173 questions from 22 interviews · updated from reports 2017-2024
About
Nokia is a Finnish telecommunications company that provides network equipment, software, and services for mobile, fixed, and cloud networks. In India, it is known to hire interns, Graduate Engineer Trainees, Grade-6 Engineer GETs, and engineers in software, testing, network, and R&D roles.
The roles that come up most are Intern, Graduate Engineer Trainee and Grade-6 Engineer GET. This covers 22 candidate interviews reported from 2017 to 2024. Most sat it at entry level (14 of 22 that recorded a level), with 8 internship interviews alongside. Among the 16 that recorded either route, arrivals split between campus drives (13, 81%) and off-campus applications (3, 19%). Most questions fall under CS fundamentals and DSA.
Ans. C++ program memory is commonly divided into code, static data, heap and stack areas. Code stores instructions, static data stores globals and static variables, including zero-initialised data. The stack holds function calls and local automatic variables. The heap holds dynamically allocated objects. The key detail is lifetime: stack objects end automatically, heap objects must be managed.
Ans. Object-Oriented Programming models software as objects that combine data and behaviour. The main concepts are encapsulation, which hides internal state; abstraction, which exposes only needed details; inheritance, which reuses and extends existing classes; and polymorphism, which lets different objects respond to the same interface in their own way.
Ans. Use a postorder DFS that returns the height of each subtree and updates a global maximum diameter at every node. For each node, the longest path through it is left height plus right height, measured in edges. Visit each node once, so the time complexity is O(n), with O(h) recursion stack space.
Ans. The diamond problem in C++ happens when a class inherits from two classes that both inherit from the same base class. The final derived class can contain two separate copies of the base, causing ambiguity when accessing base members. The usual fix is virtual inheritance, which shares one common base subobject.
Ans. Normal forms are rules for reducing redundancy and update anomalies in relational tables. 1NF requires atomic values; 2NF removes partial dependency on part of a composite key; 3NF removes transitive dependency; BCNF makes every determinant a candidate key; 4NF removes independent multivalued dependencies; 5NF removes join dependencies so decompositions are lossless.
Ans. Deadlock is a state where two or more processes are permanently blocked because each is waiting for a resource held by another process. The key point is that none can continue without external intervention. It typically requires mutual exclusion, hold and wait, no preemption, and circular wait to occur.
Ans. Use a node structure with data and a pointer to the next node, and keep a head pointer to the first node. Insert by relinking pointers, delete by finding the previous node and bypassing the target, and traverse by following next pointers. Search, traversal, and deletion by value are linear time; head insertion is constant time.
Ans. Spanning Tree Protocol is a Layer 2 protocol that prevents switching loops by creating a loop-free logical topology over a network with redundant links. Switches exchange BPDUs, elect a root bridge, choose the best paths towards it, and place some ports into blocking state so backups exist without forwarding loops.
Ans. Routers route packets by reading the destination IP address and choosing the best matching entry in their routing table. The key detail is longest prefix match, which selects the most specific route. The router then forwards the packet to the next hop or outgoing interface, decrementing TTL and dropping it if no route exists.
Ans. Use Floyd’s cycle detection with two pointers, slow and fast, starting at the head. Move slow one node at a time and fast two nodes at a time. If they ever meet, there is a loop. If fast reaches null, there is no loop. This runs in O(n) time and O(1) space.
Ans. I explain networks by tracing what happens when a client contacts a server: DNS resolves the name, TCP or UDP carries data, TLS may secure it, and routers forward packets using IP. The key detail is separating layers: application protocols define meaning, transport handles delivery behaviour, and network links move packets hop by hop.
Ans. Use the standard output stream’s print, printf, or write method instead of println. The key difference is that println appends a newline automatically, while print does not, so add a newline character manually if needed. In Java, this still writes to console output through System.out.
Ans. Process scheduling is how an operating system chooses which ready process gets the CPU next and for how long. The scheduler maintains queues of runnable processes and applies a policy such as priority, round robin, or shortest job first. Its key goal is to balance CPU utilisation, responsiveness, fairness, and throughput while minimising context-switch overhead.
