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Cracking the Coding Interview: Solve 5 Real World Problems

Amanda Fawcett

Mar 04, 2021

17 min read

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Crack the coding interview Understanding how to take on coding problems Netflix Feature: Group Similar Titles (hash maps)Practice and Solution Complexity measures Facebook Feature: Friend Circles (DFS)Practice and Solution Complexity measures Google Calendar Feature: Find Meeting Rooms (heaps)Practice and Solution Complexity Measures Amazon Feature: Products in Price Range (BST)Practice and Solution Time complexity Twitter Feature: Add Likes (strings)Practice and Solution Complexity measures Where to go from here Continue reading about coding interviews

Preparing for coding interviews is no easy task. You need the skills to break down the problem and to deploy the right tools. Educative has always been on the mission to make coding interview prep more accessible for engineers. We’ve learned firsthand that the best way to succeed is not to memorize 1500+ Leetcode problems. Understanding patterns are the key to grokking the coding interview for top tech companies.

That’s why we want to approach interview prep a bit differently today by tackling some real world problems faced by tech companies. Learning how to build real world features (like “how to merge recommendations on Amazon”) is more fun, and it’s much easier to remember what you learned. If you can understand a problem’s underlying pattern, you can apply it to just about any question.

We will dive into how to crack coding interviews, solutions for a few common real world coding problems from FAANG companies and build 5 features. We will offer our solutions in Java and Python.

Crack the coding interview#

Finding the motivation to prepare for the coding interview should be simple. Programmers everywhere are practicing coding problems and learning patterns, and you should too, if you want to compete for your dream job.

While preparing for a coding interview with a top tech company, it is essential to have a strong foundation in data structures, algorithms, dynamic programming, and problem-solving fundamentals.

Review the fundamental data structures, including arrays, linked lists, stacks, queues, trees, and graphs. Practice implementing these data structures and familiarize yourself with their operations and time complexities. Once you are comfortable with the basics, progress to more advanced data structures like hash tables, heaps, and tries.

Practice standard sorting and searching algorithms like quicksort, merge sort, binary search, and linear search. Review how to implement these algorithms and analyze the time and space complexity. Then, depending on the programming language, migrate to more complex algorithms like dynamic programming, graph algorithms, and divide and conquer algorithms. You must show your recruiter that you understand when to use each type of algorithm and how to optimize them for different technical questions.

Dynamic programming solves complex problems by breaking them down into smaller sub-problems and solving them recursively. Practice solving dynamic programming problems such as the knapsack problem and the maximum subarray problem. Identifying and solving these dynamic programming problems is vital to crack the coding interview.

In order to crack the coding interview, programmers must have certain characteristics that are a package of both problem-solving skills and technical skills. Identify the problem type, break it into smaller parts, and develop a plan for solving it. Creative problem-solving is at the core of every developer, and the interviewer wants to see you demonstrate it. Try coding your solutions on a whiteboard or paper to simulate the interview experience.

In addition to mastering these concepts, it is essential to keep up with the latest technologies and relevant trends in the industry. Read tech blogs, follow news outlets, attend meetups and hackathons, and network with other software engineers on sites like GitHub and LinkedIn. Keep your skills sharpened by working on individual projects and contributing to open-source projects.

With these tips and strategies, you can ace your coding interview with a top tech company and land your dream job as a software engineer.

For more on how to crack the coding interview with a top tech company and start collecting job offers, take a look at this article about the behavioral interview questions on the Educative blog.

Understanding how to take on coding problems#

Below are 10 tips and tricks for handling coding problems in your programming interview. Take these steps into each question during your interview preparation and the interview process to guarantee you’re giving yourself the best chance at success.

