Scenario-Based Coding Questions on Nested Loops in C++

Test Case

Input:

Enter a number: 3

Output:

3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12
3 x 5 = 15
3 x 6 = 18
3 x 7 = 21
3 x 8 = 24
3 x 9 = 27
3 x 10 = 30

Explanation:

• The program asks the user to enter a number, which is 3 in this case.
• The outer while loop runs only once because j starts at 1 and increases to 2 after the first execution, making the condition false. Inside the outer loop, the inner while loop starts with 1 and goes up to 10.
• Each time, it multiplies 3 by the current number and prints the result.
• The numbers increase one by one: first 3 × 1 = 3, then 3 × 2 = 6, then 3 × 3 = 9, and so on until 3 × 10 = 30.Once 10 calculations are printed, the inner loop stops.
• The outer loop also stops since j is now 2, and the program ends.

#include <iostream>
using namespace std;

int main() {
    int num, i = 1, j = 1;

    // Taking user input for the number
    cout << "Enter a number: ";
    cin >> num;

    // Outer while loop to iterate from 1 to 1 (just to keep it nested)
    while (j == 1) {
        // Inner while loop to iterate from 1 to 10
        while (i <= 10) {
            // Printing the multiplication result
            cout << num << " x " << i << " = " << num * i << endl;
            i++;
        }

        // Exiting the outer while loop after the inner loop completes
        j++;
    }

    return 0;
}

Test Case

Input:

Enter number of rows: 2
Enter number of columns: 4

Output:

$ $ $ $
$ $ $ $

Explanation:

• Here, the user inputs 2 rows and 4 columns.
• The outer do-while loop will run twice (once for each row) because the value of i starts at 0 and increments after each iteration.
• Inside the outer loop, the inner do-while loop runs four times for each row because there are 4 columns. It prints the $ symbol each time.
• In the first iteration, the inner loop prints $ $ $ $. After the first row is printed, the program moves to the next line and prints the second row: $ $ $ $.
• Once the second row is printed, the outer loop ends, and the program stops.

#include <iostream>
using namespace std;

int main() {
    int rows, columns;

    // Taking user input for the grid size
    cout << "Enter number of rows: ";
    cin >> rows;
    cout << "Enter number of columns: ";
    cin >> columns;

    int i = 0;
    do {
        int j = 0;
        do {
            cout << "$ ";  // Print coin symbol
            j++;
        } while (j < columns);

        cout << endl; // Move to the next row
        i++;
    } while (i < rows);

    return 0;
}

Test Case

Input:

Enter number of steps: 4

Output:

#
# #
# # #
# # # #

Explanation:

• In this case, the user enters 4 steps as input.
• The outer for loop runs 4 times, once for each step (since the number of steps is 4).
• In the first iteration, the inner for loop runs once to print one # symbol, representing the first step.
• The same applies to the other remaining loops: the second iteration prints two ## symbols, the third prints three ### symbols, and the fourth prints four #### symbols.
• Each time the inner loop completes, the outer loop moves to the next line, creating the staircase effect.

#include <iostream>
using namespace std;

int main() {
    int steps;

    // Taking user input for the number of steps
    cout << "Enter number of steps: ";
    cin >> steps;

    // Outer loop for each step
    for (int i = 1; i <= steps; i++) {
        // Inner loop for printing the '#' symbols
        for (int j = 1; j <= i; j++) {
            cout << "# ";
        }
        cout << endl; // Move to the next step
    }

    return 0;
}

Test Case

Input:

Shelves = 2
Products per shelf = 3

Output:

X X X
X X X

Explanation:

• Following the user input, there are 2 shelves, and each shelf contains 3 products.
• The outer do-while loop manages the shelves. Since there are 2 shelves, it runs 2 times.
• For each shelf, the inner do-while loop handles the products. Since there are 3 products per shelf, the inner loop runs 3 times for each shelf.
• Each time the inner loop runs, it prints the letter ‘X’ to represent a product. After printing the products for one shelf, the program moves to the next line to print the next shelf.
• Once all shelves and products are printed, the program ends.

