Landing a job at Facebook is a dream for many developers around the globe. Facebook is one of the top tech companies in the world, with a workforce of over 52,000 strong. Facebook is known for its growth-based company culture, fast promotion tracks, excellent benefits, and top salaries that few companies can match.
But competition is fierce, and with a swell of new hires, Facebook is on the lookout for the top candidates. Facebook focuses on your cultural fit, generalist knowledge, ability to build within constraints, and expert coding skills.
To help you prepare, today I will walk through everything you need to crack an Facebook interview, including coding questions and a step-by-step preparation guide.
Today we will go over:
To land a software engineering job at Facebook, you need to know what lies ahead. The more prepared you are, the more confident you will be. So, let’s break it down.
For a deeper dive into Facebook’s interview process, check out Coding Interviews’s free Facebook Coding Interview Guide.
Coding Questions: Facebook interview questions focus on generalist knowledge on algorithms, data structures, and time complexity. They also test on architecture and system design (even entry level).
Hiring Levels: Facebook normally hires at level E3 for entry level software roles with E9 behind the height of levels. E5 is considered an entry-level manager role.
Hiring Teams: Central hires for Oculus, Facebook Groups, and WhatsApp.
Programming languages: Facebook prefers most standard languages, including Java, C++, Python, Ruby, and Perl.
What’s different about Facebook interviews?
System design interview:
- At Facebook, you can expect these questions no matter what level you are interviewing for.
Structured interviewing:
- Facebook will pair you with interviewers who have either held the position you’re interviewing for or with individuals who work directly with the position you’re interviewing for.
Core values and your behavioral interview:
- Facebook interviewers will also evaluate your ability to embody their five core values: Move Fast, Be Bold, Focus on Impact, Be Open, and Build Social Value.
In this section, we’ll take a deep dive into the top 40 coding interview questions. We will discuss the answers and runtime complexities for the 15 questions you’re bound to see in an interview followed by the definitive list of 25 questions you’ll likely encounter.
As you can see to the right, Facebook focuses mostly on arrays and strings. These are essential skills to master.
Each question will be solved in Python 3. To see these solutions in C++, Ruby, Java, and JavaScript, visit here.
Given an integer array, move all elements that are 0 to the left while maintaining the order of other elements in the array. The array has to be modified in-place. Try it yourself before reviewing the solution and explanation.
def move_zeros_to_left(A):#TODO: Write - Your - Codepass
Runtime complexity: Linear, $O(n)$
Memory Complexity: Constant, $O(1)$
Keep two markers: read_index
and write_index
and point them to the end of the array. Let’s take a look at an overview of the algorithm.
While moving read_index
towards the start of the array:
read_index
points to 0
, skip.read_index
points to a non-zero value, write the value at read_index
to write_index
and decrement write_index
.write_index
and to the current position of write_index
as well.You are given an array (list) of interval pairs as input where each interval has a start and end timestamp. The input array is sorted by starting timestamps. You are required to merge overlapping intervals and return a new output array.
Consider the input array below. Intervals (1, 5), (3, 7), (4, 6), (6, 8) are overlapping so they should be merged to one big interval (1, 8). Similarly, intervals (10, 12) and (12, 15) are also overlapping and should be merged to (10, 15).
Try it yourself before reviewing the solution and explanation.
class Pair:def __init__(self, first, second):self.first = firstself.second = seconddef merge_intervals(v):# write your code hereresult = []return result
Runtime complexity: Linear, $O(n)$
Memory Complexity: Linear, $O(n)$
This problem can be solved in a simple linear scan algorithm. We know that input is sorted by starting timestamps. Here is the approach we are following:
Convert a binary tree to a doubly linked list so that the order of the doubly linked list is the same as an in-order traversal of the binary tree.
After conversion, the left pointer of the node should be pointing to the previous node in the doubly linked list, and the right pointer should be pointing to the next node in the doubly linked list. Try it yourself before reviewing the solution and explanation.
from binary_tree import *from binary_tree_node import *def convert_to_linked_list(root):# TODO: Write - Your - Codereturn root
Runtime complexity: Linear, $O(n)$
Memory Complexity: Linear, $O(h)$.
Recursive solution has $O(h)$ memory complexity as it will consume memory on the stack up to the height of binary tree
h
. It will be $O(log n)$ for balanced trees and in the worst case can be $O(n)$.
In an in-order traversal, first the left sub-tree is traversed, then the root is visited, and finally the right sub-tree is traversed.
