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TCS NQT 2025
πBatch - 2020 - 2025
Test Date : 12th May 2025
https://www.tcsion.com/hub/national-qualifier-test/
WhatsApp - https://bit.ly/3tcjxV3
CODING SOLUTION - Placement Jobs & Materials pinned Β«TCS NQT 2025 πBatch - 2020 - 2025 Test Date : 12th May 2025 https://www.tcsion.com/hub/national-qualifier-test/ WhatsApp - https://bit.ly/3tcjxV3Β»
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Navi Off Campus Drive
Job Profile : AI Solution Engineer 1
πBatch - 2022 2023 2024 2025
https://careers.navi.com/navi/jobview/ai-solutions-engineer-1-bangalore-karnataka-india-2025040217480958?source=linkedin
WhatsApp - https://bit.ly/3tcjxV3
CODING SOLUTION - Placement Jobs & Materials pinned Β«Navi Off Campus Drive Job Profile : AI Solution Engineer 1 πBatch - 2022 2023 2024 2025 https://careers.navi.com/navi/jobview/ai-solutions-engineer-1-bangalore-karnataka-india-2025040217480958?source=linkedin WhatsApp - httpβ¦Β»
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TCS NQT 2025 | Full Mock Test PDF
π₯ 30+ High-Quality Questions | β With Answers
π Numerical | βοΈ Verbal | π§ Reasoning | π» Coding
PDF Based on Real Exam Pattern
Boost your prep with the most expected questions!
Shared by: STUDY MATERIAL - Placement Jobs & Materials
Join for More Materials & Updates:
β‘οΈ https://t.me/studymaterial_IT
π₯ 30+ High-Quality Questions | β With Answers
π Numerical | βοΈ Verbal | π§ Reasoning | π» Coding
PDF Based on Real Exam Pattern
Boost your prep with the most expected questions!
Shared by: STUDY MATERIAL - Placement Jobs & Materials
Join for More Materials & Updates:
β‘οΈ https://t.me/studymaterial_IT
def update_list(val, items=[]):
items.append(val)
return items
print("Call 1:", update_list(10))
print("Call 2:", update_list(20))
def fresh_list(val, items=None):
if items is None:
items = []
items.append(val)
return items
print("Call 3:", fresh_list(30))
print("Call 4:", fresh_list(40))
Options:
A.
Call 1: [10]
Call 2: [10, 20]
Call 3: [30]
Call 4: [30, 40]
B.
Call 1: [10]
Call 2: [20]
Call 3: [30]
Call 4: [40]
C.
Call 1: [10]
Call 2: [10, 20]
Call 3: [30]
Call 4: [30]
D.
Error due to list mutation
Answer: A
CODING SOLUTION - Placement Jobs & Materials pinned Β«def update_list(val, items=[]): items.append(val) return items print("Call 1:", update_list(10)) print("Call 2:", update_list(20)) def fresh_list(val, items=None): if items is None: items = [] items.append(val) return itemsβ¦Β»
π₯ PLACEMENT PREP COMBO PACK π₯
Boost your logic + aptitude + coding skills in 5 mins!
1. Aptitude β Profit & Loss
A shopkeeper buys an item for Rs. 240 and sells it at a profit of 25%.
Selling Price = ?
Solution:
SP = CP Γ (1 + Profit%)
= 240 Γ (1 + 25/100)
= 240 Γ 1.25 = Rs. 300
2. Python Concept β List vs Tuple
List: Mutable
Example:
You can modify β
Tuple: Immutable
Example:
You cannot modify elements.
3. Mini Coding Task β Count Vowels in String
Output: 3
Boost your logic + aptitude + coding skills in 5 mins!
1. Aptitude β Profit & Loss
A shopkeeper buys an item for Rs. 240 and sells it at a profit of 25%.
Selling Price = ?
Solution:
SP = CP Γ (1 + Profit%)
= 240 Γ (1 + 25/100)
= 240 Γ 1.25 = Rs. 300
2. Python Concept β List vs Tuple
List: Mutable
Example:
my_list = [1, 2, 3]You can modify β
my_list[0] = 10Tuple: Immutable
Example:
my_tuple = (1, 2, 3)You cannot modify elements.
