web is built with HTML strings like "Yay" which draws Yay as italic text.

In this example, the "i" tag makes and which surround the word "Yay".

Given tag and word strings, create the HTML string with tags around the word,

e.g. "Yay".

make_tags('i', 'Yay') → 'Yay'make_tags('i', 'Hello') → 'Hello'make_tags('cite', 'Yay') → 'Yay'

My solution:def make_tags(tag, word): return "<" + tag + ">" + word + ""

_out_wordGiven an "out" string length 4, such as "<<>>", and a word, return a new string

where the word is in the middle of the out string, e.g. "<>".

make_out_word('<<>>', 'Yay') → '<>'make_out_word('<<>>', 'WooHoo') → '<>'make_out_word('[[]]', 'word') → '[[word]]'

My solution:def make_out_word(out, word): return out[0:2] + word + out[2:]

_endGiven a string, return a new string made of 3 copies of the last 2 chars of the

original string. The string length will be at least 2.

extra_end('Hello') → 'lololo'extra_end('ab') → 'ababab'extra_end('Hi') → 'HiHiHi'

My solution:def extra_end(str): return 3*(str[-2:])

_twoGiven a string, return the string made of its first two chars, so the String "Hello"

yields "He". If the string is shorter than length 2, return whatever there is, so "X"

yields "X", and the empty string "" yields the empty string "".

first_two('Hello') → 'He'first_two('abcdefg') → 'ab'first_two('ab') → 'ab'

My solution:def first_two(str): if len(str) < 2: return str else: return str[0:2]

_halfGiven a string of even length, return the first half. So the string "WooHoo"

yields "Woo".

first_half('WooHoo') → 'Woo'first_half('HelloThere') → 'Hello'first_half('abcdef') → 'abc'

My solution:def first_half(str): return str[0:len(str)/2]

t_endGiven a string, return a version without the first and last char, so "Hello"

yields "ell". The string length will be at least 2.

without_end('Hello') → 'ell'without_end('java') → 'av'without_end('coding') → 'odin'

My solution:def without_end(str): return str[1:-1]

_stringGiven 2 strings, a and b, return a string of the form short+long+short, with the

shorter string on the outside and the longer string on the inside. The strings will not

be the same length, but they may be empty (length 0).

combo_string('Hello', 'hi') → 'hiHellohi'combo_string('hi', 'Hello') → 'hiHellohi'combo_string('aaa', 'b') → 'baaab'

My solution:def combo_string(a, b): if len(a)

_startGiven 2 strings, return their concatenation, except omit the first char of each. The

strings will be at least length 1.

non_start('Hello', 'There') → 'ellohere'non_start('java', 'code') → 'avaode'non_start('shotl', 'java') → 'hotlava'

My solution:def non_start(a, b): return a[1:] + b[1:]

2Given a string, return a "rotated left 2" version where the first 2 chars are moved to

the end. The string length will be at least 2.

left2('Hello') → 'lloHe'left2('java') → 'vaja'left2('Hi') → 'Hi'

My solution:def left2(str): return str[2:] + str[0:2]

List-1

_last6Given an array of ints, return True if 6 appears as either the first or last element in

the array. The array will be length 1 or more.

first_last6([1, 2, 6]) → Truefirst_last6([6, 1, 2, 3]) → Truefirst_last6([3, 2, 1]) → False

My solution:def first_last6(nums): if (nums[0] == 6) or (nums[len(nums)-1] == 6): return True else: return False

_first_lastGiven an array of ints, return True if the array is length 1 or more, and the first

element and the last element are the same.

same_first_last([1, 2, 3]) → Falsesame_first_last([1, 2, 3, 1]) → Truesame_first_last([1, 2, 1]) → True

My solution:def same_first_last(nums): if len(nums)>0: if nums[0] == nums[len(nums) - 1]: return True else: return False else: return False

_piReturn an int array length 3 containing the first 3 digits of pi, {3, 1, 4}.

make_pi() → [3, 1, 4]

My solution:def make_pi(): return [3,1,4]

_endGiven 2 arrays of ints, a and b, return True if they have the same first element or

they have the same last element. Both arrays will be length 1 or more.