One simple way of solving this problem is to start with an empty doubly linked list. While doing the in-order traversal of the binary tree, keep inserting each element output into the doubly linked list.
But, if we look at the question carefully, the interviewer wants us to convert the binary tree to a doubly linked list in-place i.e. we should not create new nodes for the doubly linked list.
This problem can be solved recursively using a divide and conquer approach. Below is the algorithm specified.
Given the root of a binary tree, display the node values at each level. Node values for all levels should be displayed on separate lines. Let’s take a look at the below binary tree.
Level order traversal for this tree should look like:
# Import required functionsfrom collections import dequefrom binary_tree import *from binary_tree_node import *def level_order_traversal(root):result = ""# TODO: Write - Your - Codeprint(result, end="")
Runtime complexity: Linear, $O(n)$
Memory Complexity: Linear, $O(n)$
Here, you are using two queues: current_queue
and next_queue
. You push the nodes in both queues alternately based on the current level number.
You’ll dequeue nodes from the current_queue
, print the node’s data, and enqueue the node’s children to the next_queue
.
Once the current_queue
becomes empty, you have processed all nodes for the current level_number
. To indicate the new level, print a line break (\n
), swap the two queues, and continue with the above-mentioned logic.
After printing the leaf nodes from the current_queue
, swap current_queue
and next_queue
. Since the current_queue
would be empty, you can terminate the loop.
Reverse the order of words in a given sentence (an array of characters). Take the “Hello World” string for example:
def reverse_words(sentence): # sentence here is an array of characters#TODO: Write - Your - Codereturn sentence
Here is how the solution works:
For more on string reversal, read my article Best practices for reversing a string in JavaScript, C++, and Python.
You are given a dictionary of words and a large input string. You have to find out whether the input string can be completely segmented into the words of a given dictionary. The following example elaborates on the problem further.
def can_segment_string(s, dictionary):#TODO: Write - Your - Codereturn False
Runtime complexity: Exponential, $O(2^{n})$, if we only use recursion. With memoization, the runtime complexity of this solution can be improved to be polynomial, $O(n^{2})$.
Memory Complexity: Polynomial, $O(n^{2})$
You can solve this problem by segmenting the large string at each possible position to see if the string can be completely segmented to words in the dictionary. If you write the algorithm in steps it will be as follows:
n = length of input string
for i = 0 to n - 1
first_word = substring (input string from index [0, i] )
second_word = substring (input string from index [i + 1, n - 1] )
if dictionary has first_word
if second_word is in dictionary OR second_word is of zero length, then return true
recursively call this method with second_word as input and return true if it can be segmented
The algorithm will compute two strings from scratch in each iteration of the loop. Worst case scenario, there would be a recursive call of the second_word
each time. This shoots the time complexity up to $2^{n}$.
You can see that you may be computing the same substring multiple times, even if it doesn’t exist in the dictionary. This redundancy can be fixed by memoization, where you remember which substrings have already been solved.
To achieve memoization, you can store the second
string in a new set each time. This will reduce both time and memory complexities.
Given a list of daily stock prices (integers for simplicity), return the buy and sell prices for making the maximum profit.
We need to maximize the single buy/sell profit. If we can’t make any profit, we’ll try to minimize the loss. For the below examples, buy (orange) and sell (green) prices for making a maximum profit are highlighted.
def find_buy_sell_stock_prices(array):#TODO: Write - Your - Codereturn -1, -1 #Return a tuple with (high, low) price values
Runtime complexity: Linear, $O(n)$
Memory Complexity: Constant, $O(1)$
The values in the array represent the cost of a stock each day. As we can buy
and sell
the stock only once, we need to find the best buy
and sell
prices for which our profit is maximized (or loss is minimized) over a given span of time.
A naive solution, with runtime complexity of $O(n^{2})$, is to find the maximum gain between each element and its succeeding elements.
There is a tricky linear solution to this problem that requires maintaining current_buy_price
(which is the smallest number seen so far), current_profit
, and global_profit
as we iterate through the entire array of stock prices.
At each iteration, we will compare the current_profit
with the global_profit
and update the global_profit
accordingly.
