Crack coding interviews by building these 5 real-world features

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Preparing for coding interviews is no easy task. You need the skills to break down the problem and to deploy the right tools. Trying to memorize 1500+ Leetcode problems is not the best way to succeed.

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 the solutions for a few common real world coding problems and build 5 features. We will offer our solutions in Java. If you’d like to see them in Python, check out my original post.

This tutorial at a glance:



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.

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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.

  1. For each title, we need to compute a 26-element vector. Each vector element represents the frequency of an English letter in a title. We can represent frequency using a string that is fixed with # characters. This mapping process generates identical vectors for the strings that are anagrams. For example, we represent abbccc as #1#2#3#0#0#0...#0.

  2. 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.

  3. We then store a list of calculated character counts as a key in a Hash Map and assign a string as its value.

  4. Each value is an individual set, so we return the values of the Hash Map.

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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);
        }
  }
}
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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.



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.

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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.

  1. First we initialize an array, named visited. This will track the visited vertices of size n with 0 as the value for each index.
  2. Then, we traverse the graph using DFS if visited[v] is 0. If not, we move to the next v.
  3. Then, set visited[v] to 1 for every v that our DFS traversal runs into.
  4. 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)); 
  } 

}
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Complexity measures

Time Complexity: O(n^2) because we traverse the complete matrix of size n^2

Space Complexity: O(n) because the visited array that stores our visited users is of size n.



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.

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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.

  1. Sort the meetings by startTime.
  2. Allocate the first meeting to a room. Add the endTime as an entry in the heap.
  3. Traverse the other meetings and check if the meeting at the top has ended.
  4. 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.
  5. After processing the lis 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.
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

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