common_end([1, 2, 3], [7, 3]) → Truecommon_end([1, 2, 3], [7, 3, 2]) → Falsecommon_end([1, 2, 3], [1, 3]) → True

My solution:def common_end(a, b): if (a[0] == b[0]) or (a[len(a)-1] == b[len(b)-1]): return True else: return False

3Given an array of ints length 3, return the sum of all the elements.

sum3([1, 2, 3]) → 6sum3([5, 11, 2]) → 18sum3([7, 0, 0]) → 7

My solution:def sum3(nums): return nums[0]+nums[1]+nums[2]

_left3Given an array of ints length 3, return an array with the elements "rotated left" so

{1, 2, 3} yields {2, 3, 1}.

rotate_left3([1, 2, 3]) → [2, 3, 1]rotate_left3([5, 11, 9]) → [11, 9, 5]rotate_left3([7, 0, 0]) → [0, 0, 7]

My solution:def rotate_left3(nums): return [nums[1], nums[2], nums[0]]

e3Given an array of ints length 3, return a new array with the elements in reverse

order, so {1, 2, 3} becomes {3, 2, 1}.

reverse3([1, 2, 3]) → [3, 2, 1]reverse3([5, 11, 9]) → [9, 11, 5]reverse3([7, 0, 0]) → [0, 0, 7]

My solution:def reverse3(nums): return [nums[2], nums[1], nums[0]]

_end3Given an array of ints length 3, figure out which is larger between the first and last

elements in the array, and set all the other elements to be that value. Return the

changed array.

max_end3([1, 2, 3]) → [3, 3, 3]max_end3([11, 5, 9]) → [11, 11, 11]max_end3([2, 11, 3]) → [3, 3, 3]

My solution:def max_end3(nums): if nums[0]>nums[2]: return [nums[0],nums[0],nums[0]] else: return [nums[2],nums[2],nums[2]]

2Given an array of ints, return the sum of the first 2 elements in the array. If the

array length is less than 2, just sum up the elements that exist, returning 0 if the

array is length 0.

sum2([1, 2, 3]) → 3sum2([1, 1]) → 2sum2([1, 1, 1, 1]) → 2

My solution:def sum2(nums): if len(nums) == 0: return 0 elif len(nums) == 1: return nums[0] else: return nums[0] + nums[1]

_wayGiven 2 int arrays, a and b, each length 3, return a new array length 2 containing

their middle elements.

middle_way([1, 2, 3], [4, 5, 6]) → [2, 5]middle_way([7, 7, 7], [3, 8, 0]) → [7, 8]middle_way([5, 2, 9], [1, 4, 5]) → [2, 4]

My solution:def middle_way(a, b): return [a[1],b[1]]

_endsGiven an array of ints, return a new array length 2 containing the first and last

elements from the original array. The original array will be length 1 or more.

make_ends([1, 2, 3]) → [1, 3]make_ends([1, 2, 3, 4]) → [1, 4]make_ends([7, 4, 6, 2]) → [7, 2]

My solution:def make_ends(nums): return [nums[0], nums[-1]]

23Given an int array length 2, return True if it contains a 2 or a 3.

has23([2, 5]) → Truehas23([4, 3]) → Truehas23([4, 5]) → False

My solution:def has23(nums): if 2 in nums or 3 in nums: return True else: return False

Logic-1

_partyWhen squirrels get together for a party, they like to have cigars. A squirrel party

is successful when the number of cigars is between 40 and 60, inclusive. Unless

it is the weekend, in which case there is no upper bound on the number of cigars.

