TCS NQT 2027 Recently Asked Questions – 5th & 9th June Assessment | Coding Solutions

import java.util.HashMap;

public class Solution {
    public static int[] coinPairIndices(int[] coins, int target) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < coins.length; i++) {
            int complement = target - coins[i];
            if (map.containsKey(complement)) {
                return new int[] { map.get(complement), i };
            }
            map.put(coins[i], i);
        }
        return new int[] { -1, -1 }; // no pair found
    }

    public static void main(String[] args) {
        int[] coins = {2, 7, 11, 15};
        int[] result = coinPairIndices(coins, 9);
        System.out.println("[" + result[0] + ", " + result[1] + "]"); // [0, 1]
    }
}

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

vector<int> coinPairIndices(vector<int>& coins, int target) {
    unordered_map<int, int> mp;
    for (int i = 0; i < coins.size(); i++) {
        int complement = target - coins[i];
        if (mp.count(complement))
            return { mp[complement], i };
        mp[coins[i]] = i;
    }
    return { -1, -1 };
}

int main() {
    vector<int> coins = {2, 7, 11, 15};
    auto res = coinPairIndices(coins, 9);
    cout << "[" << res[0] << ", " << res[1] << "]" << endl; // [0, 1]
}

def coin_pair_indices(coins, target):
    seen = {}
    for i, val in enumerate(coins):
        complement = target - val
        if complement in seen:
            return [seen[complement], i]
        seen[val] = i
    return [-1, -1]

# Test
coins = [2, 7, 11, 15]
print(coin_pair_indices(coins, 9))  # [0, 1]

import java.util.Arrays;

public class MergeArrays {
    public static int[] mergeSorted(int[] A, int[] B) {
        // Reverse B to make it ascending
        int left = 0, right = B.length - 1;
        while (left < right) {
            int tmp = B[left]; B[left] = B[right]; B[right] = tmp;
            left++; right--;
        }
        // Two-pointer merge
        int[] result = new int[A.length + B.length];
        int i = 0, j = 0, k = 0;
        while (i < A.length && j < B.length)
            result[k++] = A[i] <= B[j] ? A[i++] : B[j++];
        while (i < A.length) result[k++] = A[i++];
        while (j < B.length) result[k++] = B[j++];
        return result;
    }

    public static void main(String[] args) {
        int[] A = {1,3,5,7}, B = {10,8,6,4};
        System.out.println(Arrays.toString(mergeSorted(A, B)));
        // [1, 3, 4, 5, 6, 7, 8, 10]
    }
}

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> mergeSorted(vector<int> A, vector<int> B) {
    reverse(B.begin(), B.end());
    vector<int> result;
    int i = 0, j = 0;
    while (i < A.size() && j < B.size())
        result.push_back(A[i] <= B[j] ? A[i++] : B[j++]);
    while (i < A.size()) result.push_back(A[i++]);
    while (j < B.size()) result.push_back(B[j++]);
    return result;
}

int main() {
    vector<int> A = {1,3,5,7}, B = {10,8,6,4};
    for (auto v : mergeSorted(A, B)) cout << v << " ";
    // 1 3 4 5 6 7 8 10
}

def merge_sorted(A, B):
    B_asc = B[::-1]   # reverse B to ascending
    result = []
    i = j = 0
    while i < len(A) and j < len(B_asc):
        if A[i] <= B_asc[j]:
            result.append(A[i]); i += 1
        else:
            result.append(B_asc[j]); j += 1
    result.extend(A[i:])
    result.extend(B_asc[j:])
    return result

print(merge_sorted([1,3,5,7], [10,8,6,4]))
# [1, 3, 4, 5, 6, 7, 8, 10]

public class FirstOccurrence {
    public static int firstOccurrence(int[] arr, int target) {
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == target) return i;
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] arr = {4, 2, 7, 2, 9, 2};
        System.out.println(firstOccurrence(arr, 2)); // 1
        System.out.println(firstOccurrence(arr, 3)); // -1
    }
}

#include <iostream>
#include <vector>
using namespace std;

int firstOccurrence(vector<int>& arr, int target) {
    for (int i = 0; i < arr.size(); i++)
        if (arr[i] == target) return i;
    return -1;
}

int main() {
    vector<int> arr = {4,2,7,2,9,2};
    cout << firstOccurrence(arr, 2) << endl; // 1
    cout << firstOccurrence(arr, 3) << endl; // -1
}

def first_occurrence(arr, target):
    for i, val in enumerate(arr):
        if val == target:
            return i
    return -1

print(first_occurrence([4,2,7,2,9,2], 2))  # 1
print(first_occurrence([5,8,1,6], 3))          # -1

import java.util.LinkedList;
import java.util.Queue;

public class MinDepth {
    static class TreeNode {
        int val;
        TreeNode left, right;
        TreeNode(int v) { val = v; }
    }

    public static int minDepth(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int depth = 1;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                if (node.left == null && node.right == null)
                    return depth; // leaf reached!
                if (node.left  != null) q.offer(node.left);
                if (node.right != null) q.offer(node.right);
            }
            depth++;
        }
        return depth;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        System.out.println(minDepth(root)); // 2
    }
}

