215 questions from 24 interviews · updated from reports 2017-2024
About
Media.net is an advertising technology company that provides contextual advertising and monetization tools for publishers and advertisers. In India, it is commonly seen hiring Software Engineers, Web Developers, and Associate Web Application Developers for ad platforms and web products.
The roles that come up most are Software Engineer, Web Developer and Associate Web Application Developer. This covers 24 candidate interviews reported from 2017 to 2024. Most sat it at entry level (17 of 23 that recorded a level), with 4 internship interviews alongside. Among the 14 that recorded either route, arrivals split between campus drives (11, 79%) and off-campus applications (3, 21%). Most questions fall under CS fundamentals and DSA.
Ans. A mutex is a lock for exclusive access by one thread, while a semaphore is a counter that allows a fixed number of threads to access a resource. The key difference is ownership: the thread that locks a mutex should unlock it, but a semaphore can be signalled by another thread.
Ans. A process is an independent running program with its own memory space, while a thread is a smaller unit of execution within a process that shares that process’s memory. Processes are more isolated and cost more to create or switch between. Threads are lighter, but shared memory makes synchronisation and race conditions important.
Ans. DNS lookup translates a domain name into an IP address by querying DNS servers, usually starting with a local cache or resolver. If not cached, the resolver asks root servers, then the relevant top-level domain server, then the authoritative server for the domain. The result is returned and cached for its TTL.
Ans. Use Tarjan’s DFS algorithm on an undirected graph, tracking discovery time and the lowest reachable discovery time for each vertex. For every DFS tree edge u to v, if low[v] is greater than disc[u], that edge is a bridge. Store the graph as an adjacency list. Time complexity is O(V + E).
Ans. A load balancer sits in front of multiple servers and distributes incoming traffic across them so no single server is overwhelmed. It chooses a backend using rules such as round robin, least connections, or latency. The most important detail is health checking, so failed or slow servers are removed from rotation.
Ans. Traverse the matrix by maintaining four boundaries: top, bottom, left and right. Repeatedly visit the top row, right column, bottom row and left column, then move the boundaries inward. Store values in an output list. Stop when boundaries cross. This runs in O(mn) time and uses O(1) extra space besides the result.
Ans. A semaphore is a synchronisation primitive that controls access to shared resources using an integer counter. Threads call wait to decrement it and may block if the value is unavailable, and signal to increment it and wake a waiting thread. Key properties are atomic operations, mutual exclusion or resource counting, and possible blocking.
Ans. Shared locks allow multiple transactions or threads to read the same resource at the same time, while exclusive locks allow only one transaction or thread to modify it. The key point is compatibility: shared locks can coexist with other shared locks, but an exclusive lock conflicts with both shared and exclusive locks to prevent inconsistent updates.
Ans. Kadane’s Algorithm finds the maximum sum of a contiguous subarray in linear time. It scans the array while keeping the best sum ending at the current position and the best sum seen overall. At each element, either extend the previous subarray or start a new one. It runs in O(n) time and O(1) space.
Ans. Indexing in a DBMS is a way to speed up data retrieval by keeping a separate data structure that maps column values to the locations of matching rows. Most indexes use B-trees or similar structures, allowing searches without scanning the whole table. The trade-off is extra storage and slower inserts, updates, and deletes.
Ans. Virtual memory is a memory management technique that gives each process the illusion of a large, private, continuous address space, while the operating system maps it to physical RAM and disk. Its main advantages are process isolation, simpler programming, efficient sharing, and the ability to run programs larger than available RAM using paging.
Ans. A deadlock is when two or more processes wait forever because each holds a resource another needs. It requires mutual exclusion, hold and wait, no preemption, and circular wait. Avoid it by breaking one condition, commonly using a fixed lock ordering, acquiring all resources upfront, using timeouts, or applying Banker’s algorithm for safe allocation.
Ans. The TCP three-way handshake establishes a reliable connection using SYN, SYN-ACK, and ACK messages. The client sends a SYN with its initial sequence number, the server replies with SYN-ACK and its own sequence number, then the client sends ACK. The key detail is that both sides synchronise sequence numbers before data transfer.