Return True if the party with the given values is successful, or False otherwise.

cigar_party(30, False) → Falsecigar_party(50, False) → Truecigar_party(70, True) → True

My solution:def cigar_party(cigars, is_weekend): if is_weekend: if cigars>=40: return True else: return False else: if cigars>=40 and cigars<=60: return True else: return False

_fashionYou and your date are trying to get a table at a restaurant. The parameter "you"

is the stylishness of your clothes, in the range 0..10, and "date" is the stylishness

of your date's clothes. The result getting the table is encoded as an int value with

0=no, 1=maybe, 2=yes. If either of you is very stylish, 8 or more, then the result

is 2 (yes). With the exception that if either of you has style of 2 or less, then the

result is 0 (no). Otherwise the result is 1 (maybe).

date_fashion(5, 10) → 2date_fashion(5, 2) → 0date_fashion(5, 5) → 1

My solution:def date_fashion(you, date): if you<=2 or date<=2: return 0 elif you>=8 or date>=8: return 2 else: return 1

el_playThe squirrels in Palo Alto spend most of the day playing. In particular, they play

if the temperature is between 60 and 90 (inclusive). Unless it is summer, then

the upper limit is 100 instead of 90. Given an int temperature and a boolean

is_summer, return True if the squirrels play and False otherwise.

squirrel_play(70, False) → Truesquirrel_play(95, False) → Falsesquirrel_play(95, True) → True

My solution:def squirrel_play(temp, is_summer): upper = 90 if is_summer: upper = 100 return (temp>=60 and temp<=upper)

_speedingYou are driving a little too fast, and a police officer stops you. Write code to

compute the result, encoded as an int value: 0=no ticket, 1=small ticket, 2=big

ticket. If speed is 60 or less, the result is 0. If speed is between 61 and 80

inclusive, the result is 1. If speed is 81 or more, the result is 2. Unless it is your

birthday -- on that day, your speed can be 5 higher in all cases.

caught_speeding(60, False) → 0caught_speeding(65, False) → 1caught_speeding(65, True) → 0

My solution:def caught_speeding(speed, is_birthday): gift = 0 if is_birthday: gift = 5 if speed <= 60+gift: return 0 elif speed >= 81+gift: return 2 else: return 1

_sumGiven 2 ints, a and b, return their sum. However, sums in the range 10..19

inclusive, are forbidden, so in that case just return 20.

sorta_sum(3, 4) → 7sorta_sum(9, 4) → 20sorta_sum(10, 11) → 21

My solution:def sorta_sum(a, b): total = a+b if total > 9 and total < 20: return 20 else: return total

_clockGiven a day of the week encoded as 0=Sun, 1=Mon, 2=Tue, ...6=Sat, and

a boolean indicating if we are on vacation, return a string of the form "7:00"

indicating when the alarm clock should ring. Weekdays, the alarm should be "7:00"

and on the weekend it should be "10:00". Unless we are on vacation -- then on

weekdays it should be "10:00" and weekends it should be "off".

alarm_clock(1, False) → '7:00'alarm_clock(5, False) → '7:00'alarm_clock(0, False) → '10:00'

My solution:def alarm_clock(day, vacation): weekday_alarm = "7:00" weekend_alarm = "10:00" if vacation: weekday_alarm = "10:00" weekend_alarm = "off" if day>0 and day<6: return weekday_alarm else: return weekend_alarm

6The number 6 is a truly great number. Given two int values, a and b, return True

if either one is 6. Or if their sum or difference is 6. Note: the function abs(num)

computes the absolute value of a number.

love6(6, 4) → Truelove6(4, 5) → Falselove6(1, 5) → True

My solution:def love6(a, b): if a==6 or b==6 or a+b==6 or abs(a-b)==6: return True return False

1to10Given a number n, return True if n is in the range 1..10, inclusive.

Unless "outsideMode" is True, in which case return True if the number is less or

equal to 1, or greater or equal to 10.

in1to10(5, False) → Truein1to10(11, False) → Falsein1to10(11, True) → True

My solution:def in1to10(n, outside_mode): if not outside_mode: return (n>=1 and n<=10) else: return (n<=1 or n>=10)

_tenGiven a non-negative number "num", return True if num is within 2 of a multiple of

10. Note: (a % b) is the remainder of dividing a by b, so (7 % 5) is 2.

near_ten(12) → Truenear_ten(17) → Falsenear_ten(19) → True

My solution:def near_ten(num): return (num%10==0 or num%10==1 or num%10==2 or abs(10-num%10)==2 or abs(10-num%10)==1 or abs(10-num%10)==0)