Ans. A simple example is a Dog class inheriting from an Animal class, where Animal defines shared behaviour like eat, and Dog adds specific behaviour like bark. In C++, this is usually public inheritance. The object layout stores the base part inside the derived object. Calling methods is constant time, apart from virtual dispatch overhead.
Ans. I would identify the main entities, their attributes, and relationships, then create normalised tables with primary keys, foreign keys, constraints, and indexes. For example, in an ordering system I would use Customers, Orders, OrderItems, and Products, where OrderItems links orders to products and stores quantity and price at purchase time.
Ans. Sort the heights, then consider each adjacent split where smaller elements are increased by k and larger elements are decreased by k. Keep the minimum possible difference between the current maximum and minimum. The key detail is to ignore negative adjusted heights if heights cannot be negative. Use sorting only. Time complexity is O(n log n).
Ans. The OSI model is a seven-layer framework for network communication: physical sends bits, data link frames local traffic, network routes packets, transport provides end-to-end delivery, session manages connections, presentation formats and encrypts data, and application supports user-facing protocols. The key idea is separation of responsibilities, making networks easier to design, debug, and standardise.
Ans. Deadlock is a state where two or more processes are blocked forever because each is waiting for a resource held by another. For example, process A holds a printer and waits for a file lock, while process B holds the file lock and waits for the printer. It requires circular waiting and no forced release.
Ans. Use two pointers, one starting at each list head, and advance both one node at a time. When a pointer reaches the end, redirect it to the other list’s head. If the lists intersect, the pointers meet at the shared node; otherwise both become null. This takes O(m+n) time and O(1) space.
Ans. Normalization is the process of organising database tables to reduce redundancy and avoid update, insert, and delete anomalies. Second Normal Form means a table is in First Normal Form and every non-key attribute depends on the whole primary key, not just part of a composite key. This mainly matters for tables with composite keys.
Ans. Dynamic memory allocation in C means requesting memory at run time from the heap, rather than using fixed-size stack variables. It is done with malloc, calloc, and realloc from stdlib.h, which return a pointer to the allocated block. Always check for NULL, use sizeof correctly, and release memory with free.
Ans. Count each element using a hash map, then sort the array elements using a comparator based on their frequency. The key detail is tie handling: if frequencies match, keep original order or sort by value, depending on the requirement. Counting takes O(n), sorting takes O(n log n), with O(n) extra space.
Ans. An operating system manages memory by giving each process its own virtual address space and mapping it to physical RAM. It tracks free and used memory, allocates pages to processes, enforces protection between them, and moves less-used pages to disk when RAM is full. The key idea is virtual memory through paging.
Ans. DBMS transaction management ensures a group of database operations is executed reliably as one unit, following ACID properties: atomicity, consistency, isolation and durability. Concurrency control ensures multiple transactions can run at the same time without corrupting data, using mechanisms such as locks, timestamps or MVCC to prevent conflicts like dirty reads and lost updates.
Ans. Use a static or class-level counter shared by all instances, and increment it whenever a new object is constructed. The key detail is to update it from every creation path, including overloaded constructors or factory methods. In a multithreaded program, make the counter atomic or protect it with a lock.
Ans. Model the board as a graph and use BFS to find the minimum dice throws from square 1 to the last square. Keep a queue of squares with throw count and a visited array. For each square, try moves 1 to 6, then apply any snake or ladder jump. Time complexity is O(N), where N is board size.
Ans. Use sizeof(struct_type) or sizeof(variable) to get the actual memory size of a structure. It is not usually just the sum of member sizes, because the compiler adds padding between members and at the end to satisfy alignment rules. The result can vary by compiler, platform, packing settings, and member order.
Ans. The minimum number of platforms is the maximum number of trains or buses present at the station at the same time. Sort arrival and departure times separately, then scan them with two pointers, increasing the count on an arrival and decreasing it on a departure. Track the maximum count. Time complexity is O(n log n), space is O(1) besides sorting.
Ans. The OSI model is a seven-layer framework for understanding network communication, from physical signalling up to applications. The Data Link Layer is important because it provides reliable node-to-node transfer on the same local network, using MAC addresses, framing, error detection, and media access control before data is passed to the Network Layer.
Ans. Count each value with a hash map, then sort the array elements using their frequency as the primary key. Usually, higher frequency comes first, and ties are broken by value or original order, depending on the requirement. This takes O(n log n) time for sorting and O(n) extra space for the frequency map.