Ans. Fourier Transform is used in MFCC to convert each short, windowed speech frame from the time domain into its frequency spectrum. Usually an FFT computes the power spectrum, which is then passed through Mel filter banks. This matters because MFCCs model the spectral envelope of speech, where phonetic information is most visible.
Ans. Split the string into words, then print characters column by column: for each index from 0 to the maximum word length, take that character from every word, or a space if the word is shorter. Store each row as a string and trim trailing spaces. Time complexity is O(n), where n is total characters.
Ans. I solve medium array problems by first identifying the pattern, such as two pointers, sliding window, prefix sums, sorting, or hashing. The key detail is to reduce repeated work by storing useful state, like frequencies or running sums. Most medium array solutions should aim for O(n) or O(n log n) time.
Ans. Loading brings a program from storage into memory so the operating system can start executing it, while linking combines compiled code with required libraries and resolves symbol references. Linking may happen before execution as static linking, or at load time or run time as dynamic linking, which allows shared libraries and smaller executables.
Ans. Use mobile clients connected by WebSockets to regional chat gateways, backed by messaging services, queues, user/device storage, media storage, and push notifications. The key detail is reliable delivery: assign each message an id, persist it before acknowledgement, fan it out to recipient devices, track sent, delivered and read states, and retry from queues when users reconnect.
Ans. A load balancer distributes incoming traffic across multiple servers so no single server is overloaded. It is used to improve availability, scalability and performance. The key detail is that it also detects unhealthy servers and stops sending traffic to them, helping the system continue serving users during failures.
Ans. A hard link is another directory entry pointing to the same inode, while a soft link, or symbolic link, is a separate file that stores a path to another file. The key difference is that a hard link still works if the original name is deleted, but a soft link breaks if its target path no longer exists.
Ans. Convert the numbers to strings and sort them with a custom comparator: for two strings x and y, put x before y if xy is larger than yx. Then concatenate the sorted strings. Use an array or list of strings. Sorting dominates the cost, taking O(n log n) comparisons, with extra string comparison cost. Return 0 if all values are zero.
Ans. The browser resolves google.com to an IP address using DNS, connects to that server, requests the page, receives a response, and renders it. For HTTPS, it first sets up a TCP connection and a TLS handshake, then sends an HTTP request. The browser parses HTML, fetches CSS and JavaScript, and displays the page.
Ans. Reconstruct the permutation by treating the rank array as a Lehmer code. Keep the unused values 1 to N in sorted order, then for each rank from left to right pick the rank[i] plus 1-th smallest unused value and remove it. Use an order statistic tree or Fenwick tree for O(N log N).
Ans. Sort the intervals by start time, then scan them and keep a result list of merged intervals. For each interval, compare its start with the end of the last interval in the result. If it overlaps, extend that end; otherwise append it. Sorting dominates, so time is O(n log n) and space is O(n).
Ans. Database indexing is a way to speed up reads by storing an extra data structure that maps column values to the locations of matching rows. Internally, most databases use B-trees or B+ trees, which keep keys sorted and allow logarithmic lookup, range scans, and ordered access, at the cost of extra storage and slower writes.
Ans. Model the domain with clear entities, responsibilities, and relationships, then define the public behaviours before fields or implementation details. Use interfaces for variation, composition for shared behaviour, and keep classes small. The most important detail is showing how the design handles change, such as adding a new type without rewriting existing logic.
Ans. A man-in-the-middle attack is when an attacker secretly intercepts and possibly changes communication between two parties who believe they are talking directly. It is prevented mainly by strong authentication and encryption, such as HTTPS with valid certificates, certificate validation, secure key exchange, VPNs on untrusted networks, and avoiding ignored browser security warnings.
Ans. Composition is preferred because it gives more flexible, loosely coupled designs than inheritance. With composition, an object uses other objects for behaviour, so parts can be replaced, tested, or reused independently. Inheritance creates tight parent child coupling and can make changes risky because subclasses depend on base class behaviour and hierarchy.
Ans. I would use JOINs to bring related tables together, GROUP BY to aggregate by the required key, and subqueries to filter or compare against derived results. For example, join orders to customers, group by customer, calculate total spend, then use a subquery to keep only customers above the average total.