#include <iostream>
#include <queue>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left, *right;
    TreeNode(int v) : val(v), left(nullptr), right(nullptr) {}
};

int minDepth(TreeNode* root) {
    if (!root) return 0;
    queue<TreeNode*> q;
    q.push(root);
    int depth = 1;
    while (!q.empty()) {
        int sz = q.size();
        for (int i = 0; i < sz; i++) {
            auto node = q.front(); q.pop();
            if (!node->left && !node->right) return depth;
            if (node->left)  q.push(node->left);
            if (node->right) q.push(node->right);
        }
        depth++;
    }
    return depth;
}

int main() {
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    cout << minDepth(root) << endl; // 2
}

from collections import deque

class TreeNode:
    def __init__(self, val=0):
        self.val = val
        self.left = self.right = None

def min_depth(root):
    if not root: return 0
    q = deque([(root, 1)])
    while q:
        node, depth = q.popleft()
        if not node.left and not node.right:
            return depth   # leaf!
        if node.left:  q.append((node.left,  depth + 1))
        if node.right: q.append((node.right, depth + 1))

# Build: 1 → (2→4, 3)
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left = TreeNode(4)
print(min_depth(root))  # 2

TCS NQT Python Coding Questions 2026

n = int(input())
arr = [int(input()) for _ in range(n)]
pos = 0
for x in arr:
    if x != 0:
        arr[pos] = x
        pos += 1
while pos < n:
    arr[pos] = 0
    pos += 1
print(*arr)

day = input().strip().lower()
n = int(input())
days = {"mon":0,"tue":1,"wed":2,"thu":3,"fri":4,"sat":5,"sun":6}
d = days[day]
to_sun = (6 - d) % 7
if to_sun >= n:
    print(0)
else:
    print(1 + (n - to_sun - 1) // 7)

n = int(input())
prices = [int(input()) for _ in range(n)]
min_price = float('inf')
max_profit = 0
for p in prices:
    min_price = min(min_price, p)
    max_profit = max(max_profit, p - min_price)
print(max_profit)

t = int(input())
for _ in range(t):
    n = int(input())
    print(n * (n - 1) // 2)

s = input().strip()
L = int(input())
best = 0
for i in range(0, len(s), L):
    best = max(best, s[i:i+L].count('a'))
print(best)

n = int(input())
r = int(input())
if r == 0:
    print(0)
else:
    for _ in range(r):
        n = sum(int(d) for d in str(n))
        if n < 10:
            break
    print(n)

n = int(input())
arr = [int(input()) for _ in range(n)]
result = 0
for x in arr:
    result ^= x
print(result)

n = int(input())
arr = [int(input()) for _ in range(n)]
low = mid = 0
high = n - 1
while mid <= high:
    if arr[mid] == 0:
        arr[low], arr[mid] = arr[mid], arr[low]
        low += 1; mid += 1
    elif arr[mid] == 2:
        arr[mid], arr[high] = arr[high], arr[mid]
        high -= 1
    else:
        mid += 1
print(*arr)

import math
n = int(input())
print(2 * math.factorial(n - 2))

n = int(input())
mask = (1 << n.bit_length()) - 1
print(n ^ mask)

n = int(input())
sets = [set(input().strip()) for _ in range(n)]
common = sets[0]
for s in sets[1:]:
    common &= s
print(len(common))

n = int(input())
arr = [int(input()) for _ in range(n)]
count = 1
mx = arr[0]
for x in arr[1:]:
    if x > mx:
        count += 1
        mx = x
print(count)

fname = input().strip()
valid_ext = {"exe","pdf","txt","py","java"}
if " " in fname or fname.count(".") != 1:
    print("Invalid")
else:
    name, ext = fname.split(".")
    if name and name[0].isalnum() and ext in valid_ext:
        print("Valid")
    else:
        print("Invalid")

import math
n = int(input())
print(math.factorial(n))

n = int(input())
days_list = list(map(int, input().split()))
total = 0
for d in days_list:
    if d <= 0:
        continue
    elif d <= 5:
        total += 5 * d
    elif d <= 10:
        total += 25 + 10 * (d - 5)
    else:
        total += 25 + 50 + 20 * (d - 10)
print(total)

n = int(input())
arr = [int(input()) for _ in range(n)]
unique = sorted(set(arr), reverse=True)
print(unique[1] if len(unique) >= 2 else -1)

n = int(input())
arr = [int(input()) for _ in range(n)]
print(*arr[::-1])

n = int(input())
arr = [int(input()) for _ in range(n)]
seen = set()
result = []
for x in arr:
    if x not in seen:
        seen.add(x)
        result.append(x)
print(*result)

n = int(input())
arr = [int(input()) for _ in range(n)]
k = int(input())
seen = set()
pairs = set()
for x in arr:
    comp = k - x
    if comp in seen:
        pair = (min(x,comp), max(x,comp))
        if pair not in pairs:
            print(pair)
            pairs.add(pair)
    seen.add(x)

from collections import OrderedDict
n = int(input())
arr = [int(input()) for _ in range(n)]
freq = OrderedDict()
for x in arr:
    freq[x] = freq.get(x, 0) + 1
for k, v in freq.items():
    print(f"{k}:{v}")

n = int(input())
arr = [int(input()) for _ in range(n)]
cur = max_s = arr[0]
for x in arr[1:]:
    cur = max(x, cur + x)
    max_s = max(max_s, cur)
print(max_s)