The basic algorithm is as follows:
current buy = stock_prices[0]
global sell = stock_prices[1]
global profit = global sell - current buy
for i = 1 to stock_prices.length:
current profit = stock_prices[i] - current buy
if current profit is greater than global profit
then update global profit to current profit and update global sell to stock_prices[i]
if stock_prices[i] is less than current buy
then update current buy to stock_prices[i]
return global profit and global sell
Given a double, x
, and an integer, n
, write a function to calculate x
raised to the power n
. For example:
power (2, 5) = 32
power (3, 4) = 81
power (1.5, 3) = 3.375
power (2, -2) = 0.25
def power(x, n):#TODO: Write - Your - Codereturn x
Runtime complexity: Logarithmic, $O(log n)$
Memory Complexity: Logarithmic, $O(log n)$
A simple algorithm for this problem is to multiply x
by n
times. The time complexity of this algorithm would be $O(n)$. We can use the divide and conquer approach to solve this problem more efficiently.
In the dividing step, we keep dividing n
by 2
recursively until we reach the base case i.e. n == 1
In the combining step, we get the result, r
, of the sub-problem and compute the result of the current problem using the two rules below:
We are given a set of integers and we have to find all the possible subsets of this set of integers.
def get_all_subsets(v, sets):#TODO: Write - Your - Codereturn sets
Runtime complexity: Exponential, $O(2^{n} * n)$, where $n$ is the number of integers in the given set
Memory Complexity: Constant, $O(2^{n} * n)$
There are several ways to solve this problem. We will discuss the one that is neat and easier to understand. We know that for a set of ‘n’ elements there are $2^{n}$ subsets. For example, a set with 3 elements will have 8 subsets. Here is the algorithm we will use:
n = size of given integer set
subsets_count = 2^n
for i = 0 to subsets_count
form a subset using the value of 'i' as following:
bits in number 'i' represent index of elements to choose from original set,
if a specific bit is 1 choose that number from original set and add it to current subset,
e.g. if i = 6 i.e 110 in binary means that 1st and 2nd elements in original array need to be picked.
add current subset to list of all subsets
Note that the ordering of bits for picking integers from the set does not matter; picking integers from left to right would produce the same output as picking integers from right to left.
Given the root node of a directed graph, clone this graph by creating its deep copy so that the cloned graph has the same vertices and edges as the original graph.
Let’s look at the below graphs as an example. If the input graph is G = (V, E) where V is set of vertices and E is set of edges, then the output graph (cloned graph) G’ = (V’, E’) such that V = V’ and E = E’.
Note: We are assuming that all vertices are reachable from the root vertex. i.e. we have a connected graph.
from directed_graph import *def clone(graph):# Hashmap to keep record of visited nodesnodes_completed = {}# Creating new graphclone_graph = DirectedGraph()# TODO: Write - Your - Code# Return deep copied graphreturn clone_graph
Runtime complexity: Linear $O(n)$
Memory Complexity: Logarithmic, $O(n)$
We use depth-first traversal and create a copy of each node while traversing the graph. To avoid getting stuck in cycles, we’ll use a hashtable to store each completed node and will not revisit nodes that exist in the hashtable.
The hashtable key will be a node in the original graph, and its value will be the corresponding node in the cloned graph.
Serialize a binary tree to a file and then deserialize it back to a tree so that the original and the deserialized trees are identical.
There is no restriction regarding the format of a serialized stream, therefore you can serialize it in any efficient format. However, after deserializing the tree from the stream, it should be exactly like the original tree. Consider the below tree as the input tree.
from binary_tree import *from binary_tree_node import *# Function to serialize tree into list of integer.def serialize(root):#TODO: Write - Your - Codereturn None# Function to deserialize integer list into a binary tree.def deserialize(stream):#TODO: Write - Your - Codereturn None
Runtime complexity: Linear $O(n)$
Memory Complexity: Logarithmic, $O(logn)$
There can be multiple approaches to serialize and deserialize the tree. One approach is to perform a depth-first traversal and serialize individual nodes to the stream. We’ll use a pre-order traversal here. We’ll also serialize some markers to represent a null pointer to help deserialize the tree.
Consider the below binary tree as an example. Markers (M*) have been added in this tree to represent null nodes. The number with each marker i.e. 1 in M1, 2 in M2, merely represents the relative position of a marker in the stream.
The serialized tree (pre-order traversal) from the above example would look like the below list.
When deserializing the tree we’ll again use the pre-order traversal and create a new node for every non-marker node. Encountering a marker indicates that it was a null node.
Given a sorted array of integers, return the low and high index of the given key. You must return -1 if the indexes are not found.
The array length can be in the millions with many duplicates.