#include <iostream>
using namespace std;

int main() {
    int shelves, products;
    
    // Taking input from the user for shelves and products per shelf
    cout << "Enter number of shelves: ";
    cin >> shelves;
    cout << "Enter number of products per shelf: ";
    cin >> products;
    
    // Outer do-while loop for shelves
    int i = 0;
    do {
        // Inner do-while loop for products on each shelf
        int j = 0;
        do {
            cout << "X ";  // Print product 'X'
            j++;  // Move to the next product
        } while (j < products);
        
        cout << endl;  // Move to the next shelf
        i++;  // Move to the next shelf
    } while (i < shelves);

    return 0;
}

Test Case

Input:

Enter number of rows: 3
Enter number of columns: 4

Output:

(1-1) (1-2) (1-3) (1-4)
(2-1) (2-2) (2-3) (2-4)
(3-1) (3-2) (3-3) (3-4)

Explanation:

• For this case, the theatre has 3 rows and 4 columns.
• The outer for loop takes care of the rows. Since there are 3 rows, it runs 3 times.
• For each row, the inner for loop manages the columns. With 4 columns, the inner loop runs 4 times for each row.
• Each time the inner loop runs, it prints the seat in the format (row-column), such as (1-1), (1-2), and so on.
• After printing all the seats for one row, the program moves to the next line and prints the next row’s seats.
• Once all rows and columns are processed, the seat layout is displayed, and the program finishes.

#include <iostream>
using namespace std;

int main() {
    int rows, columns;

    // Taking user input for rows and columns
    cout << "Enter number of rows: ";
    cin >> rows;
    cout << "Enter number of columns: ";
    cin >> columns;

    // Outer loop for rows
    for (int i = 1; i <= rows; i++) {
        // Inner loop for columns
        for (int j = 1; j <= columns; j++) {
            cout << "(" << i << "-" << j << ") ";
        }
        cout << endl; // Move to the next row
    }

    return 0;
}

Test Case

Input:

Enter green light duration (seconds): 5

Output:

Green light active!
Second 1…
Second 2…
Second 3…
Second 4…
Second 5…
Light turns red!

Explanation:

• In our case here, the green light stays on for 5 seconds before turning red.
• The outer do-while loop manages the countdown, starting from 1 and running until it reaches 5.
• Each time the loop runs, it prints “Second X…”, where X represents the current second.
• Inside the outer loop, the inner do-while loop simulates a delay to represent the passing of time. This loop increments a variable until it reaches 1,000,000 (though in real systems, actual timing functions are used).
• Once all 5 seconds are counted down, the outer loop stops, and the program prints “Light turns red!” to indicate the signal change.

#include <iostream>
using namespace std;

int main() {
    int duration, second = 1;

    // Taking user input for green light duration
    cout << "Enter green light duration (seconds): ";
    cin >> duration;

    cout << "Green light active!\n";

    // Outer do-while loop for counting seconds
    do {
        cout << "Second " << second << "...\n";

        // Inner do-while loop to simulate a small delay for each second (for example purposes)
        int delay = 0;
        do {
            delay++;
        } while (delay < 1000000); // Simulate delay

        second++;
    } while (second <= duration); // Outer loop for the full duration

    cout << "Light turns red!\n";
    return 0;
}

Test Case

Input:

Enter number of rows: 5

Output:

1 2 3 4 5
1 2 3 4
1 2 3
1 2
1

Explanation:

• If we look at this case, the pyramid starts with 5 rows.
• The outer while loop begins with 5 and decreases by 1 after each row, controlling the number of values printed.
• Inside this loop, the inner while loop prints numbers from 1 to the current row count.
• For the first row, numbers 1 to 5 are printed. In the next row, only 1 to 4 appear, and so on.
• This process continues until the last row, which contains only 1, after which the program ends.