n = int(input())
arr = [int(input()) for _ in range(n)]
k = int(input()) % n
if k:
    arr = arr[-k:] + arr[:-k]
print(*arr)

n = int(input())
arr = [int(input()) for _ in range(n-1)]
print(n*(n+1)//2 - sum(arr))

n = int(input())
nums = [int(input()) for _ in range(n)]
slow = fast = nums[0]
while True:
    slow = nums[slow]
    fast = nums[nums[fast]]
    if slow == fast:
        break
slow = nums[0]
while slow != fast:
    slow = nums[slow]
    fast = nums[fast]
print(slow)

n = int(input())
arr = [int(input()) for _ in range(n)]
num_set = set(arr)
best = 0
for x in num_set:
    if x - 1 not in num_set:
        cur, streak = x, 1
        while cur + 1 in num_set:
            cur += 1
            streak += 1
        best = max(best, streak)
print(best)

s = input().strip().lower()
print("Yes" if s == s[::-1] else "No")

from collections import Counter
s1 = input().strip().lower()
s2 = input().strip().lower()
print("Yes" if Counter(s1) == Counter(s2) else "No")

s = input().strip().lower()
v = sum(1 for c in s if c in 'aeiou')
con = sum(1 for c in s if c.isalpha() and c not in 'aeiou')
print(f"V:{v} C:{con}")

from collections import Counter
s = input().strip()
freq = Counter(s)
for c in s:
    if freq[c] == 1:
        print(c)
        break
else:
    print(-1)

s = input()
print(' '.join(s.split()[::-1]))

s = input().strip()
seen = set()
left = max_len = 0
for right, c in enumerate(s):
    while c in seen:
        seen.remove(s[left])
        left += 1
    seen.add(c)
    max_len = max(max_len, right - left + 1)
print(max_len)

text = input().strip()
pat = input().strip()
count = sum(1 for i in range(len(text)-len(pat)+1)
            if text[i:i+len(pat)] == pat)
print(count)

s = input().strip()
if not s:
    print(s)
else:
    res = []
    count = 1
    for i in range(1, len(s)):
        if s[i] == s[i-1]:
            count += 1
        else:
            res.append(s[i-1] + (str(count) if count > 1 else ""))
            count = 1
    res.append(s[-1] + (str(count) if count > 1 else ""))
    compressed = "".join(res)
    print(compressed if len(compressed) < len(s) else s)

s1 = input().strip()
s2 = input().strip()
print("Yes" if len(s1)==len(s2) and s2 in s1+s1 else "No")

n = int(input())
strs = [input().strip() for _ in range(n)]
strs.sort()
prefix = []
for a, b in zip(strs[0], strs[-1]):
    if a == b:
        prefix.append(a)
    else:
        break
print(''.join(prefix) if prefix else "")

n = int(input())
def is_prime(n):
    if n < 2: return False
    for i in range(2, int(n**0.5)+1):
        if n % i == 0: return False
    return True
print("Prime" if is_prime(n) else "Not Prime")

from math import gcd
a, b = int(input()), int(input())
g = gcd(a, b)
print(f"GCD:{g} LCM:{(a*b)//g}")

n = int(input())
digits = str(n)
k = len(digits)
print("Yes" if n == sum(int(d)**k for d in digits) else "No")

n = int(input())
a, b = 0, 1
fib = []
for _ in range(n):
    fib.append(a)
    a, b = b, a + b
print(*fib)

n = int(input())
result = 1
for i in range(2, n+1):
    result *= i
print(result)

n = int(input())
total = sum(i for i in range(1, n) if n % i == 0)
print("Yes" if total == n else "No")

n = int(input())
print(bin(n)[2:], oct(n)[2:], hex(n)[2:])

n = int(input())
factors = []
d = 2
while d * d <= n:
    while n % d == 0:
        factors.append(d)
        n //= d
    d += 1
if n > 1:
    factors.append(n)
print(*factors)

b, e = int(input()), int(input())
result = 1
base = b
while e > 0:
    if e % 2 == 1:
        result *= base
    base *= base
    e //= 2
print(result)

n = int(input())
print(0 if n == 0 else 1 + (n - 1) % 9)

n = int(input())
arr = [int(input()) for _ in range(n)]
for i in range(n):
    for j in range(0, n-i-1):
        if arr[j] > arr[j+1]:
            arr[j], arr[j+1] = arr[j+1], arr[j]
print(*arr)

n = int(input())
arr = [int(input()) for _ in range(n)]
t = int(input())
low, high, result = 0, n-1, -1
while low <= high:
    mid = (low + high) // 2
    if arr[mid] == t:
        result = mid; break
    elif arr[mid] < t:
        low = mid + 1
    else:
        high = mid - 1
print(result)

n = int(input())
arr = [int(input()) for _ in range(n)]
k = int(input())
arr.sort(reverse=True)
print(arr[k-1])

m = int(input())
a = [int(input()) for _ in range(m)]
n = int(input())
b = [int(input()) for _ in range(n)]
i = j = 0
res = []
while i < m and j < n:
    if a[i] <= b[j]:
        res.append(a[i]); i += 1
    else:
        res.append(b[j]); j += 1
res.extend(a[i:])
res.extend(b[j:])
print(*res)

