Overview of TCS NQT 2026
The **TCS National Qualifier Test (NQT)** for the 2026 batch continues to be India's largest campus recruitment drive. Conducted across TCS iON test centers in physical proctored environments, the exam filters candidates into three distinct compensation tiers:
- **Ninja Cadre**: ₹3.36 LPA – ₹3.60 LPA (Base Developer / Support Track)
- **Digital Cadre**: ₹7.00 LPA – ₹7.50 LPA (Advanced Engineering Track)
- **Prime Cadre**: ₹9.00 LPA – ₹11.50 LPA (High-Value R&D & Cloud Engineering)
1. Exam Architecture & Sectional Breakdown
The examination consists of two broad stages executed sequentially without pause:
Part A: Foundation Cognitive Section (Mandatory for All) 1. **Numerical Ability**: 20 Questions | 25 Minutes (Arithmetic, Speed Math, Number Systems) 2. **Reasoning Ability**: 20 Questions | 25 Minutes (Syllogisms, Clocks & Calendars, Analytical Reasoning) 3. **Verbal Ability**: 25 Questions | 25 Minutes (Sentence Correction, Paragraph Completion, Vocabulary)
*Marking Scheme*: +1 mark for correct answers. No negative marking in Part A.
Part B: Advanced Technical Section (Unlocks Digital & Prime) 1. **Advanced Quantitative & Reasoning**: 15 Questions | 25 Minutes (Calculus basics, Permutations, Matrix Logic) 2. **Advanced Coding Assessment**: 2 Hands-on Problems | 60 Minutes (Languages: C++, Java, Python, C)
2. Cadre Qualification Matrix & Cutoffs
| Cadre | Foundation Cognitive | Advanced Section | Coding Performance | Interview Track |
| :--- | :--- | :--- | :--- | :--- |
| **Ninja** | 65% – 70% | 40% – 50% | 1 Problem partially passed | Single HR + Tech Round |
| **Digital** | 80%+ | 75%+ | Both problems passed all test cases | Advanced Tech + Project Interview |
| **Prime** | 90%+ | 85%+ | Both problems passed with optimal time complexity (O(N) or O(N log N)) | System Architecture + Leadership Round |
3. Real Memory Coding Problem (Asked in Recent Shift)
Problem: Minimize Subarray XOR Sum *Given an array of integers `arr` of size `N`, find the minimum XOR sum obtainable by partitioning the array into two non-empty contiguous subarrays.*
pythonLive Solved
def min_xor_partition(arr):
n = len(arr)
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] ^ arr[i]
total_xor = prefix[n]
min_xor_sum = float('inf')
for split in range(1, n):
part1 = prefix[split]
part2 = total_xor ^ part1
min_xor_sum = min(min_xor_sum, part1 + part2)
return min_xor_sum