In the following example, according to the the key
, the low
and high
indices would be:
key
: 1, low
= 0 and high
= 0key
: 2, low
= 1 and high
= 1key
: 5, low
= 2 and high
= 9key
: 20, low
= 10 and high
= 10For the testing of your code, the input array will be:
1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 6
def find_low_index(arr, key):#TODO: Write - Your - Codereturn -2def find_high_index(arr, key):#TODO: Write - Your - Codereturn -2
Runtime complexity: Logarithmic $O(log n)$
Memory Complexity: Constant, $O(1)$
Linearly scanning the sorted array for low
and high
indices are highly inefficient since our array size can be in millions. Instead, we will use a slightly modified binary search to find the low
and high
indices of a given key.
We need to do binary search twice:
low
index.high
index.Let’s look at the algorithm for finding the low
index:
low
and high
indices and calculate the mid
index.mid
index is less than the key
, low
becomes mid + 1
(to move towards the start of range).key
, the high
becomes mid - 1
. Index at low
remains the same.low
is greater than high
, low
would be pointing to the first occurrence of the key
.low
does not match the key
, return -1
.Similarly, we can find the high
index by slightly modifying the above condition:
low
index to mid + 1
when the element at mid
index is less than or equal to the key
.high
index to mid - 1
when the element at mid
is greater than the key
.Search for a given number in a sorted array, with unique elements, that has been rotated by some arbitrary number. Return -1
if the number does not exist.
Assume that the array does not contain duplicates
Below is an original array before rotation.
After performing rotation on this array 6 times it changes to:
The task is to find a given number in this array.
def binary_search_rotated(arr, key):#TODO: Write - Your - Codereturn -1
Runtime complexity: Logarithmic, $O(log n)$
Memory Complexity: Logarithmic, $O(log n)$
The solution is essentially a binary search but with some modifications. If we look at the array in the example closely, we notice that at least one half of the array is always sorted. We can use this property to our advantage.
If the number n
lies within the sorted half of the array, then our problem is a basic binary search. Otherwise, discard the sorted half and keep examining the unsorted half. Since we are partitioning the array in half at each step, this gives us $O(log n)$ runtime complexity.
StrStr
(string search)Now that you have a sense of what to expect from an interview and know what kinds of questions to expect, let’s learn some preparation strategies based on Facebook’s unique interview process.
The first thing you should do is update your resume to be metrics/deliverables driven. It’s also a good idea to show how the work you’ve done can translate into their five core values: Move fast, Be bold, Focus on impact, Be open, and Build social value.
I recommend at least three months of self-study to be successful. This includes choosing a programming language, reviewing the basics, and studying algorithms, data structures, system design, object-oriented programming, and more.
It’s important to practice coding using different tools:
For a robust, 12-week interview guide, check out our article, the 3 Month Coding Interview Preparation Bootcamp
The design interview usually doesn’t involve any coding, so you’ll need to learn how to answer these questions. This will be done on a whiteboard during the interview, so practice your designs by hand. Study up on system design and product design.
The best way to master system design questions is not by memorizing answers but by learning the anatomy of a system design question. You need to train yourself to think from the ground up while also considering scaling and requirements.
Pro tip: If you want to stand out in the system design interview, you’ll need to discuss how Machine Learning can be implemented in your design.
Facebook wants next-gen engineers, and they focus heavily on artificial intelligence. Consider brushing up on ML concepts and ML system design principles.
Once you get the basics down and progress through the interview prep roadmap, master the best practices.
When you practice, learn how to articulate your process out loud. Facebook cares a lot about how you think. When you code, explain your thought process as if another person were in the room.
You also want to start timing yourself to learn how to manage your time effectively. It’s always better to take time planning your answer than to just jump it with brute force.
Facebook cares that you fit with their company, so you need to be prepared to talk about yourself. For each of Facebook’s values, brainstorm how you fit and why these values matter to you.
You should also think about your 2 to 4 year career aspirations, interests, and strengths as an engineer, as they will likely come up in the interview.
To learn how to prepare, check out my article Behavioral Interviews: how to prepare and ace interview questions
Facebook values self-starters, so it’s important that you come prepared with questions for your interviewers. You’ll have time during every interview to ask your own questions. This is also an opportunity to determine if Facebook is a good fit for your lifestyle and needs.
Cracking the Facebook coding interview comes down to the time you spend preparing, such as practicing coding questions, studying behavioral interviews, and understanding Facebook’s company culture.
There is no golden ticket, but more preparation will surely make you a more confident and desirable candidate. The essential resources below will help you prepare and build confidence for Facebook interviews.
Keep learning and studying!
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