n = int(input())
arr = [int(input()) for _ in range(n)]
passes = 0
for i in range(n):
    swapped = False
    for j in range(0, n-i-1):
        if arr[j] > arr[j+1]:
            arr[j], arr[j+1] = arr[j+1], arr[j]
            swapped = True
    if swapped:
        passes += 1
    else:
        break
print(passes)

n = int(input())
for i in range(1, n+1):
    print("*" * i)

n = int(input())
for i in range(1, n+1):
    print(*range(1, i+1))

n = int(input())
for i in range(n, 0, -1):
    print("*" * i)

n = int(input())
for i in range(1, n+1, 2):
    print(" " * ((n-i)//2) + "*" * i)
for i in range(n-2, 0, -2):
    print(" " * ((n-i)//2) + "*" * i)

def hanoi(n, src, dest, aux):
    if n == 1:
        print(f"{src}->{dest}")
        return
    hanoi(n-1, src, aux, dest)
    print(f"{src}->{dest}")
    hanoi(n-1, aux, dest, src)

n = int(input())
hanoi(n, "A", "C", "B")

s1 = input().strip()
s2 = input().strip()
m, n = len(s1), len(s2)
dp = [[0]*(n+1) for _ in range(m+1)]
for i in range(1, m+1):
    for j in range(1, n+1):
        if s1[i-1] == s2[j-1]:
            dp[i][j] = dp[i-1][j-1] + 1
        else:
            dp[i][j] = max(dp[i-1][j], dp[i][j-1])
print(dp[m][n])

n, c = int(input()), int(input())
w = list(map(int, input().split()))
v = list(map(int, input().split()))
dp = [[0]*(c+1) for _ in range(n+1)]
for i in range(1, n+1):
    for cap in range(c+1):
        dp[i][cap] = dp[i-1][cap]
        if w[i-1] <= cap:
            dp[i][cap] = max(dp[i][cap],
                             dp[i-1][cap-w[i-1]] + v[i-1])
print(dp[n][c])

def count_subsets(arr, n, s):
    if s == 0: return 1
    if n == 0: return 0
    if arr[n-1] > s:
        return count_subsets(arr, n-1, s)
    return (count_subsets(arr, n-1, s) +
            count_subsets(arr, n-1, s - arr[n-1]))

n = int(input())
arr = list(map(int, input().split()))
s = int(input())
print(count_subsets(arr, n, s))

text = input()
k = int(input()) % 26
result = []
for c in text:
    if c.isalpha():
        base = ord('A') if c.isupper() else ord('a')
        result.append(chr((ord(c) - base + k) % 26 + base))
    else:
        result.append(c)
print(''.join(result))

y = int(input())
if (y % 4 == 0 and y % 100 != 0) or (y % 400 == 0):
    print("Leap Year")
else:
    print("Not Leap Year")

Top 200 Infosys Verbal Ability Questions SE, DSE & SP Roles — With Answers

Top 100 Questions Asked in Cognizant Online Assessment

TCS Interview Questions 2026 – Real HR, Technical & Managerial Round Questions

Deloitte 2025 Placement Syllabus: Complete Guide for Engineering and MCA Students

Deloitte 2025 Placement Syllabus: Complete Guide for Engineering and MCA Students

If you are preparing for Deloitte’s placement drive in 2025, this guide will help you understand the latest syllabus, exam structure, and how to prepare section-by-section. Whether you’re from an Engineering background or pursuing an MCA, this article simplifies the Deloitte assessment process with accurate information and easy-to-understand tips.

Deloitte 2025 Placement Overview

Deloitte conducts a structured placement process that includes multiple assessments. These tests evaluate your aptitude, technical understanding, and problem-solving skills. The syllabus has recently been updated to include dedicated sections with time limits.

Each section has a defined duration and number of questions, and all sections are equally important for selection.

Aptitude Test Breakdown

These topics are common in most aptitude exams, but Deloitte’s test often emphasizes speed and accuracy.

Technical MCQ Round

This section evaluates your programming basics and technical knowledge. The questions are based on:

• Programming concepts (C, C++, Java, Python)
• Data structures and algorithms
• OOPs concepts
• Database management
• Operating systems and networking fundamentals

Tip: Choose your programming language wisely. Answering questions in the language you’re most comfortable with can improve your accuracy and speed.

Coding Round (Technical Skills)

In this round, you will be tested on your coding proficiency. Candidates are required to solve 1–2 coding questions within a given time frame.

Key topics to prepare:

• Arrays and Strings
• Sorting algorithms
• Matrix manipulation
• Hashing techniques
• Recursion and backtracking (basic)

Most problems will test your logical approach and the efficiency of your solution.

Time Management Strategy

Since each section has a fixed duration, managing your time is key. Here’s a quick overview:

Final Thoughts

The Deloitte 2025 syllabus is designed to test your ability to think critically, solve problems efficiently, and apply technical knowledge under pressure. The best way to prepare is to stay consistent, identify your weak areas early, and give regular mock tests.

Approach the test with a clear strategy, and keep practicing questions from each topic. With regular effort and the right preparation plan, cracking Deloitte’s placement process in 2025 is completely achievable.

Stay focused, stay sharp, and give it your best shot.

Capgemini on campus Hiring Process 2025: Full Overview and Preparation Plan

If you’re preparing for Capgemini’s recruitment drive in 2025, it’s important to know how the selection process works. This article explains each step involved, the roles being offered, and how you can get ready for each round.

