Python

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

def diff21(n): if (n>21): return ((21-n)*-2) else: return(21-n)

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'

def front3(str): if len(str)<3: return str*3 mid = str[0:3] return mid*3

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'

def front_back(str): if len(str)<=1: return str mid = str[1:len(str)-1] #last + mid + first return str[len(str)-1]+mid + str[0]

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

def makes10(a, b): return(a==10 or b==10 or a+b==10)

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'

def missing_char(str, n): start=str[:n] end=str[n+1:] return start + end

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

def monkey_trouble(a_smile, b_smile): return(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

def near_hundred(n): return((abs(100-n)<=10) or (abs(200-n)<=10))

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'

def not_string(str): if len(str) >=3 and str[:3] == "not": return str else: return("not "+ str)

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

def parrot_trouble(talking, hour): return(talking and(hour <7 or hour >20))

Given 2 int values, return True if one is negative and one is positive. Except if the parameter "negative" is True, then return True only if both are negative. pos_neg(1, -1, False) → True pos_neg(-1, 1, False) → True pos_neg(-4, -5, True) → True

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))

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

def sleep_in(weekday, vacation): return(not weekday or vacation)

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

def sum_double(a, b): if (a==b): return(a*2)+(b*2) else: return(a+b)