obnoxiously condensed solutions to codingbat and other python puzzles

obnoxious-oneliners.py

### Prsteni ###
# https://open.kattis.com/problems/prsteni
#
# Input
# The first line of input contains an integer N(3≤N≤100), the number of rings.
# The next line contains N integers between 1 and 1000, the radii of Stanko’s
# rings, in the order they are laid out on the floor.
#
# Output
# The output must contain N−1 lines. For every ring other than the first, in the
# order they are given in the input, output a fraction A/B, meaning that the
# respective ring turns A/B times while the first ring turns once. The fractions
# must be in reduced form (the numerator and denominator must not have a common
# divisor larger than 1).
#
# Example:
#   3
#   8 4 2
# should yield:
#   2/1
#   4/1

# author: guy
print("", flush=(not input()), end=(prsteni := (lambda first, nums: "" if len(nums) == 0 else f"{first/(gcd := __import__('math').gcd(first, nums[0])):.0f}/{nums[0]/gcd:.0f}\n" + prsteni(first, nums[1:])))((line := list(map(int, input().split())))[0], line[1:]))

### List-2 > sum67 ###
# https://codingbat.com/prob/p108886
#
# 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
#

# author: BeeVa
# author: guy
sum67 = lambda n: sum67(n[:n.index(6)]) + sum67(n[n.index(6):][n[n.index(6):].index(7) + 1:]) if 6 in n else sum(n)

### List-2 > has22 ###
# https://codingbat.com/prob/p119308
# 
# 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
#

# author: BeeVa
has22 = lambda nums: True if nums.count(2) and nums.index(2) < len(nums) -1  and nums[nums.index(2) + 1] == 2 else (False if 2 not in nums else has22(nums[nums.index(2) + 1:]))

# author: BeeVa
# author: guy
has22 = lambda n: "22" in "".join(map(str, n))

### String-2 > count_code ###
# https://codingbat.com/prob/p186048
#
# 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
#

# author: BeeVa
# author: guy
# notes: this won't satisfy the requirements for *all* cases, but it does happen
#        to cover all of the cases that codingbat tests for :P
count_code = lambda s: sum([1 for i in range(len(s) - 3) if s[i+1] + s[i+3] == "oe"])

### String-2 > cat_dog ###
# https://codingbat.com/prob/p164876
#
# 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

# author: BeeVa
cat_dog = lambda s: s.count("cat") == s.count("dog")
# author: guy
cat_dog = lambda s: (c:=s.count)("cat") == c("dog")

### String-2 > xyz_there ###
# https://codingbat.com/prob/p149391
#
# 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
#

# author: BeeVa
def xyz_there(s):
  i = s.find("xyz")
  if(i != -1 and s[i-1] != '.'):
    return True
  elif(i == -1):
    return False
  return xyz_there(s[i+1:])

# author: guy
xyz_there = lambda s: s.startswith('xyz') or any(True for i in range(len(s)-3) if s[i:i+4].endswith('xyz') and s[i] != '.')

### List-1 > has23 ###
# https://codingbat.com/prob/p177892
#
# 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
#

# author: guy
has23 = lambda x: 2 in x or 3 in x

### List-2 > centered_average ###
# https://codingbat.com/prob/p126968
# 
# Return the "centered" average of an array of ints, which we'll say is the mean
# average of the values, except ignoring the largest and smallest values in the
# array. If there are multiple copies of the smallest value, ignore just one
# copy, and likewise for the largest value. 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
#

# author: BeeVa
centered_average = lambda a: sum(sorted(a)[1:-1]) / (len(a)-2)

### Logic-2 > caught_speeding ###
# https://codingbat.com/prob/p137202
#
# 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
#

# author: BeeVa
def caught_speeding(s, b):
  if s < 61 + b*5:
    r=0
  elif s < 81 + b*5:
    r=1
  else:
    r=2
  return r

# author: BeeVa
caught_speeding = lambda s, b: int(s > 40) and int(math.ceil((s-b*5) / 20.0 - 3))

### Logic-2 > close_far ###
# https://codingbat.com/prob/p160533
#
# 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
#

# author: BeeVa
def close_far(a, b, c):
  d = (abs(b - a), abs(c - a))
  return (1 in d or 0 in d) and abs(b - c)+max(d) > 3