def getX(x, nums):
write your code here
if x > len(nums) or len(nums) == 0:
return 0
return sorted(nums)[x1]
print(getX(2, [6, 3, 1, 5]))
tried to do 1 line
def getX(x, nums):
return sorted(nums)[x  1] if x in range(1, len(nums) + 1) else 0
print(getX(3, [1, 2, 4, 3, 5]))
def getX(x, nums):
sorted_list = sorted(nums)
if len(sorted_list) >= x and len(sorted_list) != 0:
return sorted_list[x1]
else:
return 0
print(getX(4, [5, 10, 3 , 3, 7, 9]))
Almost every solution here uses sort which is average O(nlogn).
I don’t see anyone posting the better solution (and the one the problem hints at) that is O(n) by keeping track of the x smallest values and doing one pass of the nums. For example:
As it says in the comment there are better ways of managing the “smallest” items, a max heap being probably the best. Note you can’t use a set in case of nonunique numbers. Could use a dict of num>count though.
Using a minheap is an optimized way to be close to O(n). After heapfying the array, we should extract the top x items to find the xth number
from heapq import heapify, heappop
def getX(x, nums):
if not nums or x > len(nums):
return 0
heapify(nums)
for i in range(x):
xth = heappop(nums)
return xth
print(getX(2, [6, 3, 1, 5]))
def getX(x, nums):
nums.sort()
print(nums)
if x > len(nums) or len(nums) == 0:
return 0
else:
m = nums[x1]
return m
print(getX(2, [6, 3, 1, 5,]))
def getX(x, nums):
nums = sorted(nums,)
try:
return nums[x1]
except IndexError:
return 0
print(getX(4, [5, 10, 3, 3,7,9]))
This was my solution
how do you post it so that you can see and run the code???
def getX(x, nums):
nums.sort()
if x > len(nums) or nums == [] or x == 0 or x < len(nums) or type(x)==float:
return 0
else:
if x > 0:
return nums[x1]
else:
return nums[x]
# write your code here
print(getX(2.4, [6, 3, 1, 5]))
Paste in your code. Select it all at once and click the </>
button in the tool tray of the post editor. That will display it in code sample format.
As for running it, that can be done, but I rather doubt anybody will actually run it so it is just a drain on the page performance of the topic.
Thank you that, looks much better
Hi all,
I just completed this challenge with the following code. I think it’s a good solution, but I’m open to any recommendations on making this code more efficient. Thanks!!
def getX(x, nums):
nums_sorted = sorted(nums)
print(nums_sorted)
if x <= 0:
return 0
elif x <= len(nums_sorted):
return nums_sorted[x1]
elif nums_sorted == None:
return 0
else:
return 0
print(getX(2, [6, 3, 1, 5]))```
Hi there, generally a good code. Some minor suggestions though:
 In the challenge description it is said that function should return 0 if a list of numbers is empty. I don’t think an empty list is equal to None (line 8 in your code), the line shall look like
elif nums_sorted == [ ]
It is not a big deal as you have a “safe net” of else statement which will return 0 for an empty list, but it means that lines 8 and 9 can be omitted alltogether. Please others correct me if I am wrong.

In the case of negative x, that is probably a way of interpretation, but i thought of it as an opposite action to positive x: if positive x will return Xth smalles, than negative x will return Xth largest number

As an additional challenge, you can add function behaviour if x was by mistake not an integer or nums not a list
Thank you for the input and the recommended challenge!
I updated my code, but for some reason, the logic doesn’t flow the way it should:
def getX(x, nums):
nums_sorted = sorted(nums)
print(nums_sorted)
if x == type(str):
print("Input not an integer")
elif x <= 0:
return 0
elif x <= len(nums_sorted):
return nums_sorted[x1]
else:
return 0
print(getX("a", [6, 3, 1, 5]))
def getX(x, nums):
sorted_list = sorted(nums)
if (abs(x) > len(sorted_list)) or (bool(nums) is False):
return 0
if abs(x) <= len(sorted_list):
return sorted_list[x  1]
print(getX(0, ))