| Given 3 int values, a b c, return their sum. However, if any of the values is a teen -- in the range 13..19 inclusive -- then that value counts as 0, except 15 and 16 do not count as a teens. Write a separate helper "def fix_teen(n):"that takes in an int value and returns that value fixed for the teen rule. In this way, you avoid repeating the teen code 3 times (i.e. "decomposition"). Define the helper below and at the same indent level as the main no_teen_sum(). no_teen_sum(1, 2, 3) → 6 no_teen_sum(2, 13, 1) → 3 no_teen_sum(2, 1, 14) → 3...Save, Compile, Run |
| Given 3 int values, a b c, return their sum. However, if any of the values is a teen -- in the range 13..19 inclusive -- then that value counts as 0, except 15 and 16 do not count as a teens. Write a separate helper "def fix_teen(n):"that takes in an int value and returns that value fixed for the teen rule. In this way, you avoid repeating the teen code 3 times (i.e. "decomposition"). Define the helper below and at the same indent level as the main no_teen_sum(). no_teen_sum(1, 2, 3) → 6 no_teen_sum(2, 13, 1) → 3 no_teen_sum(2, 1, 14) → 3...Save, Compile, Run def fix_teen(num): |
| The 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) → True love6(4, 5) → False love6(1, 5) → True...Save, Compile, Run def love6(a, b): |
| Given 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'...Save, Compile, Run def first_half(str): |
| Given 3 int values, a b c, return their sum. However, if one of the values is 13 then it does not count towards the sum and values to its right do not count. So for example, if b is 13, then both b and c do not count. lucky_sum(1, 2, 3) → 6 lucky_sum(1, 2, 13) → 3 lucky_sum(1, 13, 3) → 1...Save, Compile, Run def lucky_sum(a, b, c): |
| Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 7 (every 6 will be followed by at least one 7). Return 0 for no numbers. sum67([1, 2, 2]) → 5 sum67([1, 2, 2, 6, 99, 99, 7]) → 5 sum67([1, 1, 6, 7, 2]) → 4...Save, Compile, Run def sum67(nums): |
| Given an array of ints, return True if one of the first 4 elements in the array is a 9. The array length may be less than 4. array_front9([1, 2, 9, 3, 4]) → True array_front9([1, 2, 3, 4, 9]) → False array_front9([1, 2, 3, 4, 5]) → False...Save, Compile, Run def array_front9(nums): |
| Given a string, return a new string made of every other char starting with the first, so "Hello" yields "Hlo". string_bits('Hello') → 'Hlo' string_bits('Hi') → 'H' string_bits('Heeololeo') → 'Hello'...Save, Compile, Run def string_bits(str): |
| Return an int array length 3 containing the first 3 digits of pi, {3, 1, 4}. make_pi() → [3, 1, 4]...Save, Compile, Run def make_pi(): |
| Given 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!'...Save, Compile, Run def hello_name(name): |
| Given 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) → 7 sorta_sum(9, 4) → 20 sorta_sum(10, 11) → 21...Save, Compile, Run def sorta_sum(a, b): |
| Given a non-empty string like "Code" return a string like "CCoCodCode". string_splosion('Code') → 'CCoCodCode' string_splosion('abc') → 'aababc' string_splosion('ab') → 'aab'...Save, Compile, Run def string_splosion(str): |
| We want to make a row of bricks that is goal inches long. We have a number of small bricks (1 inch each) and big bricks (5 inches each). Return True if it is possible to make the goal by choosing from the given bricks. This is a little harder than it looks and can be done without any loops. make_bricks(3, 1, 8) → True make_bricks(3, 1, 9) → False make_bricks(3, 2, 10) → True...Save, Compile, Run def make_bricks(small, big, goal): |
| Given an array of ints, return True if the array contains a 2 next to a 2 somewhere. has22([1, 2, 2]) → True has22([1, 2, 1, 2]) → False has22([2, 1, 2]) → False...Save, Compile, Run def has22(nums): |
| Given 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'...Save, Compile, Run def alarm_clock(day, vacation): |
| We 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) → True monkey_trouble(False, False) → True monkey_trouble(True, False) → False...Save, Compile, Run def monkey_trouble(a_smile, b_smile): |
| Given 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) → True near_hundred(90) → True near_hundred(89) → False...Save, Compile, Run def find(num,n): |
| Given 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]...Save, Compile, Run def make_ends(nums): |
| Given 2 ints, a and b, return True if one if them is 10 or if their sum is 10. makes10(9, 10) → True makes10(9, 9) → False makes10(1, 9) → True...Save, Compile, Run def makes10(a, b): |
| Return the "centered" average of an array of ints, which we'll say is the mean average of the values, except not counting the largest and smallest values in the array. Use int division to produce the final average. You may assume that the array is length 3 or more. centered_average([1, 2, 3, 4, 100]) → 3 centered_average([1, 1, 5, 5, 10, 8, 7]) → 5 centered_average([-10, -4, -2, -4, -2, 0]) → -3...Save, Compile, Run def centered_average(nums): |
| Given 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'...Save, Compile, Run def non_start(a, b): |