Job Profiles and Compensation

Capgemini is hiring fresh graduates for the following roles:

Your profile will be based on your scores during the selection rounds.

Selection Rounds

Capgemini’s hiring process includes five rounds. Each round is compulsory, and students are shortlisted based on performance at every step.

Round Details

Round 1: Cognitive and Technical Questions

This part checks your ability to solve problems using logic, language, and math. Technical questions may cover:

• Data structures
• Programming basics
• DBMS and networking
• Questions based on your academic field

Round 2: Essay Writing

This writing round includes one topic. You’ll have about 20–30 minutes to write an essay that includes:

• An opening paragraph
• A middle section explaining your views
• A closing statement

Sample Topics:

• Are we becoming too reliant on machines?
• Benefits and downsides of reading on screens

Round 3: Coding (For Analyst Star & Senior Analyst)

You’ll be asked to write programs using languages like C, C++, Java, or Python. Questions test your logic and coding efficiency.

Topics to practice:

• Arrays
• Strings
• Loops and conditions
• Sorting and searching

Round 4: Spoken English

This is an AI-based round that checks how well you speak and understand English. You will be tested on:

• Word clarity
• Sentence structure
• Speaking speed and confidence

Round 5: Interview

This session is led by experienced professionals. They will ask about:

• Your academic subjects
• Your final year or major project
• How you solve real-world problems

How to Prepare

Final Thoughts

Capgemini’s hiring steps are clear and well-structured. With consistent preparation, clear understanding, and regular practice, you can improve your chances of getting selected. Start early, focus on one section at a time, and keep improving day by day.

Stay calm during the tests, give your best, and believe in your preparation.

How to Prepare for Cognizant? Step by Step RoadMap

Introduction

Cognizant has officially started its hiring for the 2025 batch, opening doors for lakhs of students across the country. Whether you are a fresher targeting your first job or a final-year student preparing for placements, this is your chance to land a dream role at a reputed IT company.

This article gives you the complete roadmap, leaked syllabus breakdown, and clear insights on how to prepare for Cognizant hiring effectively. 

Watch the complete video guide here:
Cognizant Complete Roadmap | Syllabus Leaked | Which Cluster to Select

Job Roles Offered by Cognizant

Cognizant hires freshers primarily for the following roles:

1. GenC

Entry-level profile for candidates with basic technical knowledge. It involves minimal coding.

2. GenC Elevate

For students with solid programming and problem-solving abilities.

3. GenC Next

Premium role with advanced-level expectations, including project experience and strong system-level understanding.

Eligibility Criteria

To apply for Cognizant Hiring 2025:

• Eligible Degrees: B.E., B.Tech, M.E., M.Tech, MCA, M.Sc. (IT/CS)
• Batch: 2025
• Academic Scores: Minimum 60% in X, XII, and graduation
• Backlogs: No active backlogs
• Education Gap: Maximum of 2 years allowed

Selection Process (Step-by-Step)

1. Communication Test
   • Grammar, reading comprehension, sentence correction
   • Email writing and vocabulary-based questions
2. Aptitude & Logical Reasoning Test
   • Topics: Percentages, Ratios, Time-Speed-Distance, Data Interpretation, Series, Puzzles
3. Technical Assessment
   • MCQs on programming concepts, basic CS subjects (DBMS, OS, OOPs)
   • 1–2 coding questions (for Elevate and Next roles)
4. Interviews
   • Technical Interview: Projects, programming, CS fundamentals
   • HR Interview: Personality, communication, willingness to relocate

Roadmap to Crack Cognizant 2025

Here’s a simplified preparation roadmap based on the video content and latest trends:

Phase 1: Understand the Hiring Flow

• Know which role you are targeting: GenC, GenC Elevate, or GenC Next
• Each role has a different level of difficulty and salary package

Phase 2: Focused Preparation for Each Round

Phase 3: Practice and Mock Tests

• Take mock tests on platforms like PrepInsta, HackerRank, and GeeksforGeeks
• Join WhatsApp/Telegram groups for updates and peer learning
• Solve previous Cognizant questions available online

Phase 4: Interview Readiness

• Revise academic projects and how you contributed
• Brush up CS fundamentals like DBMS, OS, CN
• Practice common HR questions like strengths, goals, and relocation

Syllabus Breakdown (Leaked Syllabus 2025)

Below is a compiled and updated syllabus breakdown based on the current hiring pattern:

Communication Section:

• Grammar (tenses, articles, prepositions)
• Reading comprehension
• Email writing
• Vocabulary-based questions

Aptitude Section:

• Time, Speed & Distance
• Profit & Loss
• Percentages & Ratios
• Number series
• Data Interpretation

Logical Reasoning Section:

• Blood relations
• Coding-decoding
• Seating arrangements
• Syllogisms
• Pattern-based questions

Technical Section:

• Basic C/C++/Java questions
• OOPs concepts
• Pointers and memory
• DBMS & SQL
• OS Basics
• 1–2 Coding Questions (for Elevate & Next only)

Salary Package Overview

Note: Actual salary may vary based on location and performance in interview rounds.

Final Thoughts

Cognizant’s 2025 hiring process is a golden opportunity for students to start their careers in one of the top IT companies in India. With a clear roadmap, consistent practice, and smart preparation, you can crack any of the roles.

