CODING SOLUTION - Placement Jobs & Materials
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πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. β€œ

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TCS NQT 2025

πŸŽ–Batch -        2020 - 2025

Test Date : 12th May 2025
          
https://www.tcsion.com/hub/national-qualifier-test/

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CODING SOLUTION - Placement Jobs & Materials pinned Β«TCS NQT 2025 πŸŽ–Batch -        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


πŸŽ–Batch -        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 πŸŽ–Batch -        2022  2023  2024  2025            https://careers.navi.com/navi/jobview/ai-solutions-engineer-1-bangalore-karnataka-india-2025040217480958?source=linkedin WhatsApp -   http…»
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TCS NQT 2025 | Full Mock Test PDF
πŸ”₯ 30+ High-Quality Questions | βœ… With Answers
πŸ“Š Numerical | ✍️ Verbal | 🧠 Reasoning | πŸ’» Coding

PDF Based on Real Exam Pattern
Boost your prep with the most expected questions!

Shared by: STUDY MATERIAL - Placement Jobs & Materials
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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 items…»
πŸ”₯ PLACEMENT PREP COMBO PACK πŸ”₯
Boost your logic + aptitude + coding skills in 5 mins!

1. Aptitude – 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 – List vs Tuple

List: Mutable
Example: my_list = [1, 2, 3]
You can modify β†’ my_list[0] = 10

Tuple: Immutable
Example: my_tuple = (1, 2, 3)
You cannot modify elements.


3. Mini Coding Task – 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«πŸ”₯ PLACEMENT PREP COMBO PACK πŸ”₯ Boost your logic + aptitude + coding skills in 5 mins! 1. Aptitude – 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)…»
πŸ”₯Problem StatementπŸ”₯


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«πŸ”₯Problem StatementπŸ”₯ 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…»
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