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List-2

Basic python list problems -- no loops.. Use a[0], a[1], ... to access elements in a list, len(a) is the length.

  • first_last6
  • same_first_last
  • make_pi
  • common_end
  • sum3
  • rotate_left3
  • reverse3
  • max_end3
  • sum2
  • middle_way
  • make_ends
  • has23

###first_last6 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([13, 6, 1, 2, 3]) → False

My solution

def first_last6(nums):
  return(nums[0] == 6 or nums[len(nums)-1] == 6)

Codingbat solution

def first_last6(nums):
  return (nums[0]==6 or nums[-1]== 6)

###same_first_last Given an array of ints, return True if the array is length 1 or more, and the first element and the last element are equal.

  • same_first_last([1, 2, 3]) → False
  • same_first_last([1, 2, 3, 1]) → True
  • same_first_last([1, 2, 1]) → True

My solution

def same_first_last(nums):
  return(len(nums) > 0 and nums[0] == nums[len(nums)-1])

Codingbat solution

def same_first_last(nums):
  return (len(nums) >= 1 and nums[0] == nums[-1])

###make_pi Return 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])

Codingbat solution

None given

###common_end 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

My solution

def common_end(a, b):
  return(a[0] == b[0] or a[len(a)-1] == b[len(b)-1])

Codingbat solution

None given

###sum3 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

My solution

def sum3(nums):
  return(sum(nums[0:3]))

Codingbat solution

None given

###rotate_left3 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]

My solution

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

Codingbat solution

None given

###reverse3 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]

My solution

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

Codingbat solution

None given

###max_end3 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]

My solution

def max_end3(nums):
  return([max(nums[0],nums[2])]*3)

Codingbat solution

def max_end3(nums):
  big = max(nums[0], nums[2])
  nums[0] = big
  nums[1] = big
  nums[2] = big
  return nums

###sum2 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

My solution

def sum2(nums):
  if len(nums) == 0: return 0
  elif len(nums) < 2: return (nums[0])
  else: return(nums[0] + nums[1])

Codingbat solution

None given

###middle_way 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]

My solution

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

Codingbat solution

None given

###make_ends

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]

My solution

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

Codingbat solution

None given

###has23 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

My solution

def has23(nums):
  return(2 in nums or 3 in nums)

Codingbat solution

None given