old_contents: """Leetcode 26. Remove Duplicates from Sorted Array
  Easy
  
  URL: https://leetcode.com/problems/remove-duplicates-from-sorted-array/
  
  Given a sorted array nums, remove the duplicates in-place such that each element
  appear only once and return the new length.
  
  Do not allocate extra space for another array, you must do this by modifying the
  input array in-place with O(1) extra memory.
  
  Example 1:
  Given nums = [1,1,2],
  Your function should return length = 2, with the first two elements of nums 
  being 1 and 2 respectively.
  It doesn't matter what you leave beyond the returned length.
  
  Example 2:
  Given nums = [0,0,1,1,1,2,2,3,3,4],
  Your function should return length = 5, with the first five elements of nums
  being modified to 0, 1, 2, 3, and 4 respectively.
  It doesn't matter what values are set beyond the returned length.
  
  Clarification:
  Confused why the returned value is an integer but your answer is an array?
  Note that the input array is passed in by reference, which means modification
  to the input array will be known to the caller as well.
  
  Internally you can think of this:
  
  // nums is passed in by reference. (i.e., without making a copy)
  int len = removeDuplicates(nums);
  
  // any modification to nums in your function would be known by the caller.
  // using the length returned by your function, it prints the first len elements.
  for (int i = 0; i < len; i++) {
      print(nums[i]);
  }
  """
  
  class Solution(object):
      def removeDuplicates(self, nums):
          """
          :type nums: List[int]
          :rtype: int
          """
          # Two pointer method: tail for updated array & i.
          tail = 0
          for i in range(1, len(nums)):
              if nums[i] != nums[tail]:
                  # Increment tail and replace new tail by element i.
                  tail += 1
                  nums[tail] = nums[i]
  
          return tail + 1
  
  
  def main():
      nums = [1, 1, 2]
      print Solution().removeDuplicates(nums)
  
      nums = [0, 0, 1, 1, 1, 2, 2, 3, 3, 4]
      print Solution().removeDuplicates(nums)
  
  
  if __name__ == '__main__':
      main()
  
new_contents: """Leetcode 26. Remove Duplicates from Sorted Array
  Easy
  
  URL: https://leetcode.com/problems/remove-duplicates-from-sorted-array/
  
  Given a sorted array nums, remove the duplicates in-place such that each element
  appear only once and return the new length.
  
  Do not allocate extra space for another array, you must do this by modifying the
  input array in-place with O(1) extra memory.
  
  Example 1:
  Given nums = [1,1,2],
  Your function should return length = 2, with the first two elements of nums 
  being 1 and 2 respectively.
  It doesn't matter what you leave beyond the returned length.
  
  Example 2:
  Given nums = [0,0,1,1,1,2,2,3,3,4],
  Your function should return length = 5, with the first five elements of nums
  being modified to 0, 1, 2, 3, and 4 respectively.
  It doesn't matter what values are set beyond the returned length.
  
  Clarification:
  Confused why the returned value is an integer but your answer is an array?
  Note that the input array is passed in by reference, which means modification
  to the input array will be known to the caller as well.
  
  Internally you can think of this:
  
  // nums is passed in by reference. (i.e., without making a copy)
  int len = removeDuplicates(nums);
  
  // any modification to nums in your function would be known by the caller.
  // using the length returned by your function, it prints the first len elements.
  for (int i = 0; i < len; i++) {
      print(nums[i]);
  }
  """
  
  class Solution(object):
      def removeDuplicates(self, nums):
          """
          :type nums: List[int]
          :rtype: int
  
          Time complexity: O(n), where n is the lenght of nums.
          Space complexity: O(1).
          """
          # Two pointer method: tail for updated array & i.
          tail = 0
          for i in range(1, len(nums)):
              if nums[i] != nums[tail]:
                  # Increment tail and replace new tail by element i.
                  tail += 1
                  nums[tail] = nums[i]
  
          return tail + 1
  
  
  def main():
      nums = [1, 1, 2]
      print Solution().removeDuplicates(nums)
  
      nums = [0, 0, 1, 1, 1, 2, 2, 3, 3, 4]
      print Solution().removeDuplicates(nums)
  
  
  if __name__ == '__main__':
      main()
  
subject: Add time/space complexity