For a detailed walkthrough, syllabus discussion, and cluster selection tips, don’t miss this video:

🎥 Cognizant Complete Roadmap | Syllabus Leaked | Which Cluster to Select

TCS NQT 2025 Actual Coding Questions with Shift-wise (Java Solution)

TCS NQT – October 3rd, 2024 (Shift 1)

Question 1: Weekly Exercise Summary

Problem Statement:
You have to design a weekly exercise summary by taking the number of minutes of daily exercise for 7 consecutive days.
The exercise duration for all days will be entered by the user.

Your task is to:

1. Calculate the total exercise duration for the week
2. Calculate the average daily workout duration (rounded to nearest integer)

Input Format:

Day 1 exercise duration: <minutes> Day 2 exercise duration: <minutes> ... Day 7 exercise duration: <minutes> 

Output Format:

Exercise summary Total exercise duration : <total> (minutes) Average daily exercise duration: <average> minutes 

Example Input:

Day 1 exercise duration: 45 Day 2 exercise duration: 15 Day 3 exercise duration: 30 Day 4 exercise duration: 15 Day 5 exercise duration: 5 Day 6 exercise duration: 10 Day 7 exercise duration: 20 

Example Output:

Exercise summary Total exercise duration : 140 (minutes) Average daily exercise duration: 20 minutes 

Solution (Java):

import java.util.Scanner;

public class ExerciseSummary {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int[] exerciseDuration = new int[7];
        int total = 0;
        
        for (int i = 0; i < 7; i++) {
            System.out.print("Day " + (i + 1) + " exercise duration: ");
            exerciseDuration[i] = scanner.nextInt();
            total += exerciseDuration[i];
        }
        
        double average = total / 7.0;
        average = Math.round(average * 100.0) / 100.0;
        
        System.out.println("Exercise Summary");
        System.out.println("Total exercise duration: " + total + " minutes");
        System.out.println("Average daily exercise duration: " + average + " minutes");
        
        scanner.close();
    }
}

Question 2: Problem Statement:
You are given a range of integers from M to N (inclusive), where:

• M is the lower limit
• N is the upper limit

Your task is to find all palindrome numbers in the range and return the count of such numbers.

Input Format:

Enter M : <lower limit integer> Enter N : <upper limit integer> 

Output Format:

<number of palindrome numbers> 

Definition:
A palindrome number is a number that reads the same backward as forward.
Example: 121, 131, 11, 9

Example Input:

Enter M : 10 Enter N : 20 

Example Output:

1 

Explanation:
Only one palindrome exists between 10 and 20 → 11

Solution (C++):

#include <iostream>
using namespace std;

bool isPalindrome(int number) {
    if (number < 0) return false;
    int original = number, reversed = 0;
    while (number > 0) {
        reversed = reversed * 10 + number % 10;
        number /= 10;
    }
    return original == reversed;
}

int main() {
    int M, N;
    cout << "Enter M: ";
    cin >> M;
    cout << "Enter N: ";
    cin >> N;
    int count = 0;
    for (int i = M; i <= N; i++) {
        if (isPalindrome(i)) count++;
    }
    cout << count << endl;
    return 0;
}

TCS NQT – October 3rd, 2024 (Shift 2)

Question 3: Train Travel Time
Problem: A train covers 800 meters (400m track + 400m bridge) at a given speed (km/hr). Calculate the time taken in seconds using the formula: (Total distance / Speed) * (18/5).

Solution (C++):

#include <iostream>
using namespace std;

int main() {
    double speed;
    cin >> speed;
    int totalDistance = 800;
    double speedMs = speed * 5.0 / 18.0; // Convert km/hr to m/s
    int time = totalDistance / speedMs;
    cout << time << endl;
    return 0;
}

Question 4: Split Array with Equal Averages
Problem: Given an array of integers, check if it can be split into two subarrays with equal averages. Output true or false.

Solution (Java):

import java.util.Scanner;

public class ArraySplit {
    public static boolean canSplit(int[] arr, int n) {
        int totalSum = 0;
        for (int i = 0; i < n; i++) totalSum += arr[i];
        int leftSum = 0;
        for (int i = 0; i < n - 1; i++) {
            leftSum += arr[i];
            int rightSum = totalSum - leftSum;
            if (leftSum * (n - i - 1) == rightSum * (i + 1)) return true;
        }
        return false;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = sc.nextInt();
        System.out.println(canSplit(arr, n) ? "true" : "false");
        sc.close();
    }
}

TCS NQT – October 4th, 2024 (Shift 1)

Question 5: Perfect Donation Amount
Problem: Check if a donation amount is a perfect number (sum of its proper divisors equals the number). Return true if perfect, false otherwise.

Solution (Java):

import java.util.Scanner;

public class PerfectNumber {
    public static boolean isPerfect(int num) {
        int sum = 0;
        for (int i = 1; i <= Math.sqrt(num); i++) {
            if (num % i == 0) {
                sum += i;
                if (num / i != i && num / i != num) sum += num / i;
            }
        }
        return sum == num;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        System.out.println(isPerfect(num));
        sc.close();
    }
}

Question 6: Inventory Frequency Counter
Problem: Given a string of space-separated item names, count the frequency of each item. If the input contains digits, output “Invalid input.”

