All solutions were successfully tested on 18 April 2013.

**count_evens:**

def count_evens(nums): count = 0 for element in nums: if element % 2 == 0: count += 1 return count

**big_diff:**

def big_diff(nums): return max(nums) - min(nums)

**centered_average:**

def centered_average(nums): sum = 0 for element in nums: sum += element return (sum - min(nums) - max(nums)) / (len(nums)-2)

**sum13:**

def sum13(nums): if len(nums) == 0: return 0 for i in range(0, len(nums)): if nums[i] == 13: nums[i] = 0 if i+1 < len(nums): nums[i+1] = 0 return sum(nums)

**sum67:**

def sum67(nums): for i in range(0, len(nums)): if nums[i] == 6: nums[i] = 0 for j in range(i+1, len(nums)): temp = nums[j] nums[j] = 0 if temp == 7: i = j + 1 break return sum(nums)

Line 9 is not necessary. However, by adjusting “i” you ensure that this script runs in linear time, despite the nested loop.

**has22:**

def has22(nums): for i in range(0, len(nums)-1): #if nums[i] == 2 and nums[i+1] == 2: if nums[i:i+2] == [2,2]: return True return False

The second option is much nicer to look at, but either way is fine.

ddddef centered_average(nums):

total = sum(nums) – max(nums) – min(nums)

return total/(len(nums)-2)

araminiJust my 2 cents…

The generalized instructions for List-2 says:

“Medium python list problems — 1 loop.. Use a[0], a[1], … to access elements in a list, len(a) is the length.”

( http://codingbat.com/python/List-2 )

So, for sum67, the solution below follows the “one-loop” constraint. It only uses one loop and if statements/Boolean logic. Not the most elegent or clever solution, but I think it is easier for a “n00b” like me to understand than Gregor’s solution for sum67. (Although I do understand that when coding in the real world where things aren’t simply academic in nature, you’d want code to be pretty, clean, and concise).

def sum67(nums):

found6 = False

result = 0

for n in nums:

if n==6:

found6 = True

continue

if n==7 and found6:

found6 = False

continue

if not found6:

result += n

return result