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March 2026 — Actual Exam

TCS NQT March 2026 Coding Questions
with Java & Python Solutions

8 real questions from TCS NQT Shift 1 & 2 (March 20–21, 2026) — fully solved, explained, and ready for your next attempt.

March 20–21, 2026

Java + Python

All Test Cases Covered

70 min Coding Round

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TCS NQT March 2026 Coding Questions

If you’re preparing for TCS NQT 2026, stop scrolling through random LeetCode problems.
This page contains the actual coding questions that appeared in TCS NQT on March 20 and March 21, 2026 — compiled from student reports immediately after the exam. Every solution is tested and passes all given test cases in both Java and Python.

Real Questions

Languages (Java + Python)

100%

Test Cases Passing

TCS NQT 2026 Exam Structure — What You Need to Know

Before diving into questions, understand the exam format. Many students waste time on the wrong sections. Here’s the confirmed breakdown for the March 2026 exam:

Section	Questions	Time	Part
Numerical Ability	20	25 min	Foundation (Part A)
Verbal Ability	25	25 min	Foundation (Part A)
Reasoning Ability	20	25 min	Foundation (Part A)
Advanced Quantitative & Reasoning	15	—	Advanced (Part B)
Coding (2 problems)	2	70 min	Advanced (Part B)

Pro Tip: The Advanced section (Part B) is mandatory if you’re targeting Digital or Prime cadre (₹7–12 LPA). Most students skip it — that’s your competitive advantage.

Questions asked in TCS NQT March 2026 -Check out

Gym Membership Cost
Easy

A gym offers membership plans based on the number of months a customer enrolls. Write a program that prints the total membership cost or an appropriate message for edge cases.

Pricing Rules

Duration	Cost (₹)
1 month	₹2,000
2 or 3 months	₹5,000
4 to 6 months	₹9,000
9 months	₹12,000
12 months	₹15,000
Input < 0	`invalid input`
Input = 0	`0`
Any other value	`Error`

Test Cases

Input

Output

Input

-1

Output

invalid input

Java Solution

Java

import java.util.Scanner;

public class GymMembership {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int months = sc.nextInt();

        if (months < 0)         System.out.println("invalid input");
        else if (months == 0)   System.out.println(0);
        else if (months == 1)   System.out.println(2000);
        else if (months <= 3)  System.out.println(5000);
        else if (months <= 6)  System.out.println(9000);
        else if (months == 9)  System.out.println(12000);
        else if (months == 12) System.out.println(15000);
        else                    System.out.println("Error");
    }
}

Python Solution

Python 3

months = int(input())

if   months < 0:    print("invalid input")
elif months == 0:   print(0)
elif months == 1:   print(2000)
elif months <= 3:   print(5000)
elif months <= 6:   print(9000)
elif months == 9:   print(12000)
elif months == 12:  print(15000)
else:               print("Error")

Key Insight: Classic if-else ladder. Watch the edge cases — months=0 returns 0, negative returns “invalid input”. Both are different outputs!

Transaction Monitoring System
Medium

Given N transactions (Sender, Receiver, Timestamp, Amount), validate them against fraud and duplication rules.

Rules

Same Sender + Receiver seen before → error duplication transaction
Consecutive timestamp difference > 60 seconds → fraud detected
All valid → all transaction are valid

Test Cases

Input

2
A B 100 500
A B 110 600

Output

error duplication transaction

Input

2
A B 100 500
C D 200 300

Output

fraud detected

Java Solution

Java

import java.util.*;

public class TransactionMonitor {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        Set<String> seen = new HashSet<>();
        int prevTime = -1;

        for (int i = 0; i < n; i++) {
            String sender   = sc.next();
            String receiver = sc.next();
            int timestamp  = sc.nextInt();
            sc.nextInt(); // amount — not used

            String key = sender + "-" + receiver;
            if (seen.contains(key)) {
                System.out.println("error duplication transaction");
                return;
            }
            seen.add(key);

            if (prevTime != -1 && (timestamp - prevTime) > 60) {
                System.out.println("fraud detected");
                return;
            }
            prevTime = timestamp;
        }
        System.out.println("all transaction are valid");
    }
}

Python Solution

Python 3

n = int(input())
seen = set()
prev_time = -1

for _ in range(n):
    s, r, t, a = input().split()
    t = int(t)
    key = s + '-' + r

    if key in seen:
        print("error duplication transaction")
        exit()
    seen.add(key)

    if prev_time != -1 and (t - prev_time) > 60:
        print("fraud detected")
        exit()
    prev_time = t

print("all transaction are valid")

Key Insight: Always check duplicate first, then fraud. HashSet lookup is O(1). String key = sender+”-“+receiver uniquely identifies a pair.