Solution (Java):

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class InventoryManager {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        Map<String, Integer> freq = new HashMap<>();
        String word = "";
        
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                System.out.println("Invalid input");
                return;
            } else if (s.charAt(i) == ' ') {
                if (!word.isEmpty()) freq.put(word, freq.getOrDefault(word, 0) + 1);
                word = "";
            } else {
                word += s.charAt(i);
            }
        }
        if (!word.isEmpty()) freq.put(word, freq.getOrDefault(word, 0) + 1);
        
        for (Map.Entry<String, Integer> entry : freq.entrySet()) {
            System.out.println(entry.getKey() + " " + entry.getValue());
        }
        sc.close();
    }
}

TCS NQT – October 4th, 2024 (Shift 2)

Question 7: Calculate Speed in km/hr
Problem: Given a distance of 1000 meters and time in seconds, calculate the speed in km/hr.

Solution (C++):

#include <iostream>
using namespace std;

int main() {
    double time;
    cin >> time;
    double distance = 1000.0; // meters
    double speedMs = distance / time; // m/s
    double speedKmh = speedMs * 18.0 / 5.0; // Convert m/s to km/hr
    cout << speedKmh << endl;
    return 0;
}

Question 8: Jump Game
Problem: Given an array where each element is the maximum jump length from that index, check if you can reach the last index from the first index.

Solution (Java):

import java.util.Scanner;

public class JumpGame {
    public static boolean canJump(int[] nums) {
        int maxReach = 0;
        for (int i = 0; i < nums.length && i <= maxReach; i++) {
            maxReach = Math.max(maxReach, i + nums[i]);
            if (maxReach >= nums.length - 1) return true;
        }
        return false;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String input = sc.nextLine();
        String[] s = input.split(",");
        int[] nums = new int[s.length];
        for (int i = 0; i < s.length; i++) {
            nums[i] = Integer.parseInt(s[i].trim());
        }
        System.out.println(canJump(nums));
        sc.close();
    }
}

Tips for TCS NQT Success

• Practice Input Handling: Many questions use Scanner (Java) or cin (C++). Test with sample inputs to avoid errors.
• Master Arrays and Strings: Most problems involve arrays or string manipulation. 
• Understand Math Basics: Questions like Train Travel Time and Speed Calculation require unit conversions. Brush up on formulas!
• Learn HashMaps: The Inventory Frequency Counter uses a HashMap, which is super common in coding tests.
• Test Edge Cases: For example, in the Jump Game, check if an array with one element or zeros works correctly.
• Time Management: TCS NQT has tight time limits. Solve easier problems first, then tackle complex ones like Jump Game.

Actually TCS CODING QUESTION Asked Shift wise 2024 (Video Solution)

TCS NQT 2024 Coding Questions Shift-wise with Video Solution

Q1: Problem Statement:
A device named ABC is placed on a rectangular grid with M rows and N columns.
The device starts at the top-left corner (1, 1) and must reach the bottom-right corner (M, N).
The device can only move East (→) or South (↓) at any step.

However, some cells are blocked (obstacles), and the device cannot pass through them.

Your task is to calculate the total number of unique paths from (1, 1) to (M, N) while avoiding all blocked cells.

Input Format:

• First line: Two integers M and N (number of rows and columns)
• Second line: An integer K (number of blocked cells)
• Next K lines: Each contains two integers r and c (row and column of a blocked cell)

Output Format:

• Print the total number of unique valid paths from (1,1) to (M,N)

Constraints:

• 1 ≤ M, N ≤ 100
• 0 ≤ K ≤ M×N

Example Input:

3 3 1 2 2 

Example Output:

2 

Explanation:

The possible paths from (1,1) to (3,3) are:

• (1,1) → (1,2) → (1,3) → (2,3) → (3,3)
• (1,1) → (2,1) → (3,1) → (3,2) → (3,3)

The path that goes through (2,2) is blocked.

https://www.youtube.com/embed/IfOcCTwUqqQ?si=dM1eRFSKtW1CbnzZ

Q2. Problem Statement:
You are given a list of n items sold in a store.
Each item has the following details:

• Name (string)
• Unit price (float)
• Quantity sold (integer)

Your task is to compute the following:

1. Total Sales Price — sum of (unit price × quantity) for all entries.
2. Average Sales Price — (Total Sales Price) ÷ n
3. Top Selling Product — item name with the highest total quantity sold (combine duplicates).

Input Format:

• First line: An integer n — number of items
• Next n lines: Each contains an item in the format:
  name unit_price quantity

Output Format:

• First line: Total Sales Price (rounded to 2 decimal places)
• Second line: Average Sales Price (rounded to 2 decimal places)
• Third line: Name of the top selling product

Example Input:

4 Apple 1.0 10 Orange 5.0 5 Banana 4.5 8 Apple 1.0 5 

Example Output:

87.5 21.88 Apple

Q3. Problem Statement:
You are given an integer N.
Your task is to print the sum of the multiplication table of N, up to N × 10.

Input Format:

• A single integer N (1 ≤ N ≤ 100)

Output Format:

• A single integer representing the sum of the first 10 multiples of N

Example Input:

10 

Example Output:

550 

Explanation:

Multiplication table of 10 up to 10:

10 × 1 = 10 10 × 2 = 20 10 × 3 = 30 ... 10 × 10 = 100 

Sum = 10 + 20 + 30 + … + 100 = 550

Q4. Problem Statement:
You are given an array of integers and a number K.
Your task is to print the maximum element in every contiguous subarray of size K.