Ans. Use SELECT to choose columns, FROM to choose tables, WHERE to filter rows, JOIN to combine related tables, GROUP BY for aggregation, HAVING to filter groups, and ORDER BY to sort results. The key detail is applying row conditions in WHERE and aggregate conditions in HAVING, because they run at different stages.
Ans. OSI has seven layers: physical transmits bits, data link handles frames and MAC, network routes packets using IP, transport provides TCP or UDP delivery, session manages conversations, presentation formats and encrypts data, and application provides user protocols like HTTP, DNS, SMTP and FTP. The key idea is encapsulation, where each layer adds its own header.
Ans. I would scale it by making chat servers stateless behind a load balancer, using WebSockets for live delivery, and a pub-sub layer like Kafka or Redis Streams to route messages between servers. The most important detail is separating connection handling from durable message storage, so reconnects, retries and horizontal scaling remain reliable.
Ans. Cookies are small pieces of data stored in the user’s browser, while sessions store user state on the server and usually identify it using a session ID kept in a cookie. Web applications use them for login state, preferences, baskets, and tracking. The key point is that sensitive data should stay server side.
Ans. The browser resolves www.google.com to an IP address, connects to it, requests the page, receives a response, and renders it. The key detail is DNS lookup first, often using caches. Then the browser opens a TCP connection, negotiates TLS for HTTPS, sends an HTTP request, downloads resources, and builds the page.
Ans. Use two passes with a result array: first store each index’s prefix product, then multiply by the suffix product while scanning from the right. This avoids division and handles zeros correctly. The data structure is the output array plus two running products. Time complexity is O(n), with O(1) extra space excluding the output.
Ans. A vtable is a compiler-created table of function addresses for a class’s virtual functions, and a vptr is a hidden pointer in each polymorphic object that points to that table. When a virtual function is called through a base pointer or reference, the vptr selects the correct function at runtime.
Ans. Use a greedy single pass. Track current surplus fuel from a tentative start, and total surplus over all pumps. If current surplus becomes negative at pump i, no earlier pump in that segment can start the tour, so set start to i + 1 and reset current surplus. If total surplus is non-negative, return start, otherwise no tour exists.
Ans. In linear probing, search and insertion start at the key’s hash index and then check the next table slots one by one, wrapping around if needed. Search stops when the key is found or an empty slot proves it is absent. Insertion places the key in the first empty or deleted slot. Average time is O(1), worst case O(n).
Ans. Take the last value in postorder as the root, find it in inorder, and recursively build the left and right subtrees from the two inorder partitions. The key detail is to process postorder backwards, building the right subtree before the left. Use a hash map for inorder indices. Time is O(n), space is O(n).
Ans. Use breadth first search, treating each reachable number as a node and each allowed operation as an edge of cost one. Start from M, generate valid next numbers, and stop when N is first reached. Store visited numbers to avoid loops. Time is O(V plus E) over the explored state space.
Ans. Dynamic binding means the method to run is chosen at runtime based on the actual object type, not the reference type. For example, a Shape reference can point to a Circle or Square, and calling draw runs the correct overridden method. This enables polymorphism. Method dispatch is typically constant time using a virtual table.
Ans. Delta and Double Delta features add temporal dynamics to MFCCs, capturing how the spectral shape changes over time. Delta represents the first-order change, similar to velocity, while Double Delta represents the second-order change, similar to acceleration. They improve speech recognition by including context and transitions between neighbouring frames, not just static spectral information.
Ans. Use a query that selects the distinct values, sorts them in descending order, skips the first n minus 1 rows, and returns the next row. In MySQL or PostgreSQL this is done with ORDER BY descending plus LIMIT 1 and OFFSET n minus 1. Sorting dominates the cost, usually O(m log m).
Ans. Use dynamic programming to check if the target can be formed by preserving the order of characters from both strings. First ensure the lengths add up. Let dp[i][j] mean the first i and j characters can form the first i plus j target characters. Fill it by matching from either string. Time is O(mn), space can be O(n).
Ans. TCP uses a three way handshake: client sends SYN, server replies SYN-ACK, and client sends ACK. This proves both sides can send and receive, agrees initial sequence numbers, and establishes the connection state. A two way handshake is not sufficient because the server cannot know its SYN-ACK reached the client, causing half-open or stale connections.