| You 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) → 2 date_fashion(5, 2) → 0 date_fashion(5, 5) → 1...Save, Compile, Run def date_fashion(you, date): |
| Given 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. "<<word>>". make_out_word('<<>>', 'Yay') → '<<Yay>>' make_out_word('<<>>', 'WooHoo') → '<<WooHoo>>' make_out_word('[[]]', 'word') → '[[word]]'...Save, Compile, Run def make_out_word(out, word): |
| The web is built with HTML strings like "<i>Yay</i>" which draws Yay as italic text. In this example, the "i" tag makes <i> and </i> which surround the word "Yay". Given tag and word strings, create the HTML string with tags around the word, e.g. "<i>Yay</i>". make_tags('i', 'Yay') → '<i>Yay</i>' make_tags('i', 'Hello') → '<i>Hello</i>' make_tags('cite', 'Yay') → '<cite>Yay</cite>'...Save, Compile, Run def make_tags(tag, word): |
| Given 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]...Save, Compile, Run def max_end3(nums): |
| The 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) → True squirrel_play(95, False) → False squirrel_play(95, True) → True...Save, Compile, Run def squirrel_play(temp, is_summer): |
| You 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) → 0 caught_speeding(65, False) → 1 caught_speeding(65, True) → 0...Save, Compile, Run def caught_speeding(speed, is_birthday): |
| Given 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'...Save, Compile, Run def without_end(str): |
| Given two int values, return their sum. Unless the two values are the same, then return double their sum. sum_double(1, 2) → 3 sum_double(3, 2) → 5 sum_double(2, 2) → 8...Save, Compile, Run def sum_double(a, b): |
| Given 3 int values, a b c, return their sum. However, if one of the values is the same as another of the values, it does not count towards the sum. lone_sum(1, 2, 3) → 6 lone_sum(3, 2, 3) → 2 lone_sum(3, 3, 3) → 0...Save, Compile, Run def lone_sum(a, b, c): |
| Given a string, return the count of the number of times that a substring length 2 appears in the string and also as the last 2 chars of the string, so "hixxxhi" yields 1 (we won't count the end substring). last2('hixxhi') → 1 last2('xaxxaxaxx') → 1 last2('axxxaaxx') → 2...Save, Compile, Run def last2(str): |
| Given 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]) → True common_end([1, 2, 3], [7, 3, 2]) → False common_end([1, 2, 3], [1, 3]) → True...Save, Compile, Run def common_end(a, b): |
| Given 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'...Save, Compile, Run def front3(str): |
| Given 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]...Save, Compile, Run def rotate_left3(nums): |
| Given 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'...Save, Compile, Run def extra_end(str): |
| Return True if the given string contains an appearance of "xyz" where the xyz is not directly preceeded by a period (.). So "xxyz" counts but "x.xyz" does not. xyz_there('abcxyz') → True xyz_there('abc.xyz') → False xyz_there('xyz.abc') → True...Save, Compile, Run def xyz_there(str): |
| Given 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'...Save, Compile, Run def missing_char(str, n): |
| Given 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'...Save, Compile, Run def front_back(str): |
| Given 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) → True in1to10(11, False) → False in1to10(11, True) → True...Save, Compile, Run def in1to10(n, outside_mode): |
| Given three ints, a b c, return True if one of b or c is "close" (differing from a by at most 1), while the other is "far", differing from both other values by 2 or more. Note: abs(num) computes the absolute value of a number. close_far(1, 2, 10) → True close_far(1, 2, 3) → False close_far(4, 1, 3) → True...Save, Compile, Run def close_far(a, b, c): |
| Given 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'...Save, Compile, Run def left2(str): |
| Given 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) → True pos_neg(-1, 1, False) → True pos_neg(1, 1, False) → False...Save, Compile, Run def pos_neg(a, b, negative): |
| Return True if the string "cat" and "dog" appear the same number of times in the given string. cat_dog('catdog') → True cat_dog('catcat') → False cat_dog('1cat1cadodog') → True...Save, Compile, Run def cat_dog(str): |
| Given a string and a non-negative int n, we'll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front; front_times('Chocolate', 2) → 'ChoCho' front_times('Chocolate', 3) → 'ChoChoCho' front_times('Abc', 3) → 'AbcAbcAbc'...Save, Compile, Run def front_times(str, n): |
| Given 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) → True near_ten(17) → False near_ten(19) → True...Save, Compile, Run def near_ten(num): |
| Given an array of ints, return the number of 9's in the array. array_count9([1, 2, 9]) → 1 array_count9([1, 9, 9]) → 2 array_count9([1, 9, 9, 3, 9]) → 3...Save, Compile, Run def array_count9(nums): |
| We 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) → True parrot_trouble(True, 7) → False parrot_trouble(False, 6) → False...Save, Compile, Run def parrot_trouble(talking, hour): |
| Return the number of times that the string "hi" appears anywhere in the given string. count_hi('abc hi ho') → 1 count_hi('ABChi hi') → 2 count_hi('hihi') → 2...Save, Compile, Run def count_hi(str): |
