old_contents: """Leetcode 350. Intersection of Two Arrays II
  Easy
  
  URL: https://leetcode.com/problems/intersection-of-two-arrays-ii/
  
  Given two arrays, write a function to compute their intersection.
  
  Example 1:
  Input: nums1 = [1,2,2,1], nums2 = [2,2]
  Output: [2,2]
  
  Example 2:
  Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
  Output: [4,9]
  
  Note:
  - Each element in the result should appear as many times as it shows in both arrays.
  - The result can be in any order.
  
  Follow up:
  - What if the given array is already sorted? How would you optimize your algorithm?
  - What if nums1's size is small compared to nums2's size? Which algorithm is better?
  - What if elements of nums2 are stored on disk, and the memory is limited such that
    you cannot load all elements into the memory at once?
  """
  
  class Solution(object):
      def intersect(self, nums1, nums2):
          """
          :type nums1: List[int]
          :type nums2: List[int]
          :rtype: List[int]
          """
          pass
  
  
  def main():
      pass
  
  
  if __name__ == '__main__':
      main()
  
new_contents: """Leetcode 350. Intersection of Two Arrays II
  Easy
  
  URL: https://leetcode.com/problems/intersection-of-two-arrays-ii/
  
  Given two arrays, write a function to compute their intersection.
  
  Example 1:
  Input: nums1 = [1,2,2,1], nums2 = [2,2]
  Output: [2,2]
  
  Example 2:
  Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
  Output: [4,9]
  
  Note:
  - Each element in the result should appear as many times as it shows in both arrays.
  - The result can be in any order.
  
  Follow up:
  - What if the given array is already sorted? How would you optimize your algorithm?
  - What if nums1's size is small compared to nums2's size? Which algorithm is better?
  - What if elements of nums2 are stored on disk, and the memory is limited such that
    you cannot load all elements into the memory at once?
  """
  
  class SolutionNaiveIter(object):
      def intersect(self, nums1, nums2):
          """
          :type nums1: List[int]
          :type nums2: List[int]
          :rtype: List[int]
  
          Time complexity: O(n1*n2), where ni is the length of numsi.
          Space complexity: O(1).
          """
          if len(nums1) > len(nums2):
              nums1, nums2 = nums2, nums1
  
          intersect = []
          for n1 in nums1:
              for i2, n2 in enumerate(nums2):
                  if n1 == n2:
                      intersect.append(n1)
                      nums2[i2] = None
                      break
  
          return intersect
  
  def main():
      # Output: [2,2]
      nums1 = [1,2,2,1]
      nums2 = [2,2]
      print SolutionNaiveIter().intersect(nums1, nums2)
  
      # Output: [4,9]
      nums1 = [4,9,5]
      nums2 = [9,4,9,8,4]
      print SolutionNaiveIter().intersect(nums1, nums2)
  
  
  if __name__ == '__main__':
      main()
  
subject: Complete naive iter sol w/ time/space complexity