old_contents: # Remove Element https://oj.leetcode.com/problems/remove-element/
  # Given an array and a value, remove all instances of that value in place and return the new length.
  # The order of elements can be changed. It doesn't matter what you leave beyond the new length.
  
  #Arrays
  
  # Xilin SUN
  # Jan 8 2015
  
  # Don't forget to return 0 if A == [].
  
  class Solution:
      # @param    A       a list of integers
      # @param    elem    an integer, value need to be removed
      # @return an integer
      def removeElement(self, A, elem):
          if len(A) == 0:
      	    return 0
          else:
              k=0
              for i in range(0, len(A)):
                  if A[i] != elem:
                      A[k] = A[i]
                      k += 1
              return k
  
new_contents: # Remove Element https://oj.leetcode.com/problems/remove-element/
  # Given an array and a value, remove all instances of that value in place and return the new length.
  # The order of elements can be changed. It doesn't matter what you leave beyond the new length.
  
  #Arrays
  
  # Xilin SUN
  # Jan 8 2015
  
  # Don't forget to return 0 if A == [].
  
  class Solution:
      # @param    A       a list of integers
      # @param    elem    an integer, value need to be removed
      # @return an integer
      def removeElement(self, A, elem):
          if len(A) == 0:
      	    return 0
          else:
              k=0
              for i in range(0, len(A)):
                  if A[i] != elem:
                      if i!= k:
                          A[k] = A[i]
                      k += 1
              return k
  
subject: Add one more if to speed up