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Sample Coding Problem

This question was recently asked by Google.
Given two sorted linked lists, merge them in order.
def merge(self, list1, list2):
  # Fill this in.

# Definition for a Linked List.
class ListNode(object):
  def __init__(self, x):
    self.val = x
    self.next = None
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 singly-linked list.
class ListNode(object):
  def __init__(self, x):
    self.val = x
    self.next = None

class Solution:
  def mergeTwoLists(self, l1, l2):
    if l1 is None:
      return l2
    elif l2 is None:
      return l1
    elif l1.val < l2.val:
      l1.next = self.mergeTwoLists(l1.next, l2)
      return l1
    else:
      l2.next = self.mergeTwoLists(l1, l2.next)
      return l2

  def mergeTwoListsIterative(self, l1, l2):
    current = None
    root = None
    while True:
      if l1 is None:
        nextNode = l2
      elif l2 is None:
        nextNode = l1
      elif l1.val < l2.val:
        nextNode = l1
      else:
        nextNode = l2

      if nextNode == l1:
        l1 = l1.next if l1 else None
      if nextNode == l2:
        l2 = l2.next if l2 else None

      if nextNode is None:
        break
      if not current:
        current = nextNode
        root = current
      else:
        current.next = nextNode
      current = nextNode
    return root

# Test program
a = ListNode(1)
a.next = ListNode(3)
a.next.next = ListNode(5)

b = ListNode(2)
b.next = ListNode(4)
b.next.next = ListNode(6)

c = Solution().mergeTwoListsIterative(a, b)
while c:
  print c.val
  c = c.next

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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Have a great day!

TechLead and Joma

Meet your instructors

TechLead

"TechLead" is Patrick Shyu - ex-Google tech lead, ex-Facebook Staff Software Engineer, multi-millionaire app entrepreneur, digital nomad, and software engineer. He has conducted over 100 interviews at Google.

Joma

"Joma" is Jonathan Ma - he has done internships and full-time at Citadel, Linkedin, Facebook and Microsoft. He has worked as a data scientist, software engineer and PM. He is currently working as a software engineer in FANG.

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