3. Mini Coding Task β Count Vowels in String
def count_vowels(s): return sum(1 for ch in s.lower() if ch in 'aeiou') # Example: count_vowels("placement") β Output: 3
CODING SOLUTION - Placement Jobs & Materials pinned Β«π₯ PLACEMENT PREP COMBO PACK π₯ Boost your logic + aptitude + coding skills in 5 mins! 1. Aptitude β Profit & Loss A shopkeeper buys an item for Rs. 240 and sells it at a profit of 25%. Selling Price = ? Solution: SP = CP Γ (1 + Profit%) = 240 Γ (1 + 25/100)β¦Β»
π₯Problem Statementπ₯
A robot is stuck in a maze represented by an NxN grid. The grid contains:
'S': Starting point
'E': Ending point
'.': Empty cell (can move)
'#': Wall (cannot move)
'P': Portal (teleports to the next portal clockwise)
Rules:
Robot can move up, down, left, right.
If robot steps on 'P', it is instantly teleported to the next portal (clockwise).
You need to find the minimum number of steps to reach 'E' from 'S'.
C++
Telegram : http://t.me/codingsolution_IT
A robot is stuck in a maze represented by an NxN grid. The grid contains:
'S': Starting point
'E': Ending point
'.': Empty cell (can move)
'#': Wall (cannot move)
'P': Portal (teleports to the next portal clockwise)
Rules:
Robot can move up, down, left, right.
If robot steps on 'P', it is instantly teleported to the next portal (clockwise).
You need to find the minimum number of steps to reach 'E' from 'S'.
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef pair<int, int> pii;
int N;
vector<vector<char>> grid;
vector<pii> portals;
bool isValid(int x, int y, vector<vector<bool>>& visited) {
return x >= 0 && x < N && y >= 0 && y < N && !visited[x][y] && grid[x][y] != '#';
}
pii getNextPortal(pii current) {
for (int i = 0; i < portals.size(); i++) {
if (portals[i] == current) {
return portals[(i + 1) % portals.size()];
}
}
return current;
}
int shortestPathWithPortals() {
pii start, end;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (grid[i][j] == 'S') start = {i, j};
else if (grid[i][j] == 'E') end = {i, j};
else if (grid[i][j] == 'P') portals.push_back({i, j});
}
}
vector<vector<bool>> visited(N, vector<bool>(N, false));
queue<pair<pii, int>> q;
q.push({start, 0});
visited[start.first][start.second] = true;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
while (!q.empty()) {
pii curr = q.front().first;
int steps = q.front().second;
q.pop();
if (curr == end) return steps;
for (int d = 0; d < 4; d++) {
int nx = curr.first + dx[d];
int ny = curr.second + dy[d];
if (isValid(nx, ny, visited)) {
if (grid[nx][ny] == 'P') {
pii tele = getNextPortal({nx, ny});
if (!visited[tele.first][tele.second]) {
visited[tele.first][tele.second] = true;
q.push({tele, steps + 1});
}
} else {
visited[nx][ny] = true;
q.push({{nx, ny}, steps + 1});
}
}
}
}
return -1;
}
int main() {
cin >> N;
grid = vector<vector<char>>(N, vector<char>(N));
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
cin >> grid[i][j];
cout << shortestPathWithPortals() << endl;
return 0;
}
C++
Telegram : http://t.me/codingsolution_IT
CODING SOLUTION - Placement Jobs & Materials pinned Β«π₯Problem Statementπ₯ A robot is stuck in a maze represented by an NxN grid. The grid contains: 'S': Starting point 'E': Ending point '.': Empty cell (can move) '#': Wall (cannot move) 'P': Portal (teleports to the next portal clockwise) Rules: Robotβ¦Β»
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Cognizant Test Preparation β Study Material
import heapq
def max_recipes_cooked(N, C, T):
C.sort() # Sort recipes by cooking time
