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Sample Coding Problem

This question was recently asked by Google.
Given two sorted linked lists, merge them in order.
# Definition for a linked-list node.
class Node(object):
  def __init__(self, val, next=None):
    self.val = val
    self.next = next

class Solution:
  def mergeTwoLists(self, a, b):
    #
    # Fill this in.
    #

# Test program
# 1 -> 3 ->5
a = Node(1)
a.next = Node(3)
a.next.next = Node(5)

# 2 -> 4 -> 6
b = Node(2)
b.next = Node(4)
b.next.next = Node(6)

c = Solution().mergeTwoLists(a, b)
while c:
  print(c.val)
  c = c.next
# 1 2 3 4 5 6
This can be done recursively and iteratively. See if you can get both solutions.

This problem can be solved recursively or iteratively. We traverse the two linked lists in parallel, advancing both pointers simultaneously. If the first list's value is smaller we advance that one, otherwise we advance the second list. If either list is shorter, then we take values from the longer list.

The time complexity is linear O(n) since both lists are traversed just once. The space complexity of the recursive algorithm is linear O(n), since it builds up a recursive stack that may be as deep as the length of both lists. The space complexity of the iterative solution is constant O(1), since only a few variables are used.

# Definition for a linked-list node.
class Node(object):
  def __init__(self, val, next=None):
    self.val = val
    self.next = next

class Solution:
  def mergeTwoLists(self, a, b):
    if a is None:
      return b
    elif b is None:
      return a
    elif a.val < b.val:
      a.next = self.mergeTwoLists(a.next, b)
      return a
    else:
      b.next = self.mergeTwoLists(a, b.next)
      return b

  def mergeTwoListsIterative(self, a, b):
    node = None
    head = None
    while True:
      if a is None:
        nextNode = b
      elif b is None:
        nextNode = a
      elif a.val < b.val:
        nextNode = a
      else:
        nextNode = b

      if nextNode == a:
        a = a.next if a else None
      if nextNode == b:
        b = b.next if b else None

      if nextNode is None:
        break
      if not node:
        node = nextNode
        head = node
      else:
        node.next = nextNode
      node = nextNode
    return head

# Test program
# 1 -> 3 ->5
a = Node(1)
a.next = Node(3)
a.next.next = Node(5)

# 2 -> 4 -> 6
b = Node(2)
b.next = Node(4)
b.next.next = Node(6)

c = Solution().mergeTwoLists(a, b)
while c:
  print(c.val)
  c = c.next
# 1 2 3 4 5 6

We use the language Python.  Why Python?  It's a fairly simple language to read (and very similar to Javascript, Java, or C).  So, even if you don't know it well you'll still be able to understand it and adapt it to any other language like Javascript, C, PHP, Java, etc., Python is widely used at Google/YouTube, Facebook/Instagram, Netflix, Uber, Dropbox, and many more.  It is a modern language and its conciseness makes it an ideal choice for interviews.  Python is a worthwhile language to add to your resume if you don't know it already, and it is simple enough that you can pick it up as we go along.

Daily, consistent practice is the best way to get good at solving interview programming questions. In this field, it is a lifelong skill whether you are a beginner or senior software engineer.

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TechLead 

Meet your instructor

TechLead

TechLead is an ex-Google/ex-Facebook Tech Lead, multi-millionaire app entrepreneur, digital nomad, Silicon Valley native, and senior software engineer.  He's held roles in full-stack web development and mobile engineering. He has conducted over 100 interviews at Google, and has worked in the tech industry for over a decade from startups to Fortune 500 companies.

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