The first step to solving any coding problem is to understand it completely. Ensure you read the problem statement carefully, ask clarifying questions, and understand the inputs and outputs.
Feel free to ask questions if you need more information. Technical interviewers want to see that you can communicate and ask insightful questions. Especially for the trickiest algorithm problems, don’t handcuff yourself by neglecting to ask clarifying questions.
Don’t shy away from thinking out loud. As you work through the problem, explain your thought process out loud. Thinking out loud helps the interviewer understand your approach and thought process. It also acts as a fail-safe to catch any mistakes or oversights you may make.
Break the problem down into more manageable pieces. Doing this will help you tackle the problem more effectively and stay calm. It is also why understanding the overarching patterns are so helpful when solving coding problems.
Try to devise multiple solutions, even if you only implement some. This practice shows your ability to think creatively with different approaches.
Write code that is the easiest to read and understand. This includes using descriptive variable names, proper formatting, and commenting on your code when necessary.
Make sure to test your code thoroughly before submitting it. Mainly, this refers to testing edge cases and ensuring your code handles errors.
Be aware of your time constraints. Time management is absolutely vital in any coding interview. Make sure to pace yourself and spend only a little time on any one part of a coding problem. Solving a coding problem yourself is much different than solving it in front of an audience and with a time restraint. Take yourself through mock interviews in order to better prepare yourself for the actual interview.
Practice more than you think you should. Although it may appear straightforward, this learning strategy is excellent for learning anything, not just coding problems. The more you practice coding problems, the more comfortable and confident you’ll become in a coding interview. While practicing, you must focus on practicing the coding patterns behind each question instead of trying to memorize over a thousand unique coding problems.
Try to stay calm and focused during the coding interview. Take deep breaths and stay hydrated. If you’re feeling especially nervous, remember that it’s more than okay to make mistakes or not know an answer. The interviewer is more interested in your problem-solving skills and critical thought process than in getting a perfectly optimized solution.

Now, let’s start with a tutorial on five programming interview questions you’ll likely see in interviews from big tech companies!

This tutorial at a glance:

Netflix Feature: Group Similar Titles (hash maps)
Facebook Feature: Friend Circles (DFS)
Google Calendar Feature: Find Meeting Rooms (heaps)
Amazon Feature: Products in Price Range (BST)
Twitter Feature: Add Likes (strings)
Where to go from here

Netflix Feature: Group Similar Titles (hash maps)#

Netflix is one of the biggest video streaming platforms out there. The Netflix engineering team is always looking for new ways to display content. For this first problem, imagine you’re a developer on these teams.

Task: Our task here is to improve search results by enabling users to see relevant search results without being hindered by typos, which we are calling the “Group Similar Titles” feature.

First, we need to determine how to individually group any character combination for a given title. Let’s imagine that our library has the following titles: "duel", "dule", "speed", "spede", "deul", "cars". You are tasked to design a functionality so that if a user misspells a word (for example speed as spede), they will still be shown the correct title.

First, we need to split our titles into sets of words so that the words in a set are anagrams. We have three sets: {"duel", "dule", "deul"},{"speed", "spede"}, and {"cars"}. The search results should include all members of the set that the string is found in.

Note: It’s best to pre-compute our sets rather than forming them when a user searches.

The members of each set have the same frequency of each alphabet, so the frequency of each alphabet in words in the same group is equal. For example, in our set {{"speed", "spede"}}, the frequency of the characters are the same in each word: s, p, e, and d.

So, how do we design and implement this functionality? Let’s break it down.

For each title, we need to compute a 26-element vector. Each vector element represents the frequency of an English letter in a title. In Java, we can represent frequency using a string that is fixed with # characters. For Python, we can represent the count as a tuple. This mapping process generates identical vectors for the strings that are anagrams. So, for Java, we represent abbccc as #1#2#3#0#0#0...#0, and for Python, abbccc will be represented as (1, 2, 3, 0, 0, ..., 0).
We then use this vector as a key for inserting titles into a Hash Map. Our anagrams will all be mapped to the same entry. When a user searches a title or word, it should compute the 26-element English letter frequency vector based on input. It will then search the Hash Map and return all the map entries using the vector.
We then store a list of calculated character counts as a key in a Hash Map and assign a string as its value.
Each value is an individual set, so we return the values of the Hash Map.

import java.util.*;
class Solution {
    public static List<List<String>> groupTitles(String[] strs){
      if (strs.length == 0) 
            return new ArrayList<List<String>>();
        Map<String, List<String>> res = new HashMap<String, List<String>>();
        int[] count = new int[26];
        for (String s : strs) {
            Arrays.fill(count, 0);
            for (char c : s.toCharArray()){
                int index = c - 'a';
                count[index]++;
            }
            StringBuilder delimStr = new StringBuilder("");
            for (int i = 0; i < 26; i++) {
                delimStr.append('#');
                delimStr.append(count[i]);
            }
            