Purchase Discount Calculator
Easy

Given a purchase amount, apply a tiered discount and print the final amount rounded to 2 decimal places.

Amount Range	Discount
Amount < 1000	5%
1000 ≤ Amount < 5000	10%
Amount ≥ 5000	15%

Input → Output

800  → 760.00
3000 → 2700.00
6000 → 5100.00

Formula

final = amount - (amount × discount)
rounded to 2 decimals

Java Solution

Java

import java.util.Scanner;
public class Discount {
    public static void main(String[] args) {
        double amt = new Scanner(System.in).nextDouble();
        double d = amt < 1000 ? 0.05 : amt < 5000 ? 0.10 : 0.15;
        System.out.printf("%.2f%n", amt - amt * d);
    }
}

Python Solution

Python 3

a = float(input())
d = 0.05 if a < 1000 else 0.10 if a < 5000 else 0.15
print(f"{a - a*d:.2f}")

Arrange the King’s Army
Medium

Count valid sequences of length N using soldiers 1 to R where: first = 1, last = END, no two adjacent soldiers are equal.

Input

N=3, R=3, END=2

Output

1  →  [1,3,2]

Python Solution

Python 3

def solve(arr, pos, N, R, END):
    if pos == N:
        return 1 if arr[-1] == END else 0
    total = 0
    for s in range(1, R+1):
        if s != arr[pos-1]:
            arr[pos] = s
            total += solve(arr, pos+1, N, R, END)
    return total

N, R, END = int(input()), int(input()), int(input())
arr = [0]*N
arr[0] = 1
print(solve(arr, 1, N, R, END))

Parking Fee Calculation
Easy

Calculate parking fee by tiered rate. Handle non-integer input with “ERROR”.

Hours	Rate
First 2 hours	₹100/hr → total ₹200 max
Next 3 hours (hrs 3–5)	₹50/hr → adds ₹150 max
Beyond 5 hours	₹20/hr
Non-integer input	ERROR

Input → Output

2   → 200
6   → 370
SIX → ERROR

Breakdown for 6

2×100 = 200
3×50  = 150
1×20  =  20
─────────
Total = 370

Python Solution

Python 3

try:
    h = int(input())
    fee = (min(h, 2) * 100 +
           max(0, min(h, 5) - 2) * 50 +
           max(0, h - 5) * 20)
    print(fee)
except ValueError:
    print("ERROR")

Balloon Capacity — Greedy
Easy

Given weights of people and max balloon capacity, find the maximum number of people that can fit. Always pick the lightest person first.

Input

weights=[20,30,40,50]
maxWeight=100

Output

3  (20+30+40=90 ≤ 100)

Python Solution

Python 3

n = int(input())
weights = list(map(int, input().split()))
cap = int(input())

weights.sort()  # greedy: smallest first
count = total = 0
for w in weights:
    if total + w <= cap:
        total += w
        count += 1
    else:
        break
print(count)

Why Greedy Works: Sorting ensures we maximize the count of people by always adding the minimum weight possible, leaving room for more.

Maximum Subset Sum ≤ K
Medium

Find the maximum possible sum from any subset of the array such that the sum does not exceed K.

Input

ARR=[2,3,5,7], K=10

Output

10  (2+3+5=10)

Python Solution

Python 3

from itertools import combinations

n = int(input())
arr = list(map(int, input().split()))
k = int(input())

best = 0
for r in range(1, n+1):
    for combo in combinations(arr, r):
        s = sum(combo)
        if s <= k:
            best = max(best, s)
print(best)

Minimum Cost to Connect All Nodes (MST)
Hard

Given a graph with V vertices and E edges (each with a weight), find the Minimum Spanning Tree (MST) cost using Kruskal’s Algorithm.

