Python. <Remove Element> in a list - python-2.7

Question

Title:
172. Remove Element
Description:
Given an array and a value, remove all occurrences of that value in place and return the new length.
The order of elements can be changed, and the elements after the new length don't matter.
My Answer:
On LintCode
def removeElement(self, A, elem):
# write your code here
if A == []:
return A
if elem in A:
sortedA = sorted(A)
li = []
for i in xrange(len(sortedA)):
if sortedA[i] == elem:
li += [i]
newLength = sortedA[:min(li)] + sortedA[max(li)+1:]
return newLength
else:
return A
On my mac
A = [0,4,4,0,0,2,4,4]
elem = 4
def sss(A, elem):
if A == []:
return A
if elem in A:
print A
sortedA = sorted(A)
print sortedA
li = []
for i in xrange(len(sortedA)):
# print type(i)
if sortedA[i] == elem:
li += [i]
print li
newLength = sortedA[:min(li)] + sortedA[max(li)+1:]
print newLength
return newLength
else:
return A
print sss(A, elem)
This answer On my mac work's good, but On LintCode doesn't accept.

Answer 1

Why not use a list comprehension to filter out the unwanted elements?
class Solution:
def removeElement(self, A, elem):
A[:] = [item for item in A if item != elem]
return len(A)
The key here is the slice notation on the left hand side of the assignment. This makes it an "in place" operation, so the original list A is mutated, rather than a copy being made.
Example usage:
>>> l = [1, 2, 3, 4, 4, 5, 4, 10, 11, 4]
>>> len(l)
10
>>> Solution().removeElement(l, 4)
6
>>> l
[1, 2, 3, 5, 10, 11]
>>> len(l)
6

Answer 2

Although this is an old post, yet would like to add the different approaches I tried. Time Complexity is O(n) in all cases.
Method I: Override the elements if they are not val (Logical Approach)
def removeElement(self, nums: List[int], val: int) -> int:
i : int = 0 # This will be an index pointer & will provide the sixe at the end
for j in range(len(nums)):
if nums[j] != val: # if the values are not same
nums[i] = nums[j] # Override the value at location i
i+=1
return i
Method II: Override the val with last array element (Logical Approach)
def removeElement(self, nums: List[int], val: int) -> int:
i: int = 0
n: int = len(nums)
while(i<n):
if(nums[i] == val):
# Replace the element with last array element & reduce the array size by 1
nums[i] = nums[n-1]
n -=1
else:
i +=1
return n
Method III -- Improved form of Method 2 (Python way)
def removeElement(self, nums: List[int], val: int) -> int:
i : int = 0
n : int = len(nums)
while i< n :
print(i, nums, n)
if val in nums[:n]:
i = nums.index(val,0,n)
nums[i] = nums[n-1]
n -=1
else:
break
return n
Method IV -- Python way
def removeElement(self, nums: List[int], val: int) -> int:
nums[:] = [x for x in nums if x !=val]
return len(nums)
All of them have an average runtime of around 36ms and memory utilization around 13.8MB.