            String key = delimStr.toString();
            if (!res.containsKey(key)) 
                res.put(key, new ArrayList<String>());
            
            res.get(key).add(s);
        }
        return new ArrayList<List<String>>(res.values());
    }
    public static void main(String[] args) {
        // Driver code
        String titles[] = {"duel","dule","speed","spede","deul","cars"};
        List<List<String>> gt = groupTitles(titles);
        String query = "spede"; 
        // Searching for all titles
        for (List<String> g : gt){
            if (g.contains(query))
                System.out.println(g);
        }
  }
}

Solution in Java and Python

Complexity measures#

$n$ is the size of the list of strings, and $k$ is the maximum length of a single string.

Time Complexity: $O(n*k)$ , since we are counting each letter for each string in a list

Space Complexity: $O(n*k)$ , since every string is stored as a value in the dictionary, and the size of the string can be $k$ .

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Facebook Feature: Friend Circles (DFS)#

Facebook is the biggest social media company in the world. They also own and operate Instagram. Pretend you are a Facebook engineer team, and you are tasked to improve integration among their sister platforms.

Task: Our task here is to find all the people on Facebook that are in a user’s friend circle, which we are calling the “Friend Circles” feature.

We need to first identify the people that are in each user’s friends circle, which includes users that are directly or indirectly friends with another user. Let’s assume there are $n$ users on Facebook. The friendship connection is transitive.

Example: If Nick is a direct friend of Amy, and Amy is a direct friend of Matt, then Nick is an indirect friend of Matt.

We will use an $n*n$ square matrix. For example, cell [i,j] will hold value 1 if these users are friends. If not, the cell will hold the value 0. In the illustration below, there are two friend circles in the above example. Nick is only friends with Amy, but Amy is friends with Nick and Matt. This forms a friend circle. Mario makes another friend circle on his own.

Think of our symmetric input matrix as an undirected graph. Both the indirect and direct friends who are in one friend circle also exist in one connected component in our graph. This means that the number of connected graph components will give us how many friend circles we have.

So, our task is to find the number of connected components. We treat the input matrix as an adjacency matrix. So, how do we design and implement this? Let’s break it down.

First we initialize an array, named visited. This will track the visited vertices of size n with 0 as the value for each index.
Then, we traverse the graph using DFS if visited[v] is 0. If not, we move to the next v.
Then, set visited[v] to 1 for every v that our DFS traversal runs into.
When the DFS traversal is done, we should increment the circles counter by 1. This means that a connected component has been fully traversed.

class Solution {
  public static void DFS (boolean[][] friends, int n, boolean[] visited, int v) {
    for (int i = 0; i < n; ++i) {
        // A user is in the friend circle if he/she is friends with the user represented by
        // user index and if he/she is not already in a friend circle
        if (friends[v][i] == true && !visited[i] && i != v) {
            visited[i] = true;
            DFS(friends, n, visited, i);
        }
    }
  }
  public static int friendCircles(boolean[][] friends, int n) {
    if (n == 0) {
        return 0;
    }
 
    int numCircles = 0;     //Number of friend circles
    
    //Keep track of whether a user is already in a friend circle
    boolean visited[] = new boolean[n];
    for (int i=0;i < n; i++){
      visited[i] = false;
    }
    
    //Start with the first user and recursively find all other users in his/her
    //friend circle. Then, do the same thing for the next user that is not already
    //in a friend circle. Repeat until all users are in a friend circle. 
    for (int i = 0; i < n; ++i) {
        if (!visited[i]) {
            visited[i] = true;
            DFS(friends, n, visited, i); //Recursive step to find all friends
            numCircles = numCircles + 1;
        }
    }
    
    return numCircles;
  }
  public static void main(String args[]) 
  { 
      int n = 4;
      boolean[][] friends = {
        {true,true,false,false},
        {true,true,true,false},
        {false,true,true,false},
        {false,false,false,true}
      };
     System.out.println("Number of friends circles: " + friendCircles(friends, n)); 
  } 
}

Solution in Java and Python

Google Calendar Feature: Find Meeting Rooms (heaps)#

The Google Calendar tool is part of the GSuite used to manage events and reminders. Imagine you are a developer on the Google Calendar application team, and you’re tasked with implementing some productivity enhancing features.