#include <iostream>
using namespace std;

int main() {
    int rows;

    // Taking user input for the number of rows
    cout << "Enter number of rows: ";
    cin >> rows;

    int i = rows;
    while (i >= 1) { // Loop runs from the largest row down to 1
        int j = 1;
        while (j <= i) { // Print numbers from 1 to current row count
            cout << j << " ";
            j++;
        }
        cout << endl; // Move to the next row
        i--;
    }

    return 0;
}

Test Case

Input:

Enter number of floors: 3
Enter number of rooms per floor: 4

Output:

Floor 3: Room 304 Room 303 Room 302 Room 301
Floor 2: Room 204 Room 203 Room 202 Room 201
Floor 1: Room 104 Room 103 Room 102 Room 101

Explanation:

• In this case, the number of floors is 3 and rooms per floor are 4.
• The outer for loop begins with the top floor (3) and decreases to 1, controlling the floor sequence.
• For each floor, the inner for loop prints the room numbers in reverse, from Room 4 to Room 1.
• On Floor 3, the output is Room 304 Room 303 Room 302 Room 301. The same pattern follows for Floor 2 and Floor 1.
• Once all floors and rooms are displayed, the program ends.

#include <iostream>
using namespace std;

int main() {
    int floors, rooms;

    // Taking user input for floors and rooms
    cout << "Enter number of floors: ";
    cin >> floors;
    cout << "Enter number of rooms per floor: ";
    cin >> rooms;

    // Outer loop for floors (reverse order)
    for (int i = floors; i >= 1; i--) {
        cout << "Floor " << i << ": ";
        
        // Inner loop for rooms (reverse order)
        for (int j = rooms; j >= 1; j--) {
            cout << "Room " << i << j << " ";
        }
        
        cout << endl;
    }

    return 0;
}

Test Case

Input:

Enter bell interval in minutes: 15
Enter total school hours: 3

Output:

8:0 AM
8:15 AM
8:30 AM
8:45 AM
9:0 AM
9:15 AM
9:30 AM
9:45 AM
10:0 AM
10:15 AM
10:30 AM
10:45 AM
School day over!

Explanation:

• The test case begins with a 15-minute interval and a 3-hour school duration as input.
• The outer do-while loop runs through each hour, ensuring the bell rings throughout the school day. Inside it, the inner do-while loop starts from 0 minutes and increases by 15 until the hour is up.
• The output begins with 8:00 AM, then 8:15 AM, 8:30 AM, and so on. When the minutes reach 60, the hour increments, moving to 9:00 AM and continuing the pattern.
• This process repeats until the final bell at 10:45 AM, after which the program announces “School day over!” and stops.

#include <iostream>
using namespace std;

int main() {
    int interval, hours;
    int time = 0, currentHour = 8; // Start at 8 AM

    // Taking user input
    cout << "Enter bell interval in minutes: ";
    cin >> interval;
    cout << "Enter total school hours: ";
    cin >> hours;

    cout << "Bell Ringing Schedule:\n";

    do {
        int minutes = 0;

        do {
            cout << currentHour << ":";

            // Print minutes with leading zero without using if-else
            cout << (minutes / 10) * 10 + (minutes % 10) << " AM\n";

            minutes += interval;
            time += interval;
        } while (minutes < 60);

        currentHour += 1;
    } while (time < hours * 60);

    cout << "School day over!\n";
    return 0;
}

Test Case

Input:

Enter the number of customers: 2
Enter the number of cups for customer 1: 2
Enter the number of cups for customer 2: 4

Output:

Customer 1:
Cup 1 – $5
Cup 2 – $5
Total: $10

Customer 2:
Cup 1 – $5
Cup 2 – $5
Cup 3 – $5
Cup 4 – $5
Total: $20

Explanation:

• In this case, there are 2 customers. The first customer orders 2 cups, while the second orders 4 cups.
• The outer for loop runs once per customer, prompting for the number of cups they order. Inside this loop, the inner for loop iterates based on the cup count, printing the cup number along with its price.
• Each cup is priced at $5, so for Customer 1, the program prints Cup 1 – $5 and Cup 2 – $5, followed by the total price of $10.
• For Customer 2, the process repeats, displaying 4 cups and a total price of $20. Once all customer orders are processed, the program ends.