Algorithm: Kruskal’s + Union-Find

Sort all edges by weight (ascending)
For each edge, add it if it doesn’t form a cycle (use Union-Find)
Stop when V-1 edges are added

Input

Output

Minimum Cost: 19
(edges: 3-4=4, 1-4=5, 1-2=10)

Java Solution

Java — Kruskal’s Algorithm

import java.util.*;
public class MST {
    static int[] parent, rank;
    static int find(int x) {
        if (parent[x] != x) parent[x] = find(parent[x]);
        return parent[x];
    }
    static boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) return false;
        if (rank[pa] < rank[pb]) { int t=pa; pa=pb; pb=t; }
        parent[pb] = pa;
        if (rank[pa] == rank[pb]) rank[pa]++;
        return true;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int V = sc.nextInt(), E = sc.nextInt();
        int[][] edges = new int[E][3];
        for (int i=0; i<E; i++) {
            edges[i][0]=sc.nextInt(); edges[i][1]=sc.nextInt(); edges[i][2]=sc.nextInt();
        }
        Arrays.sort(edges, (a,b) -> a[2]-b[2]);
        parent = new int[V+1]; rank = new int[V+1];
        for (int i=0; i<=V; i++) parent[i]=i;
        int cost=0, used=0;
        for (int[] e : edges) {
            if (union(e[0],e[1])) { cost+=e[2]; if (++used==V-1) break; }
        }
        System.out.println("Minimum Cost: " + cost);
    }
}

Python Solution

Python 3 — Kruskal’s Algorithm

def find(p, x):
    if p[x] != x: p[x] = find(p, p[x])
    return p[x]

def union(p, r, a, b):
    pa, pb = find(p,a), find(p,b)
    if pa==pb: return False
    if r[pa]<r[pb]: pa,pb=pb,pa
    p[pb]=pa
    if r[pa]==r[pb]: r[pa]+=1
    return True

V,E = map(int,input().split())
edges = [tuple(map(int,input().split())) for _ in range(E)]
edges.sort(key=lambda x: x[2])
parent=[*range(V+1)]; rank=[0]*(V+1)
cost=used=0
for u,v,w in edges:
    if union(parent,rank,u,v):
        cost+=w; used+=1
        if used==V-1: break
print(f"Minimum Cost: {cost}")

Key Insight: Kruskal’s is greedy over edges. Union-Find (DSU) detects cycles in O(α(n)) ≈ O(1). Time complexity: O(E log E).

How to Crack TCS NQT 2026 — Strategy

Week 1: Master conditionals, loops, arrays — 80% of Easy questions come from here
Week 2: HashMap, HashSet, String manipulation — Medium question territory
Week 3: Recursion, backtracking, greedy — covers Medium-Hard
Week 4: Graphs (BFS, DFS, MST), DP basics — Advanced round preparation

Output Format is Critical: TCS uses exact string matching. “Error” ≠ “error” ≠ “ERROR”. Match the output format shown in the problem — character for character.

Frequently Asked Questions

What coding questions were asked in TCS NQT March 2026? +

TCS NQT March 2026 included: Gym Membership Cost, Transaction Monitoring, Purchase Discount Calculator, King’s Army Sequences, Parking Fee, Balloon Capacity (Greedy), Maximum Subset Sum ≤ K, and Minimum Cost to Connect All Nodes (Kruskal’s).

How many coding questions are in TCS NQT? +

The Foundation section has 2 coding questions in 70 minutes. The Advanced section has 1-2 harder problems. Total coding time is approximately 90 minutes.

Java or Python — which should I use? +

Both are accepted. Python is faster to write for medium problems. Java is preferred for strong OOP structure. Python is recommended for speed during the exam.

What is the TCS NQT 2026 cutoff for Digital/Prime? +

TCS doesn’t officially publish cutoffs. Based on student reports, scoring above 70% in Foundation + completing at least one coding problem fully is sufficient for Ninja. For Digital/Prime, aim to solve the hard coding problem.

PlacementDrive Team

Placement Coaching & Tech Interview Preparation

We’ve been tracking TCS NQT patterns since 2020 and have helped 10,000+ students crack placements at TCS, Infosys, Wipro, and Cognizant. Follow us on Instagram @Placementdrive for daily prep content.

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TCS NQT March 2026 Coding Questions with Java & Python Solutions

TCS NQT March 2026 Coding Questions with Java & Python Solutions

TCS NQT 2026 Exam Structure — What You Need to Know

Questions asked in TCS NQT March 2026 -Check out

Pricing Rules

Test Cases

Java Solution

Python Solution

Rules

Test Cases

Java Solution

Python Solution

Java Solution

Python Solution

Python Solution

Python Solution

Python Solution

Python Solution

Algorithm: Kruskal’s + Union-Find

Java Solution

Python Solution

How to Crack TCS NQT 2026 — Strategy

Frequently Asked Questions

Keep Preparing — Related Posts

Useful Links:

Posted by Neeraj kumar

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TCS NQT March 2026 Coding Questions
with Java & Python Solutions