Ans. Use two binary searches: one to find the leftmost occurrence and one to find the rightmost occurrence. In the first, move left when the target is found; in the second, move right when it is found. No extra data structure is needed. Time complexity is O(log n), space is O(1).
Ans. One network is connected to another network through a router, also called a gateway, which has an interface in each network. Devices send packets for outside addresses to their default gateway. The router examines the destination IP address, uses its routing table to choose the next hop, and forwards the packet between networks.
Ans. Implement LRU by evicting the page that has not been used for the longest time when a page fault occurs and memory is full. Use a hash map from page to node plus a doubly linked list ordered by recent use. Access, insert, move-to-front, and eviction are all O(1).
Ans. Data moves between network layers by encapsulation on the sender side and decapsulation on the receiver side. Each layer adds its own control information, such as headers, before passing data down to the next layer. At the destination, each layer reads and removes its header, then passes the remaining data upward.
Ans. Use a greedy scan, keeping the farthest index reachable in the current jump range and the farthest index reachable for the next range. Each time the scan reaches the end of the current range, take one jump and extend the range. If the current range cannot move further, the end is unreachable. This is O(n) time and O(1) space.
Ans. Machine Learning builds models that learn patterns from data to make predictions or decisions, while Data Analytics examines data to understand what happened, why it happened, and what actions to take. The key difference is purpose: analytics focuses on insight and reporting, whereas machine learning focuses on automated prediction and adaptation from data.
Ans. Fill the 5-litre jug and pour into the 3-litre jug, leaving 2 litres in the 5-litre jug. Empty the 3-litre jug. Pour the 2 litres into it. Fill the 5-litre jug again, then pour into the 3-litre jug until full. Exactly 4 litres remain in the 5-litre jug.
Ans. A voice call is established by signalling between the caller’s phone, the network, and the receiver’s phone. The caller’s network recognises the number, finds the receiving network and handset, checks availability, and sends a ringing signal. When the receiver answers, the network connects a voice path, either as a circuit or packet stream.
Ans. Choose a real example where leadership was absent or unclear and you helped create structure without taking over. Emphasise agreeing priorities, clarifying ownership, setting communication routines, and keeping progress visible. Interviewers listen for initiative, collaboration, respect for peers, decision-making under ambiguity, and evidence that the work was delivered, not just coordinated.
Ans. The TCP/IP model has four practical layers, while the OSI model has seven conceptual layers. TCP/IP combines OSI’s application, presentation and session layers into one application layer, and maps transport, internet, and link layers roughly to OSI transport, network, and data link plus physical layers. TCP/IP is used in real networks.
Ans. Create a custom exception by defining a class that extends Exception for checked errors or RuntimeException for unchecked errors. Throw this type from your code, and catch it using a catch block for that custom class. To handle different failures through it, wrap the original exception as the cause so details are not lost.
Ans. Mobile communication evolved from 2G digital voice and SMS, to 3G mobile internet, to 4G high speed IP data, and then to 5G ultra fast, low latency, high capacity networks. The key shift is from voice-centric circuit switching towards packet-based broadband supporting streaming, cloud services, IoT, automation, and real-time applications.
Ans. A static variable is stored for the whole lifetime of the program, so its value persists between function calls or object uses. A global variable is declared outside functions and is usually accessible across the file or program. The key difference is scope and linkage: static can limit visibility, while global is broadly visible.
Ans. Traverse the string from right to left and build the output, capitalising every alternate character in the reversed order. Use a counter to decide whether the current output position should be uppercase, usually starting with uppercase at position zero. A StringBuilder is suitable. Time complexity is O(n), space complexity is O(n).
Ans. The most suitable language is the one that best matches the problem’s constraints, ecosystem and team skills. For example, Python suits data analysis because of its libraries, Java suits large backend systems because of tooling and portability, and C++ suits performance-critical code because it gives low-level control.
Ans. Use two pointers, slow and fast, both starting at the head. Move slow one node at a time and fast two nodes at a time until fast reaches the end. Slow then points to the middle element. This uses only linked list pointers, runs in O(n) time and O(1) space.
