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.
def centered_average(nums):
total = sum(nums) – max(nums) – min(nums)
return total/(len(nums)-2)
Just 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
Armani,
Your is a much better solution – no need at all for the nested loops, and no need for list indexes.
I did it slightly differently:
“`
def sum67_loop(nums):
“””
A basic loop method. the key is to keep a flag
set to know whether a 6 has been encountered
“””
total = 0
is6 = False
for num in nums:
if num == 6 or is6:
is6 = True
else:
total += num
if num == 7:
is6 = False
return total
“`
But the logic is the same.
For sum67, here is my solution:
def sum67(nums):
sum = 0
ignore = False
for i in nums:
if i == 6:
ignore = True
if not ignore:
sum += i
if i == 7:
ignore = False
return sum
Your code for sum13 will fail in the following test case: nums = [13, 13, 1]
This should return 0, your code returns 1. CodingBat is missing that test case :-/
Here’s a link to my solution:
https://gist.github.com/WorryingWonton/777fa30f854d23cae198af12cacd3ea9
a little comment in your solution of sum13 exercise.
Your solution works for most of the cases but not for all. I may not know a lot of python just learning it, but this solution works better:
def sum13(nums):
if len(nums)==0:
return 0
total=0
for i in range(len(nums)):
total+=nums[i]
if nums[i]==13:
total-=nums[i]
if i+1<len(nums):
if nums[i+1]==13:
continue
total-=nums[i+1]
return total
My solution passes all test cases on the CodingBat website, so I view it as correct. What test does your code pass which mine does not? Your approach is suboptimal because you add and then remove elements. A more elegant approach would be to simply skip over every element after a ’13’. Here’s how you’d do it:
def sum13(nums): total = 0 i = 0 while i < len(nums): val = nums[i] if val == 13: i += 2 else: total += val i += 1 return total