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HCL_C++_Study_Material.pdf0.03 KB

CGI Interview ExperienceTechnical Round: 1. Write a program to check palindrome string. 2. Implement a binary search algorithm. 3. Write a code to find the factorial of a number. 4. Find the duplicate elements in an array. 5. Basic SQL queries like: •Find second highest salary •Count number of employees in each department

from collections import deque

def heat_spread(grid):
    n = len(grid)
    m = len(grid[0])
    new_grid = [row[:] for row in grid]
    directions = [(-1,0), (1,0), (0,-1), (0,1)]

    for i in range(n):
        for j in range(m):
            if grid[i][j] == 'H':
                for dx, dy in directions:
                    ni, nj = i + dx, j + dy
                    if 0 <= ni < n and 0 <= nj < m and grid[ni][nj] == '.':
                        new_grid[ni][nj] = 'H'
    return new_grid

def simulate_heat(grid, start_x, start_y, end_x, end_y):
    n = len(grid)
    m = len(grid[0])
    queue = deque()
    visited = set()
    queue.append((start_x, start_y, 0, grid))
    visited.add((start_x, start_y))

    while queue:
        x, y, days, curr_grid = queue.popleft()
        if (x, y) == (end_x, end_y):
            return days

        next_grid = heat_spread(curr_grid)
        directions = [(-1,0), (1,0), (0,-1), (0,1)]

        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < m and next_grid[nx][ny] == '.' and (nx, ny) not in visited:
                visited.add((nx, ny))
                queue.append((nx, ny, days + 1, next_grid))
    return -1

# Input Format
n = int(input())
grid = []
for _ in range(n):
    grid.append(list(input().strip()))

start_x, start_y = map(int, input().split())
end_x, end_y = map(int, input().split())

print(simulate_heat(grid, start_x, start_y, end_x, end_y))
City Heatwave – Codevita PYTHON

📌 Here is a Previous Year Coding Question asked in Wipro / Cognizant placement test: ### Question 1: Count Frequency of Words in a Sentence Problem:  Write a program that takes a sentence and prints the frequency of each word, sorted alphabetically. Input:  "the quick brown fox jumps over the lazy dog" Expected Output: brown: 1 dog: 1 fox: 1 jumps: 1 lazy: 1 over: 1 quick: 1 the: 2 Python Code:
def count_word_frequency(sentence):
    # Remove punctuation and convert to lowercase
    sentence = sentence.lower()
    
    # Split the sentence into words
    words = sentence.split()
    
    # Create a dictionary to store frequency
    frequency = {}
    
    for word in words:
        if word in frequency:
            frequency[word] += 1
        else:
            frequency[word] = 1
    
    # Sort dictionary by key (alphabetically)
    for word in sorted(frequency):
        print(f"{word}: {frequency[word]}")

# Example
input_sentence = "the quick brown fox jumps over the lazy dog the"
count_word_frequency(input_sentence)
--- ### Question 2: Check if a Matrix is Symmetric Problem:  Write a program to check if a square matrix is symmetric (i.e., matrix[i][j] == matrix[j][i]). Input:
[
[1, 2, 3],
[2, 4, 5],
[3, 5, 6]
]
Output: The matrix is symmetric. Python Code:
def is_symmetric(matrix):
    n = len(matrix)
    for i in range(n):
        for j in range(n):
            if matrix[i][j] != matrix[j][i]:
                return False
    return True

# Example matrix
matrix = [
    [1, 2, 3],
    [2, 4, 5],
    [3, 5, 6]
]

if is_symmetric(matrix):
    print("The matrix is symmetric.")
else:
    print("The matrix is not symmetric.")

// 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);
    }
}

4

Input: N = 10 A = [5, 4, 7, 2, 7, 4, 4, 7, 7, 2] Output: 4 This means smallest array [5, 4, 7, 2] can be used to form the final array by repeating some parts.`

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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 × T × (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

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 × (1 + 2 + 3 + ... + t) = a × (t × (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 × 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

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): #
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

import heapq def max_recipes_cooked(N, C, T): &nbsp;&nbsp;&nbsp; C.sort()&nbsp; # Sort recipes by cooking time &nbsp;&nbsp;&n
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

Cognizant Test Preparation – Study Material

🔥Problem Statement🔥 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

🔥 PLACEMENT PREP COMBO PACK 🔥 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 × (1 + Profit%) = 240 × (1 + 25/100) = 240 × 1.25 = Rs. 300 2. Python Concept – List vs Tuple List: Mutable Example: my_list = [1, 2, 3] You can modify → 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") → 
Output: 3

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

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