Task: Your goal is to create a functionality for scheduling meetings. You need to determine and block the minimum number of meeting rooms for these meetings.

To do this, we are given a some meeting times. We need to find a way to identify the number of meeting rooms needed to schedule them all. Each meeting will contain positive integers for a startTime and an endTime.

Our meeting times can be listed as follows: {{2, 8}, {3, 4}, {3, 9}, {5, 11}, {8, 20}, {11, 15}}. We could schedule each meeting in a separate room, but we want to use the minimum amount of rooms. We observe that three meetings overlap: {2, 8}, {3, 4}, and {3, 9}. Therefore, at least these three will require separate rooms.

To solve this, we use either a heap or priority queue for storing meeting timings, using end_time of each meeting as a key. The room at the top of our heap would become free earliest. If the meeting room from the top of the heap is not free, then no other room is free yet.

So, how do we build this functionality? Let’s break it down.

Sort the meetings by startTime.
Allocate the first meeting to a room. Add the endTime as an entry in the heap.
Traverse the other meetings and check if the meeting at the top has ended.
If the room is free, extract this element, add it to the heap again with the ending time of the current meeting we want to process. If it is not free, allocate a new room and add it to our heap.
After processing the list of meetings, the size of the heap will tell us how many rooms are allocated. This should be the minimum number of rooms we need.

Practice and Solution#

import java.util.Arrays; 
import java.util.PriorityQueue;
class Solution {
    public static int minMeetingRooms(int[][] meetingTimes){
        
        if(meetingTimes.length == 0){
            return 0;
        }
        Arrays.sort(meetingTimes, (a, b) -> Integer.compare(a[0], b[0]));       
        //min Heap keeps track of earliest ending meeting:
        PriorityQueue<Integer> minHeap = new PriorityQueue<>((A, B) -> A - B);
        
        minHeap.add(meetingTimes[0][1]);
        
        for (int i = 1; i < meetingTimes.length; i++) {
            int currStart = meetingTimes[i][0];
            int currEnding = meetingTimes[i][1];
            int earliestEnding = minHeap.peek();
            
            if (earliestEnding <= currStart) {
                minHeap.remove();
            } 
            //update heap with curr ending
            minHeap.add(currEnding);
        }
        return minHeap.size();
    }
    public static void main( String args[] ) {
        // Driver code
        int[][] meetingTimes = {{2, 8}, {3, 4}, {3, 9}, {5, 11}, {8, 20}, {11, 15}};
        System.out.print(minMeetingRooms(meetingTimes));
    }
}

Solution in Java and Python

Amazon Feature: Products in Price Range (BST)#

Amazon is the largest online store around the world, and they are always on the lookout for better ways to recommend products to their customers. Imagine you are a developer for Amazon’s store.

Task: Implement a search filter to find products in a given price range. The product data is in the form of a binary search tree. The values are the prices of products.

The parameters we are working with are low and high, which represent a user’s price range. The example list of products below are mapped to their prices.

We can assume that the selected price range is low = 7 and high = 20, so our function solution should only return the prices {9, 8, 14, 20, 17}. So, how do we implement this? Let’s break it down.

We will solve this problem using a variation of a preorder traversal on a binary tree, but other binary tree traversals would work as well.

Implement a recursive helper function for a preorder traversal.
Check heck if the current node value is in our given range. If so, add it to the output array.
Recursively call the preorder function on the left child of the node if the current node value greater than or equal to low.
If the current node value is less than or equal to high, recursively traverse the right child of the node.
When traversal is completed, the output array will be returned.