#include <iostream>
using namespace std;

int main() {
    int customers, cups;
    const int price = 5; // Price per cup
  
    cout << "Enter the number of customers: ";
    cin >> customers;
	
  	//outer loop for no. of cups a customer ordered
    for (int i = 1; i <= customers; i++) {
        cout << "Enter the number of cups for customer " << i << ": ";
        cin >> cups;
        int total = 0;

        cout << "Customer " << i << ":" << endl; //customer no.
		
      	//inner loop for calculating the price of cups
        for (int j = 1; j <= cups; j++) {
            cout << "  Cup " << j << " - $" << price << endl;
            total += price;
        }
        cout << "  Total: $" << total << "\n" << endl;
    }
    return 0;
}

Test Case

Input:

Enter number of books returned late: 1

Output:

Book 1:
Day 1 – Rs. 5
Day 2 – Rs. 10
Day 3 – Rs. 20
Day 4 – Rs. 40
Day 5 – Rs. 80
Day 6 – Rs. 160
Day 7 – Rs. 320
Day 8 – Rs. 640
Max fine reached: Rs. 1000

Explanation:

• The no. of books is 1 in this case.
• The outer while loop runs for each book in this code, ensuring every overdue book is processed individually.
• The inner while loop handles the daily fine calculation.
• For Book 1, the fine increases from Rs. 5 to Rs. 640 on Day 8.
• Since the next fine would exceed Rs. 1000, the program stops and displays “Max fine reached: Rs. 1000” before moving to the next book.

#include <iostream>
using namespace std;

int main() {
    int books;

    // Taking user input for number of late books
    cout << "Enter number of books returned late: ";
    cin >> books;

    int book = 1;
    while (book <= books) {
        cout << "Book " << book << ":\n";
        
        int day = 1;
        int fine = 5; // Initial fine
        
        while (fine <= 1000) { // Stops at Rs. 1000
            cout << "Day " << day << " - Rs. " << fine << endl;
            fine *= 2; // Fine doubles each day
            day++;
        }

        cout << "Max fine reached: Rs. 1000\n\n";
        book++;
    }

    return 0;
}

Test Case

Input:

Enter the number of customers: 2
Enter the items for customer 1: 3
Enter the items for customer 2: 5

Output:

Total number of items ordered: 8

Explanation:

• Here, the number of customers is 2.
• The outer while loop runs for each customer, ensuring every customer’s order is processed individually.
• Inside the outer loop, the program asks how many items each customer ordered.
• For Customer 1, he ordered 3 items, so the inner while loop runs 3 times, increasing the totalItems count by 1 each time.
• For Customer 2, he ordered 5 items, so the inner while loop runs 5 times, and the totalItems count increases by 1 for each item.
• Once all customers are processed, the program displays the total number of items ordered, which is 8 in this case.

#include <iostream>
using namespace std;

int main() {
    int customers, totalItems = 0;

    cout << "Enter the number of customers: ";
    cin >> customers;

    int i = 1;
  
  	//outer loop asking how many items a customer bought
    while (i <= customers) {
        int items;
        cout << "Enter the items for customer " << i << ": ";
        cin >> items;
        
        // inner loop to accumulate items for the current customer
        int j = 1; 
        while (j <= items) {
            totalItems++; // Increase the total item count
            j++; // Move to the next item for this customer
        }

        i++; // Move to the next customer
    }

    cout << "Total number of items ordered: " << totalItems << endl;

    return 0;
}

Test Case

Input:

Enter a word: WOW
Enter repetition count: 3

Output:

