658. Find K Closest Elements - for loop 'comp' comparison variable not updating as expected - python-3.x

Question

So I am working on problem 658 'Find K Closest Elements'(https://leetcode.com/problems/find-k-closest-elements/), which asks to return a list from the given list, 'arr'. This return list will be the length of 'k' and will contain the 'k' # of values closest to the given 'x' value. I've created all the base cases and constraints, and am now stuck on the comparison part below:
I've created an empty list, 'a'. While the length of 'a' is not 'k', the function will go through the list and compare abs(i - x) < abs(comp - x), 'comp' starting at 'arr[0]' and updating to 'i' if the comparison is true. The problem is that the comparison is not working correctly. Here is an example case I'm trying to figure out:
arr = [1,1,1,10,10,10], k = 4, x = 9
Below is the portion of the code I am focusing on:
a = []
comp = arr[0]
iteration = 0
i_index = 0
while len(a) != k:
for i in arr:
comp_1 = abs(i - x)
comp_2 = abs(comp - x)
if comp_1 < comp_2:
comp == i
print(f"comp: {comp}")
arr.pop(arr.index(comp))
a.append(comp)
return a
I am including the entirety of the code just in case below:
def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
# Constraints
if k < 1 or k > len(arr):
return "k must be greater than 0 and less than the arr length"
if len(arr) < 1 or len(arr) > 10**4:
return "arr length must be greater than 0 and less than 10^4"
if x > 10**4:
return "x must be less than 10^4"
if sorted(arr) != arr:
return "arr must be sorted"
for i in arr:
if i < -10**4:
return "arr item cannot be less than -10^4"
#Variables 1
begin = arr[:k]
end = arr[-k:]
# Base cases
if len(arr) == k:
return arr
if x < arr[0]:
return begin
elif x > arr[-1]:
return end
try:
x_index = arr.index(x)
half_k = int(k/2)
#if k == x and x_index != None:
# return [x]
# Captures all other lists that begin at arr[0] or end at arr[-1]
if x_index - half_k < 0:
return begin
elif x_index + half_k > len(arr) - 1:
return end
# Create list out of interior of arr if necessary
else:
return arr[x_index - half_k : x_index + half_k]
# Means x is not in arr
except ValueError:
a = []
comp = arr[0]
iteration = 0
i_index = 0
while len(a) != k:
for i in arr:
print(f"{iteration} - {i_index}:")
print(f"i: {i}")
print(f"comp_1: {abs(i - x)}")
print(f"comp_2: {abs(comp - x)}")
comp_1 = abs(i - x)
comp_2 = abs(comp - x)
if comp_1 < comp_2:
comp == i
print(f"comp: {comp}")
i_index += 1
print("\n")
iteration += 1
arr.pop(arr.index(comp))
a.append(comp)
return a

Answer 1

I would take the approach of:
finding the shortest distance from x to the elements of arr.
sorting arr by the distances.
So, this is a good method:
arr = [1,1,1,10,10,10]
k = 4
x = 9
distances = [abs(x - n) for n in arr]
Z = [a for _,a in sorted(zip(distances,arr))]
print(Z[:k])
result
[10, 10, 10, 1]