| Given a string, return a string where for every char in the original, there are two chars. double_char('The') → 'TThhee' double_char('AAbb') → 'AAAAbbbb' double_char('Hi-There') → 'HHii--TThheerree'...Save, Compile, Run def double_char(str): |
| Given 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]...Save, Compile, Run def middle_way(a, b): |
| The 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) → True sleep_in(True, False) → False sleep_in(False, True) → True...Save, Compile, Run def sleep_in(weekday, vacation): |
| Given two strings, return True if either of the strings appears at the very end of the other string, ignoring upper/lower case differences (in other words, the computation should not be "case sensitive"). Note: s.lower() returns the lowercase version of a string. end_other('Hiabc', 'abc') → True end_other('AbC', 'HiaBc') → True end_other('abc', 'abXabc') → True...Save, Compile, Run def end_other(a, b): |
| Given an int array length 2, return True if it contains a 2 or a 3. has23([2, 5]) → True has23([4, 3]) → True has23([4, 5]) → False...Save, Compile, Run def has23(nums): |
| Given 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]) → False same_first_last([1, 2, 3, 1]) → True same_first_last([1, 2, 1]) → True...Save, Compile, Run def same_first_last(nums): |
| For this problem, we'll round an int value up to the next multiple of 10 if its rightmost digit is 5 or more, so 15 rounds up to 20. Alternately, round down to the previous multiple of 10 if its rightmost digit is less than 5, so 12 rounds down to 10. Given 3 ints, a b c, return the sum of their rounded values. To avoid code repetition, write a separate helper "def round10(num):" and call it 3 times. Write the helper entirely below and at the same indent level as round_sum(). round_sum(16, 17, 18) → 60 round_sum(12, 13, 14) → 30 round_sum(6, 4, 4) → 10...Save, Compile, Run def round10(num): |
| Given 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]) → True first_last6([6, 1, 2, 3]) → True first_last6([3, 2, 1]) → False...Save, Compile, Run def first_last6(nums): |
| Given 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'...Save, Compile, Run def make_abba(a, b): |
| Given 2 strings, a and b, return the number of the positions where they contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3, since the "xx", "aa", and "az" substrings appear in the same place in both strings. string_match('xxcaazz', 'xxbaaz') → 3 string_match('abc', 'abc') → 2 string_match('abc', 'axc') → 0...Save, Compile, Run def string_match(a, b): |
| Given 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'...Save, Compile, Run def first_two(str): |
| Given an array length 1 or more of ints, return the difference between the largest and smallest values in the array. Note: the built-in min(v1, v2) and max(v1, v2) functions return the smaller or larger of two values. big_diff([10, 3, 5, 6]) → 7 big_diff([7, 2, 10, 9]) → 8 big_diff([2, 10, 7, 2]) → 8...Save, Compile, Run def big_diff(nums): |
| Return the number of times that the string "code" appears anywhere in the given string, except we'll accept any letter for the 'd', so "cope" and "cooe" count. count_code('aaacodebbb') → 1 count_code('codexxcode') → 2 count_code('cozexxcope') → 2...Save, Compile, Run def count_code(str): |
| Given 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'...Save, Compile, Run def not_string(str): |
| Return the number of even ints in the given array. Note: the % "mod" operator computes the remainder, e.g. 5 % 2 is 1. count_evens([2, 1, 2, 3, 4]) → 3 count_evens([2, 2, 0]) → 3 count_evens([1, 3, 5]) → 0...Save, Compile, Run def count_evens(nums): |
| Given an array of ints length 3, return the sum of all the elements. sum3([1, 2, 3]) → 6 sum3([5, 11, 2]) → 18 sum3([7, 0, 0]) → 7...Save, Compile, Run def sum3(nums): |
| Given 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]) → 3 sum2([1, 1]) → 2 sum2([1, 1, 1, 1]) → 2...Save, Compile, Run def sum2(nums): |
| Given 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]...Save, Compile, Run def reverse3(nums): |
| Given a string and a non-negative int n, return a larger string that is n copies of the original string. string_times('Hi', 2) → 'HiHi' string_times('Hi', 3) → 'HiHiHi' string_times('Hi', 1) → 'Hi'...Save, Compile, Run def string_times(str, n): |
| Given an array of ints, return True if .. 1, 2, 3, .. appears in the array somewhere. array123([1, 1, 2, 3, 1]) → True array123([1, 1, 2, 4, 1]) → False array123([1, 1, 2, 1, 2, 3]) → True...Save, Compile, Run def array123(nums): |
| Given 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'...Save, Compile, Run def combo_string(a, b): |
| When 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) → False cigar_party(50, False) → True cigar_party(70, True) → True...Save, Compile, Run def cigar_party(cigars, is_weekend): |
| Given an int n, return the absolute difference between n and 21, except return double the absolute difference if n is over 21. diff21(19) → 2 diff21(10) → 11 diff21(21) → 0...Save, Compile, Run def diff21(n): |
