Tcs Digital Placement Papers 2026
TCS Digital Placement Papers 2026 — Questions & Preparation
Last Updated: March 2026
TCS Digital is the premium hiring track at Tata Consultancy Services, offering a significantly higher package of ₹7-7.3 LPA compared to the standard Ninja profile. This track is designed for candidates with strong technical skills and problem-solving abilities.
TCS Digital Program Overview
| Aspect | Details |
|---|---|
| Full Form | TCS Digital Hiring |
| Package | ₹7.0 - 7.3 LPA |
| Role | Digital Technology Engineer |
| Training | Advanced technical training |
| Projects | Digital transformation projects |
| Eligibility | BE/B.Tech/ME/M.Tech/MCA/M.Sc (CS/IT) with 70%+ |
Eligibility Criteria
- Academic: Minimum 70% or 7 CGPA throughout (10th, 12th, Graduation)
- Backlogs: No active backlogs
- Gap: Maximum 1 year gap between academics
- Degree: Full-time courses from recognized universities
- Additional: Strong programming and problem-solving skills required
TCS Digital vs Ninja Comparison
| Feature | TCS Digital | TCS Ninja |
|---|---|---|
| Package | ₹7-7.3 LPA | ₹3.36 LPA |
| Difficulty Level | High | Moderate |
| Coding Questions | 3-4 (Advanced) | 2 (Basic-Moderate) |
| Technical Interview | Advanced topics | Basic concepts |
| Cut-off | 80%+ | 65-70% |
| Training | Advanced | Standard |
TCS Digital Exam Pattern
| Section | Questions | Duration | Difficulty |
|---|---|---|---|
| Aptitude (Advanced) | 20 | 40 min | Hard |
| Logical Reasoning | 20 | 40 min | Hard |
| Verbal Ability | 15 | 20 min | Moderate |
| Programming MCQs | 10 | 15 min | Hard |
| Advanced Coding | 3-4 | 60-90 min | Hard |
| Total | 65-70 | 175-205 min | - |
15 Sample Questions with Solutions
Advanced Aptitude (Questions 1-4)
Q1. If log₂(log₃(log₄x)) = 0, what is the value of x?
- a) 64
- b) 81
- c) 256
- d) 4
Solution: log₂(log₃(log₄x)) = 0 → log₃(log₄x) = 2⁰ = 1 → log₄x = 3¹ = 3 → x = 4³ = 64
Q2. A and B can complete a work in 12 days. B and C in 15 days. C and A in 20 days. How long will A alone take?
- a) 20 days
- b) 25 days
- c) 30 days
- d) 35 days
Solution: 2(A+B+C) = 1/12 + 1/15 + 1/20 = (5+4+3)/60 = 12/60 = 1/5. So A+B+C = 1/10. A = (A+B+C) - (B+C) = 1/10 - 1/15 = 1/30. Days = 30.
Q3. The sum of the first n odd natural numbers is:
- a) n(n+1)
- b) n²
- c) 2n²
- d) n(n+1)/2
Explanation: 1 + 3 + 5 + ... + (2n-1) = n²
Q4. A man rows upstream at 8 km/hr and downstream at 12 km/hr. What is the speed of the stream?
- a) 2 km/hr
- b) 4 km/hr
- c) 6 km/hr
- d) 8 km/hr
Solution: Speed of stream = (Downstream - Upstream)/2 = (12-8)/2 = 2 km/hr
Advanced Logical Reasoning (Questions 5-7)
Q5. In a code language, '123' means 'bright little boy', '145' means 'tall big boy', and '637' means 'beautiful little flower'. Which digit represents 'bright'?
- a) 1
- b) 2
- c) 3
- d) 4
Solution: From 123 and 145: '1' = 'boy'. From 123 and 637: '3' = 'little'. So '2' = 'bright'.
Q6. Six friends A, B, C, D, E, and F are sitting in a circle facing the center. A is between D and F. C is between E and B. E is not next to D. Who is between A and C?
- a) B
- b) D
- c) E
- d) F
Explanation: Arrangement (clockwise): D-A-F-?-E-C-B. Since E is not next to D, the arrangement is D-A-F-B-E-C. So D is between C and A (going the other way).
Q7. Complete the pattern: 1, 2, 6, 24, 120, ?
- a) 600
- b) 720
- c) 840
- d) 5040
Solution: Pattern: ×2, ×3, ×4, ×5, ×6 → 120 × 6 = 720 (Factorials: 1!, 2!, 3!, 4!, 5!, 6!)
Programming Concepts (Questions 8-10)
Q8. What is the time complexity of the following code?
for(int i = 0; i < n; i++)
for(int j = 0; j < i; j++)
printf("*");
- a) O(n)
- b) O(n²)
- c) O(n log n)
- d) O(n³)
Explanation: Inner loop runs 0 + 1 + 2 + ... + (n-1) = n(n-1)/2 = O(n²)
Q9. Which data structure is best for implementing LRU (Least Recently Used) cache?
