TCS Digital Interview Questions: Technical Round Guide 2026

The TCS Digital track sits above Ninja on both CTC and technical bar, with the gap most visible in what happens after you clear the NQT.

Clearing the NQT Foundation section puts you in the Ninja pool. Clearing both the Foundation and the Advanced section puts you in the Digital pool. The technical interview that follows is the stage where the two tracks diverge most clearly. Digital interviewers ask about data structures, core CS theory, system design thinking, and the internals of your own projects. This guide covers each of those areas with verified, worked examples.

How TCS Digital Differs from Ninja

The table below summarises the three engineering-student TCS tracks for the 2026 hiring cycle. TCS Smart Hiring, which covers BSc/BCA/BCom graduates, runs on a separate funnel and is out of scope here.

Track	Min. CGPA	Starting CTC	NQT sections required	Key interview addition
TCS Ninja	6.0+	₹3.5 to 3.9 LPA	Foundation only	Standard technical + HR
TCS Digital	~7.0 to 8.0+	₹7.0 to 7.5 LPA	Foundation + Advanced	Higher-bar DSA + CS theory
TCS Prime	6.0+ (top NQT scorers)	₹9.0 to 11.0 LPA	Foundation + Advanced	Extended technical + AI project review

For the full NQT structure, including section-wise time splits, syllabus, and how the Advanced section is scored, the TCS NQT aptitude questions guide covers those details. Registration for NQT opens through TCS NextStep when the cycle for your graduation year begins. This article picks up at the interview round.

The Technical Interview Format

TCS Digital technical interviews typically run in one or two rounds after the NQT score is released. The standard format includes three phases, though the order and depth vary by interviewer and campus:

• A shared code editor or whiteboard problem. The interviewer watches you work live and asks about your approach as you go. Reasoning aloud is expected.
• Core CS fundamentals questions covering OS, DBMS, and networking. The exact topics depend on the interviewer; most sessions cover two of the three areas rather than all three.
• A project deep-dive. The interviewer picks one project from your resume and probes the design decisions, trade-offs, and edge cases. This phase separates Digital-track candidates from Ninja-track candidates more than any written test does.

An HR round follows if the technical round clears. For the NQT Foundation and Advanced section details, the TCS Ninja test pattern guide explains the scoring model that routes candidates into the Digital pool. The questions below are organised by the three interview phases above.

Worked DSA Problems

For Ninja-level coding practice, the TCS coding questions guide covers five worked problems with step-by-step traces. The three problems below target the higher-bar DSA questions that surface at the Digital-track level.

Q1: Reverse a Linked List (Iterative)

The task: given the head of a singly linked list, return the head of the reversed list without allocating a new list.

• Approach: maintain three pointer variables, prev (starts as None), curr (starts at head), and next_node (temporary storage for the next node).
• At each iteration: save curr.next to next_node, then point curr.next backwards to prev, then advance prev to curr, then advance curr to next_node.
• Continue until curr is None. Return prev as the new head.
• Time complexity: O(n). Each node is visited exactly once. No early exit.
• Space complexity: O(1). Exactly three pointer variables are allocated, independent of list length.

def reverse_linked_list(head):
    prev = None
    curr = head
    while curr:
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node
    return prev

• Trace for 1 -> 2 -> 3 -> None: after iteration 1, prev=node(1), curr=node(2). After iteration 2, prev=node(2), curr=node(3). After iteration 3, prev=node(3), curr=None. Return node(3), yielding 3 -> 2 -> 1 -> None. Correct.

Q2: Find the Second-Largest Element in an Array

The task: given an array of integers, find the second-largest value in a single pass.

• Approach: maintain two variables, max_val and second_max, both initialised to negative infinity.
• For each element x: if x > max_val, set second_max = max_val then set max_val = x. Otherwise if x > second_max and x != max_val, set second_max = x.
• After the loop, second_max holds the answer.
• Time complexity: O(n). One pass, constant-time operations per element.
• Space complexity: O(1). Two extra variables, regardless of array length.
• Edge case: if all elements are identical, second_max remains at negative infinity, indicating no second-largest exists.

def second_largest(arr):
    max_val = second_max = float('-inf')
    for x in arr:
        if x > max_val:
            second_max = max_val
            max_val = x
        elif x > second_max and x != max_val:
            second_max = x
    return second_max if second_max != float('-inf') else None

• Trace for [3, 1, 4, 1, 5, 9, 2, 6]: after full pass, max_val = 9, second_max = 6. Correct.

Q3: Check If Two Strings Are Anagrams

The task: given two strings s1 and s2, determine whether one is an anagram of the other (same characters, same frequency, different arrangement).

• Approach A (sort): sort both strings; if sorted(s1) == sorted(s2), they are anagrams.
  • Time complexity: O(n log n) where n is the length of the longer string. Simple to implement but slower.
• Approach B (frequency map): count character frequencies for s1. Decrement each count for characters in s2. If any count drops below zero or a character from s2 is absent, the strings are not anagrams.
  • Time complexity: O(n). Two passes, each O(n).
  • Space complexity: O(1) for ASCII inputs (at most 128 distinct entries in the map).
• Interviewers at the Digital level expect you to name both approaches and explain why Approach B is preferred for large inputs.

def are_anagrams(s1, s2):
    if len(s1) != len(s2):
        return False
    freq = {}
    for c in s1:
        freq[c] = freq.get(c, 0) + 1
    for c in s2:
        if c not in freq or freq[c] == 0:
            return False
        freq[c] -= 1
    return True

• Trace for s1 = "listen", s2 = "silent": both have the same length. After counting s1, freq = {l:1, i:1, s:1, t:1, e:1, n:1}. Processing s2 decrements each to zero without going negative. Returns True. Correct.