stove_heap = [0] * N # Time occupied per stove (min-heap)
for time in C:
# Pop stove with least time
earliest_free_time = heapq.heappop(stove_heap)
if earliest_free_time + time <= T:
# Assign recipe to this stove
heapq.heappush(stove_heap, earliest_free_time + time)
else:
# Can't cook this recipe, push back the original time
heapq.heappush(stove_heap, earliest_free_time)
# Count stoves used for cooking (total recipes completed)
return sum(t > 0 for t in stove_heap)
Example
N = 2
C = [2, 3, 4, 5, 9]
T = 10
print(max_recipes_cooked(N, C, T)) # Output will depend on input
Ini
Distance = a * (1 + 2 + 3 + ... + t) = a * t * (t + 1) / 2
Ini
t = int(S * 60)
Python
def distance_covered(N, A, X, S):
# Convert minutes to seconds
t = int(S * 60)
# Acceleration of Xth car
acceleration = A[X]
# Total distance = a * t * (t + 1) // 2
distance = acceleration * t * (t + 1) // 2
return distance
Example
Python
N = 3
A = [1, 2, 3]
X = 1
S = 1.5
print(distance_covered(N, A, X, S)) # Output: 2 * 90 * 91 // 2 = 8190
input2: Array A (acceleration of each car)
input3: X (index of car to check distance)
input4: S (time in minutes, decimal allowed)
Speed increases like: a, 2a, 3a, ..., ta
Distance = a + 2a + 3a + ... + ta = a Γ (1 + 2 + 3 + ... + t) = a Γ (t Γ (t + 1)) / 2
def total_distance(N, A, X, S):
# Convert minutes to seconds
T = int(float(S) * 60)
# Get acceleration of X-th car (0-based index)
a = A[X]
# Use formula: distance = a Γ t(t+1)/2
distance = a * T * (T + 1) // 2
return distance
# Example usage
N = 3
A = [2, 4, 3]
X = 1
S = 0.5
print(total_distance(N, A, X, S)) # Output: Distance covered by car at index 1 in 30 sec
public class CarDistanceCalculator {
public static int totalDistance(int N, int[] A, int X, double S) {
// Minutes ko seconds me convert karo
int T = (int)(S * 60);
// X-th car ki acceleration le lo (0-based index)
int a = A[X];
// Distance = a Γ T Γ (T + 1) / 2
int distance = a * T * (T + 1) / 2;
return distance;
}
public static void main(String[] args) {
// Example inputs
int N = 3;
int[] A = {2, 4, 3};
int X = 1;
double S = 0.5;
int result = totalDistance(N, A, X, S);
System.out.println("Distance covered: " + result);
}
}
Input Explanation:
N = 3 cars
Accelerations: [2, 4, 3]
X = 1 means 2nd car (0-based indexing)
S = 0.5 minutes = 30 seconds// Function to check if the array can be built using subarrays of prefix
public static boolean canBuild(int[] prefix, int[] arr) {
List<Integer> list = new ArrayList<>();
for (int num : prefix) list.add(num);
int i = prefix.length;
while (i < arr.length) {
boolean matched = false;
// Try all subarrays of prefix
for (int start = 0; start < prefix.length; start++) {
for (int end = start + 1; end <= prefix.length; end++) {
List<Integer> sub = list.subList(start, end);
if (i + sub.size() <= arr.length) {
boolean same = true;
for (int j = 0; j < sub.size(); j++) {
if (arr[i + j] != sub.get(j)) {
same = false;
break;
}
}
if (same) {
i += sub.size();
matched = true;
break;
}
}
}
if (matched) break;
}
if (!matched) return false;
}
return true;
}
public static int findMinOriginalLength(int[] arr) {
for (int len = 1; len <= arr.length; len++) {
int[] prefix = Arrays.copyOfRange(arr, 0, len);
if (canBuild(prefix, arr)) {
return len;
}
}
return arr.length;
}
public static void main(String[] args) {
int[] arr = {5, 4, 7, 2, 7, 4, 4, 7, 7, 2};
int result = findMinOriginalLength(arr);
System.out.println(result);
}
}
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