- a) Array
- b) Stack
- c) Queue
- d) Hash Map + Doubly Linked List
Q10. What is the output of the following Java code?
class Test {
public static void main(String[] args) {
String s1 = new String("Hello");
String s2 = new String("Hello");
System.out.println(s1 == s2);
System.out.println(s1.equals(s2));
}
}
- a) true true
- b) false true
- c) true false
- d) false false
Explanation: '==' compares references (different objects), equals() compares content.
Data Structures & Algorithms (Questions 11-12)
Q11. Which algorithm is most efficient for finding the shortest path in a weighted graph with negative edges?
- a) Dijkstra's
- b) Bellman-Ford
- c) Floyd-Warshall
- d) BFS
Explanation: Bellman-Ford handles negative edge weights, unlike Dijkstra's.
Q12. What is the best case time complexity of QuickSort?
- a) O(n)
- b) O(n log n)
- c) O(n²)
- d) O(log n)
Advanced Coding Questions (Questions 13-15)
Q13. Write a program to find the longest palindromic substring.
Solution:
def longest_palindrome(s):
if not s:
return ""
start, max_len = 0, 1
for i in range(len(s)):
# Odd length palindromes
l, r = i, i
while l >= 0 and r < len(s) and s[l] == s[r]:
if r - l + 1 > max_len:
start = l
max_len = r - l + 1
l -= 1
r += 1
# Even length palindromes
l, r = i, i + 1
while l >= 0 and r < len(s) and s[l] == s[r]:
if r - l + 1 > max_len:
start = l
max_len = r - l + 1
l -= 1
r += 1
return s[start:start + max_len]
s = input()
print(longest_palindrome(s))
Q14. Write a program to find the maximum sum subarray (Kadane's Algorithm).
Solution:
#include<stdio.h>
int max_subarray_sum(int arr[], int n) {
int max_so_far = arr[0];
int max_ending_here = arr[0];
for(int i = 1; i < n; i++) {
if(max_ending_here + arr[i] > arr[i])
max_ending_here = max_ending_here + arr[i];
else
max_ending_here = arr[i];
if(max_ending_here > max_so_far)
max_so_far = max_ending_here;
}
return max_so_far;
}
int main() {
int n;
scanf("%d", &n);
int arr[n];
for(int i = 0; i < n; i++)
scanf("%d", &arr[i]);
printf("%d", max_subarray_sum(arr, n));
return 0;
}
Q15. Write a program to detect a cycle in a linked list.
Solution:
class Node {
int data;
Node next;
Node(int d) { data = d; next = null; }
}
class LinkedList {
Node head;
boolean hasCycle() {
Node slow = head, fast = head;
while(fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if(slow == fast)
return true;
}
return false;
}
}
Topic-Wise Weightage
| Section | Weightage | Priority |
|---|---|---|
| Advanced Aptitude | 25% | High |
| Logical Reasoning | 25% | High |
| Programming MCQs | 15% | Medium |
| Advanced Coding | 25% | Very High |
| Verbal Ability | 10% | Low |
Preparation Tips for TCS Digital
1. Advanced Aptitude
- Focus on advanced topics like logarithms, permutations, probability
- Practice complex DI problems
- Solve CAT-level quantitative questions
- Focus on speed and accuracy
2. Logical Reasoning
- Practice advanced puzzles
- Focus on data sufficiency
- Solve seating arrangement with multiple variables
- Practice coding-decoding variations
3. Programming Concepts
- Master data structures thoroughly
- Understand time and space complexity
- Study OOPs concepts in depth
- Learn DBMS and OS fundamentals
4. Advanced Coding
- Practice on LeetCode Medium-Hard problems
- Focus on: Arrays, Strings, Linked Lists, Trees, DP
- Learn standard algorithms
- Practice writing optimized code
Recommended Study Resources
| Resource | Purpose |
|---|---|
| LeetCode | Coding practice (Medium-Hard) |
| GeeksforGeeks | Technical concepts |
| HackerRank | Problem solving |
| Indiabix | Aptitude practice |
Frequently Asked Questions (FAQs)
Q1. Can I directly apply for TCS Digital? A: Yes, if you meet the eligibility criteria (70%+ throughout).
Q2. What is the cut-off for TCS Digital? A: Usually 80-85% overall with good performance in coding sections.
Q3. Is TCS Digital tougher than Ninja? A: Yes, significantly tougher with advanced questions in all sections.
Q4. What coding topics should I focus on? A: Arrays, strings, linked lists, trees, dynamic programming, and graph algorithms.
Q5. Can I get TCS Digital if I applied for Ninja? A: Sometimes candidates who perform exceptionally well in Ninja are upgraded to Digital.
Related Resources:
All the best for your TCS Digital preparation! 🚀