#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0

# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/

# Additional basic list exercises

# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
  # intialize the reduced list
  nums_reduced = [];
  # initialize a counter
  counter = 0
  # initialize a int variable to hold the last number in the reduced list that was inspected
  last_num = 0
  # iterate through each number in the input list
  for num in nums:
    # we are processing the first number in the input list, this code block will only run once
    if counter == 0:
      # add the number to the reduced list since there cannot be an adjacent == element
      nums_reduced.append(nums[counter])
      # keep track of what the last (and first in this case) number in the reduced list is
      last_num = nums[counter]
    # we are not processing the first number in the input list, check if the current input list number is equal to the last element in the reduced list
    elif counter >= 1 and (last_num != nums[counter]):
        # the number is not adjacent == element, so add it to the reduced list
        nums_reduced.append(nums[counter])
        # record a new last reduced list number
        last_num = nums[counter]
    # increment the counter
    counter = counter + 1
  # return the reduced list
  return nums_reduced


# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
  # merge both lists into a single list
  list3 = list1 + list2
  # sort the merged list
  return sorted(list3)

# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.


# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
  if got == expected:
    prefix = ' OK '
  else:
    prefix = '  X '
  print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))


# Calls the above functions with interesting inputs.
def main():
  print 'remove_adjacent'
  test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
  test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
  test(remove_adjacent([]), [])

  print
  print 'linear_merge'
  test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
       ['aa', 'bb', 'cc', 'xx', 'zz'])
  test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
       ['aa', 'bb', 'cc', 'xx', 'zz'])
  test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
       ['aa', 'aa', 'aa', 'bb', 'bb'])


if __name__ == '__main__':
  main()