WWW OOO WWW
WWW OOO WWW
WWW OOO WWW

Explanation:

• Here, the user inputs the word “WOW” with a repetition count of 3.
• The outer for loop runs 3 times, once for each row to be printed. For each iteration of the outer loop, the program loops through each character in the word.
• Inside the inner for loop, each character is repeated 3 times, based on the repetition count. So, for ‘W’, ‘O’, and ‘W’, it prints “WWW”, “OOO”, and “WWW” respectively.
• Once all characters are printed in a row, the program moves to the next row and repeats the process. After 3 rows are printed, the program ends.

#include <iostream>
using namespace std;

int main() {
    string word;   
    int repeat;    

    cout << "Enter a word: "; //Word entered for repitition
    cin >> word;

    cout << "Enter repetition count: "; //No. of repitition
    cin >> repeat;
	
  	// Outer loop to repeat the entire word 'repeat' number of times
    for (int i = 0; i < repeat; i++) {
        
      	// Loop through each letter in the word
        for (char letter : word) {
            
          	// Inner loop to repeat the current letter 'repeat' times
            for (int j = 0; j < repeat; j++) {
                cout << letter;  
            }
            cout << " ";
        }
        
        cout << endl;
    }

    return 0; 
}

Test Case

Input:

Enter a starting number: 2
Enter the last letter (A-Z): D

Output:

2A 2B 2C 2D
3A 3B 3C 3D
4A 4B 4C 4D
5A 5B 5C 5D

Explanation:

• The user inputs a starting number 2 and the last letter D in this case.
• The outer while loop runs for 4 iterations, starting from the given number 2 up to 5.
• For each iteration, the inner while loop starts with the letter A and iterates up to the user-defined letter D. It generates combinations like 2A, 2B, 2C, 2D for the first iteration, then moves on to 3A, 3B, 3C, 3D, and so on.
• After the program has printed all the rows, it stops.

#include <iostream>
using namespace std;

int main() {
    int startNum;
    char endLetter;
    
    cout << "Enter a starting number: ";
    cin >> startNum;
    cout << "Enter the last letter (A-Z): ";
    cin >> endLetter;
    
    int i = startNum;
  
  	// Outer loop: Generates 4 rows of numbers, starting from the given number
    while (i < startNum + 4) {  // 4 rows of numbers
        char ch = 'A';
      
      	// Inner loop: Iterates through letters from 'A' to the user-defined endLetter
        while (ch <= endLetter) {  // Loop through letters
            cout << i << ch << "  ";
            ch++;
        }
        cout << endl;
        i++;
    }
    
    return 0;
}

Test Case

Input:

Enter the number of customers: 1
Enter the number of products for customer 1: 2
Enter quantity for product 1: 2
Enter quantity for product 2: 3

Output:

Customer 1 purchased 5 items.

Explanation:

• Here, there is 1 customer, and he bought 2 products.
• The outer for loop takes care of the customers. Since there is only 1 customer, it runs once.
• For this customer, the inner for loop handles the products. With 2 products, the inner loop runs 2 times.
• Each time the inner loop runs, it asks for the quantity of each product and adds it to the total quantity for that customer.
• Once all products are processed, the program displays the total number of items purchased by the customer, and the program finishes.

#include <iostream>
using namespace std;

int main() {
    int customers;
    cout << "Enter the number of customers: ";
    cin >> customers;

    // Outer loop for customers
    for (int i = 1; i <= customers; i++) {
        int products;
        cout << "Enter the number of products for customer " << i << ": ";
        cin >> products;

        int totalQuantity = 0;

        // Inner loop for products
        for (int j = 1; j <= products; j++) {
            int quantity;
            cout << "Enter quantity for product " << j << ": ";
            cin >> quantity;
            totalQuantity += quantity;
        }
        cout << "Customer " << i << " purchased " << totalQuantity << " items." << endl;
    }

    return 0;
}