Input Format:

• First line: A space-separated list of integers representing the array
• Second line: An integer K (size of the subarray)

Output Format:

• A single line containing space-separated integers — the maximum of each subarray of size K.

Constraints:

• 1 ≤ K ≤ length of array
• Array length ≤ 10⁵

Example Input:

1 4 7 7 6 8 3 

Example Output:

7 7 7 8 

Explanation:

Subarrays of size 3:

• [1, 4, 7] → 7
• [4, 7, 7] → 7
• [7, 7, 6] → 7
• [7, 6, 8] → 8

TCS Actual Coding Question | How to Take input?  | 29 April

Q5. Problem Statement:
You are given an integer N.
Your task is to calculate the sum of the first N terms of the Fibonacci series.

The Fibonacci series is defined as:
0, 1, 1, 2, 3, 5, 8, 13, ...
(i.e., F(0) = 0, F(1) = 1, F(n) = F(n-1) + F(n-2) for n ≥ 2)

Input Format:

• A single integer N (number of terms)

Output Format:

• An integer representing the sum of the first N Fibonacci numbers

Constraints:

• 1 ≤ N ≤ 100

Example Input:

5 

Example Output:

7 

Explanation:
Fibonacci terms: 0, 1, 1, 2, 3
Sum = 0 + 1 + 1 + 2 + 3 = 7

Q6: Problem Statement:
You are given an array of integers.
Your task is to perform bitwise OR operation on all possible subarrays and return the number of distinct results obtained.

Input Format:

• A single line containing space-separated integers representing the array.

Output Format:

• An integer representing the count of distinct values obtained from bitwise OR operations on all subarrays.

Constraints:

• 1 ≤ N ≤ 100 (N is the number of elements in the array)
• 1 ≤ arr[i] < 10⁹

Example:

Input:

1 2 3 

Output:

3 

Explanation:
Subarrays and their Bitwise OR values:

• [1] = 1
• [2] = 2
• [3] = 3
• [1,2] = 3
• [2,3] = 3
• [1,3] = 3
• [1,2,3] = 3
  Unique OR results: {1, 2, 3} → Count = 3

TCS Actual Coding Question | How to Take input?  | 30 April

Q7. Problem Statement:

You are given a database of student records. Each student record contains the following details:

• Name
• Age
• Grade (A, B, C, …)
• Gender (either “Male” or “Female”)

Your Task:

1. Return the names of all students whose age is greater than 20.
2. Return the average grade of all Female students in terms of their ASCII values (and then return the average as an integer).

Input Format:

First line contains an integer N — number of students Next N lines contain student records in the format: Name Age Grade Gender 

Output Format:

• First list: Names of students with age > 20 in a list format
• Second line: Average of grades (only of Female students), in ASCII value (rounded or floored as integer)

Example Input:

3 AAA 21 A Female BBB 24 B Male CCC 26 C Female 

Example Output:

['AAA', 'BBB', 'CCC'] 66 

Explanation:

• All students have age > 20, so all names are listed.
• Female students: AAA (Grade A = 65), CCC (Grade C = 67)
  → Average = (65 + 67) / 2 = 66

TCS Actual Coding Question | 03 April  | Shift 01

Q8. Problem Statement:
The organization has a data warehouse that stores 3-digit numbers.
Your task is to write a program that checks whether a given 3-digit number is divisible by 9 or not.

Input Format:

• A single integer N (a 3-digit number)

Output Format:

• Print "Number <N> is divisible by 9" if the number is divisible by 9
• Otherwise, print "Number <N> is not divisible by 9"

Constraints:

• 100 ≤ N ≤ 999

Examples:

Input:

236 

Output:

Number 236 is not divisible by 9 

Input:

162 

Output:

Number 162 is divisible by 9

Q9. Problem Statement:
You are given a list of numbers.
Your task is to return the maximum difference between a smaller number that appears before a larger number in the list.

Note:

• The smaller number must come before the larger number in the array (i.e., maintain order).
• If no such pair exists, return 0.

Input Format:

• First line: An integer n representing the number of elements
• Second line: n space-separated integers

Output Format:

• A single integer representing the maximum difference

Example Input:

7 -3 -5 1 6 -7 8 11 

Example Output:

18 

Explanation:

• Minimum value before 11 is -7
• 11 – (-7) = 18 → This is the maximum valid difference.

TCS Actual Coding Question | 03 April  | Shift 02

Q10: Problem Statement:
A person has many shoes of different sizes and orientations (Left ‘L’ or Right ‘R’).
You are given a list of shoes represented as size followed by either ‘L’ or ‘R’.
Your task is to calculate how many valid pairs of shoes can be formed.
A valid pair is made of one left and one right shoe of the same size.

Input Format:

• First line: An integer n (total number of shoes)
• Second line: n space-separated strings (e.g., 7L, 7R, etc.)

Output Format:

• A single integer representing the number of valid shoe pairs

Constraints:

• 1 ≤ n ≤ 1000

Example 1:

Input: 8 7L 7R 7L 8L 6R 7R 8R 6R Output: 3 

Explanation:

• 7L appears twice, 7R appears twice → 2 pairs
• 8L and 8R → 1 pair
• 6R appears twice, but no 6L → 0 pair
  → Total = 3 pairs