Core CS Fundamentals: OS, DBMS, and Networks

Q4: Process vs. Thread (Operating Systems)

OS concepts appear regularly in TCS Digital technical interviews. The process-vs-thread question often comes early in the CS fundamentals phase and opens into follow-ups on scheduling and synchronisation.

• A process is an independent program in execution with its own isolated memory space: a separate stack, heap, code segment, and data segment. Processes do not share memory with each other by default.
• A thread is the smallest unit of execution within a process. Threads in the same process share the heap and code segment. Each thread has its own stack and register set.
• Context-switching between processes is slower than between threads because the OS must save and restore the full memory map, file descriptor table, and page table, not just the register set.
• Communication between processes requires explicit IPC mechanisms: pipes, message queues, shared memory segments, or sockets. Threads communicate through the shared heap, which is faster but requires synchronisation primitives (mutexes, semaphores) to avoid race conditions.

A common follow-up: what is a race condition? When two threads access shared data concurrently without proper synchronisation, and the final state depends on the order of execution, a race condition exists.

Q5: Find Employees Earning Above Department Average (DBMS / SQL)

The task: given an employees table with columns name, department, and salary, return every employee whose salary exceeds the average salary in their own department.

• Approach A: correlated subquery. The inner query runs once for each row of the outer query.

SELECT e1.name, e1.department, e1.salary
FROM employees e1
WHERE e1.salary > (
    SELECT AVG(e2.salary)
    FROM employees e2
    WHERE e2.department = e1.department
);

• Approach B: JOIN with a pre-aggregated subquery. The per-department average is computed once and then joined.

SELECT e.name, e.department, e.salary
FROM employees e
JOIN (
    SELECT department, AVG(salary) AS avg_sal
    FROM employees
    GROUP BY department
) dept_avg ON e.department = dept_avg.department
WHERE e.salary > dept_avg.avg_sal;

• Approach B is preferred for large tables: the GROUP BY pass aggregates once, making the query O(n) after that step, compared to repeated inner-query execution in Approach A.
• Be ready to name the trade-off explicitly: Approach A is simpler to write; Approach B is more efficient.

Q6: TCP Three-Way Handshake (Networking)

Networking questions at the protocol level appear in TCS Digital technical interviews. The TCP handshake is the most consistently tested.

• Step 1 (SYN): the client sends a TCP segment with the SYN flag set and its initial sequence number (ISN). This signals the client’s intent to open a connection.
• Step 2 (SYN-ACK): the server responds with both SYN and ACK flags set. The ACK value is the client’s ISN plus 1 (acknowledging the client’s SYN). The server also sends its own ISN.
• Step 3 (ACK): the client sends a final ACK with a value of the server’s ISN plus 1. Both sides have synchronised their sequence numbers. The connection is ready for data transfer.

A common follow-up is the four-way termination sequence: FIN from the active closer, ACK from the passive closer, FIN from the passive closer when ready, final ACK from the active closer. Four steps rather than three because each side closes independently.

System Design at the Fresher Level

TCS Digital interviewers at the fresher level are not expecting a distributed-systems design. They want to see you decompose a problem into named components, describe the data each component holds, and reason through at least one real trade-off.

A common starter problem is: design a URL shortener.

• Component 1: a hash function or auto-increment counter that maps the original long URL to a short code, typically six to eight characters.
• Component 2: a database table with columns (short_code, long_url, created_at), where short_code is the primary key and the lookup key for redirects.
• Component 3: a redirect server that receives a request for a short URL, queries short_code in the database, and returns an HTTP 301 redirect to the stored long_url.

Expect follow-up questions:

• What happens if two long URLs produce the same short code (hash collision)?
• Why use HTTP 301 (permanent redirect) rather than HTTP 302 (temporary redirect)?
• How would you expire a short link after 30 days?

A clear, sequential answer to one of these follow-ups matters more than a perfect opening answer. The follow-up is what reveals whether you understand the system or just memorised the component list.

The Project Deep-Dive Round

The project deep-dive separates Digital-track candidates more than any written test does. Interviewers are not looking for a complex project. They are looking for someone who can explain what the system does, why the design choices were made, and where the trade-offs lie.

Expect questions like:

• Why did you choose this approach over the obvious alternative?
• What is the worst-case time or space cost of the core algorithm in your project?
• If you had two more weeks, what would you change first?
• Is this deployed or testable with real data?

Structuring the answer as context (what problem, what constraints, what scale) followed by one or two explicit design decisions with stated trade-offs is more effective than listing features. A candidate who says “I used a hash map here because lookup is O(1) versus O(n) for a list, and the extra memory cost was acceptable at this input size” creates a clearer signal than one who narrates the full feature list without reasoning about cost.

The most common gap in project deep-dives is not complexity but specificity. Saying “I built a machine learning model” is a starting point; saying “I used a decision tree because the data was tabular and interpretability mattered, and I chose Gini impurity as the split criterion” is an answer.

Where AI Skills Fit in the Digital Track

According to TCS CHRO Sudeep Kunnumal at the AI Impact Summit in March 2026, 60% of TCS’s FY26 fresher hires are AI-skilled, up from 10 to 15% three years ago. The same interview noted a 50% volume increase in the Prime and Digital cadre.

The Digital track does not test AI directly in its standard 2026 interview format. The practical implication is specific: if you list an AI or ML project on your resume, the interviewer will probe it at the same depth as the DSA problems in this guide, asking about design decisions, cost analysis, and edge cases. An AI project you built and understand is a strong asset in the deep-dive round. One you ran without understanding is a liability.

That difference, built versus ran, is what determines whether the project deep-dive works in your favour.

TinkerLLM starts at ₹299 and runs structured AI build labs that end with a deployable artifact, which is the kind of concrete, explainable project the deep-dive round rewards.