Practice and Solution#

main.java

BinarySearchTree.java

public class BinarySearchTree{
    public Node root;
    public BinarySearchTree(){
        this.root = null;
    }
    public void insert(int val){
        if (this.root == null)
            this.root = new Node(val);
        else
            this.root.insert(val);
    }
}
class Node{
    public int val;
    public Node leftChild;
    public Node rightChild;
    public Node(int val){
        this.val = val;
        this.leftChild = null;
        this.rightChild = null;
    }
    public void insert(int val){
        Node current = this;
        Node parent = current;
        while (current != null){
            parent = current;
            if (val < current.val)
                current = current.leftChild;
            else
                current = current.rightChild;
        }
        if(val < parent.val)
            parent.leftChild = new Node(val);
        else
            parent.rightChild = new Node(val);
    }
}

Solution in Java and Python

Time complexity#

Time complexity: $O(n)$

Space complexity: $O(n)$

Twitter Feature: Add Likes (strings)#

Twitter is a popular social media platform. Imagine you’re a Twitter developer, and your team must create an API that calculates the number of likes on a given person’s Tweets.

Task: Create an API that calculates the total number of likes on a person’s Tweets. Create a module that takes two numbers and returns the sum of the numbers.

The data is already extracted and stored in a simple text file for you. All of the values should remain strings and cannot be converted into integers. We must do digit-by-digit addition due these restrictions. So, how do we create this module? Let’s break it down.

Initialize an empty res variable and the carry as 0.
Then, set two pointers, p1 and p2, that will point to the end of like1 and like2.
Traverse the strings from the end using these pointers. Set it to stop when both strings are done.
Set x1 equal to a digit from string like1 at index p1. If p1 has reached the beginning of like1, set x1 to 0. Do the same for x2 with like2 at index p2.
Compute the current value using value = (x1 + x2 + carry) % 10. Then update the carry so that carry = (x1 + x2 + carry) / 10.
Append the current value to the result.
If both of the strings are traversed but the carry is still non-zero, append the carry to res.
Lastly, reverse the result and convert it to a string. Return the final string.

Practice and Solution#

class Solution {
    public static String addLikes(String like1, String like2){
        StringBuilder res = new StringBuilder();
        int carry = 0;
        int p1 = like1.length() - 1;
        int p2 = like2.length() - 1;
        while (p1 >= 0 || p2 >= 0) {
            int x1, x2;
            if(p1 >= 0)
                x1 = like1.charAt(p1) - '0';
            else 
                x1 = 0;
            if(p2 >= 0)
                x2 = like2.charAt(p2) - '0';
            else 
                x2 = 0;
            int value = (x1 + x2 + carry) % 10;
            carry = (x1 + x2 + carry) / 10;
            res.append(value);
            p1--;
            p2--;    
        }
        
        if (carry != 0)
            res.append(carry);
        
        return res.reverse().toString();
    }
    public static void main(String args[]) {
        System.out.println(addLikes("1545", "67"));
    }
}

Solution in Java and Python

Complexity measures#

Time complexity: $O(max(n_1,n_2))$ , where $n_1$ and $n_2$ are the lengths of the input.

Space complexity: $O(max(n_1,n_2))$ because the length of the result is $\max(N_1, N_2) + 1$

Where to go from here#

Congrats on making it to the end! You just build 5 real-world features, using skills like DFS, BST, heaps, and more. As you can see, this is a powerful way to prepare for coding interviews that tends to be more effective. If you can understand a problem’s underlying pattern, you can solve just about any question.

There are still dozens of other practice problems like this out there, like:

Merge Tweets in Twitter Feed
AT&T Determine Location
Zoom: Display Lobby Meeting
Search Engine: Fetch Words

If you want to learn more about coding patterns, check out the following Educative courses:

They’ll teach you the 24 patterns behind every coding interview question, so you can be prepared to answer any problem you might face.

Happy learning!

Continue reading about coding interviews#

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Cracking the Coding Interview: Solve 5 Real World Problems

Crack the coding interview#

Understanding how to take on coding problems#

Netflix Feature: Group Similar Titles (hash maps)#

Practice and Solution#

Complexity measures#

Facebook Feature: Friend Circles (DFS)#

Practice and Solution#

Complexity measures#

Google Calendar Feature: Find Meeting Rooms (heaps)#

Practice and Solution#

Complexity Measures#

Amazon Feature: Products in Price Range (BST)#

Practice and Solution#

Time complexity#

Twitter Feature: Add Likes (strings)#

Practice and Solution#

Complexity measures#

Where to go from here#

Continue reading about coding interviews#