Ans. One network connects to another through a router or gateway that forwards packets between different IP networks. A device sends traffic to its default gateway, the router checks the destination IP against its routing table, chooses the next hop, rewrites link-layer details, and forwards the packet until it reaches the target network.
Ans. A static function is a function that belongs to a class rather than to an object of that class. We can call it using the class name, without creating an instance. For example, utility methods are often static. The key point is that it can access only static data directly, not instance fields.
Ans. Trees are used for hierarchical data like file systems, DOM parsing, and database indexes. Graphs model networks such as maps, social connections, and dependency systems. Hashmaps provide fast lookups in caches, dictionaries, and frequency counters. The key benefit is choosing structures that match access patterns, improving search, traversal, and update efficiency.
Ans. Object-Oriented Programming organises software around objects that combine data and behaviour. Its main concepts are encapsulation, abstraction, inheritance and polymorphism. In practice, a banking app might model accounts, customers and transactions as objects, making code easier to reuse, extend, test and maintain as requirements change.
Ans. Pointers in C are variables that store memory addresses, usually of typed objects, enabling indirect access and dynamic data structures. A pointer to const cannot modify the pointed value, while a const pointer cannot be reassigned. A void pointer holds any object address but must be cast before dereferencing. Function pointers store callable function addresses.
Ans. The Physical layer is OSI layer 1, responsible for sending raw bits over a physical medium as electrical, optical, or radio signals. It defines cables, connectors, voltages, frequencies, timing, data rates, and modulation. Devices at this layer include hubs, repeaters, cables, connectors, network interface transceivers, modems, and media converters.
Ans. Find the node at index n and make it the new head by reconnecting the list around that point. Keep a pointer to the node before it, find the tail, set tail.next to the old head, then set previous.next to null. This runs in O length time and O1 space.
Ans. Pick a situation where you challenged a feature respectfully using evidence, not personal opinion. Emphasise understanding the goal, raising risks early, offering alternatives, and accepting the final decision once stakeholders decide. Interviewers listen for judgement, collaboration, customer focus, and whether you can disagree constructively without blocking delivery.
Ans. Three weighings are needed if the odd ball may be heavier or lighter. Two weighings give only 3² = 9 outcomes, but there are 8 × 2 possible cases. Weigh 123 against 456. If balanced, compare 7 with a known good ball. If not, use the remaining two weighings to test the suspect heavy side and light side.
Ans. Let the first half be x and the second half be y. The condition is 100x + y = x² + y², so y = (1 + √(1 + 400x - 4x²)) / 2. Checking x from 10 to 99, the square root is integral only for x = 12 or 88, giving y = 33. Hence 1233 and 8833.
Ans. Choose a real situation where you stayed calm, assessed impact quickly, and acted with priorities. Emphasise informing your mentor early, explaining facts, risks, options, and the support needed. Show ownership, not panic or blame. Interviewers listen for judgement, transparency, problem solving, time management, and willingness to escalate before damage increases.
Ans. Use a structured approach: read the question carefully, identify the type, note key data, and choose the fastest method. Eliminate impossible options first in multiple-choice questions. For calculations, estimate before solving fully. Manage time by skipping difficult questions initially and returning later if time remains.
Ans. Identify what is being asked, list the given values, and choose the right formula or concept, such as percentage, ratio, time and work, speed, profit, or probability. Convert units if needed, form a simple equation, solve step by step, and check whether the answer is reasonable.
Ans. Identify the type of pattern first, such as sequence, analogy, coding, direction, seating, syllogism, or arrangement. Write down the given facts clearly, convert words into symbols or diagrams, and test each option against the rules. Eliminate impossible answers quickly, but check the remaining choice for every condition before selecting it.
No question matches those filters. Reset filters.
Showing 60 of 173 questions. Ranked by how often the same question came back across interviews.
A spoken interview built from these questions, scored when you finish; the feedback is yours.
Start practisingWhen you are ready, record The One: a single interview hiring teams watch, so you stop repeating first rounds.
Candidate interviews most often cover CS fundamentals (61%) and DSA (25%).
Candidate interviews show an average of 3.6 rounds per experience, with a typical sequence of Online test → Technical → Technical → HR. Individual interview paths can vary.
Among questions with a recorded difficulty, the mix is easy 52%, medium 47%, hard 1%.