Ans. Use a longest non-decreasing subsequence algorithm: maintain a tails array where tails[i] is the smallest possible ending value of a non-decreasing subsequence of length i + 1. For each number, place it after the last value less than or equal to it using upper bound. The tails length is the answer, in O(n log n) time.
Ans. Create a UDP server socket, bind it to a port, receive datagrams with the sender address, and send the same bytes back to that address. The client creates a UDP socket, sends a message to the server address, then waits for the echoed reply. Use a byte buffer. Each request is O(n) in message size.
Ans. When a process calls a function in its text segment, the CPU transfers execution to that function’s code address. The call instruction typically saves the return address on the stack, creates or uses a stack frame for local data, runs the function instructions, then returns control to the instruction after the call.
Ans. I would use a risk-based test strategy covering unit, integration, API, UI, performance, security and regression tests. The plan would define scope, test environments, data, responsibilities, entry and exit criteria, and automation targets. The most important detail is prioritising critical user journeys and business risks before testing lower-impact edge cases.
Ans. Transfer learning is used to reuse knowledge from a model trained on a large source task for a related target task with less data. It can reduce training time, improve accuracy, and work well with limited labelled data. Drawbacks include negative transfer, bias from the source data, and poor fit if tasks differ too much.
Ans. Use a subquery to find the maximum value, then find the maximum value smaller than that. In SQL terms, select the maximum element from the table where the element is less than the maximum element returned by an inner subquery. This handles duplicates correctly if you mean the second distinct largest value.
Ans. I handled complex UI scenarios by breaking them into stable page objects and using explicit waits, robust locators, and clear state checks. For frames, I identified the correct iframe, switched context before interacting, and always switched back afterwards. The key detail was isolating frame handling in reusable helper methods to avoid flaky tests.
Ans. I would use merge sort, or external merge sort if the array cannot fit in memory. It has guaranteed O(n log n) time and works well on large data because it splits the input and merges sorted runs sequentially. If memory is tight, the runs can be stored and merged from disk.
Ans. Use a rolling time-window counter: every received call records its timestamp, and GET /calls/count returns how many timestamps are within the last 5 minutes. Store timestamps in a queue or ring buffer, evict entries older than now minus 300 seconds on each write or read. Writes are O(1), reads are O(k) evicted.
Ans. Encode each card as a unique number 0 to 51. XOR all 52 possible codes, then XOR the 51 given card codes. Equal values cancel, leaving the missing card’s code. This uses one accumulator, so constant memory. It needs 51 input iterations, plus 52 setup iterations if not precomputed, and no comparisons.
Ans. Draw the new square at 45 degrees to the original. Through each corner of the given square, draw a line parallel to one of its diagonals, choosing the four lines so they enclose it. These lines form a larger square. Its side equals the original diagonal, so its area is s² + s² = 2s².
Ans. The second meeting is after 33/14 hours, or about 2 hours 21 minutes. For such problems, use relative speed. The circumference is 2πr = 44 km. They start half a circumference apart, so meetings occur after closing 22 km, then another 44 km. Total for second meeting is 66 km at 28 km/hr.
Ans. The second meeting occurs after 7π/12 seconds, about 1.83 seconds. Use arc distance, not diameter. The first gap is half the circumference, 7π/2 metres, and closing speed is 18 m/s. First meeting takes 7π/36 seconds. Each later meeting needs one full circumference more in relative motion, so add 7π/18.
Ans. Treat circles as closed obstacles. Union circles with centre distance at most 2R, and record which sides each component touches: x<=R, l-x<=R, y<=R, b-y<=R. If start or finish is inside a circle, fail. A component touching left-bottom, top-right, left-right, or bottom-top forms a continuous barrier, so fail. Otherwise succeed.
Ans. Identify the rule or relationship before trying to answer. Separate facts from assumptions, look for patterns, categories, sequences, cause and effect, or exclusions. Work step by step, eliminate impossible options, and check that the remaining answer fits every condition. If stuck, test simple examples rather than guessing.
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Candidate interviews most often cover CS fundamentals (51%) and DSA (36%).
Candidate interviews show an average of 3.2 rounds per experience, with a typical sequence of Online test → Technical → Technical. Individual interview paths can vary.
Among questions with a recorded difficulty, the mix is easy 30%, medium 55%, hard 15%.
