C++ Loops Exercises: 30+ Coding Problems with Solutions

#include <iostream>

int main() {
    std::cout << "Numbers between 1 and 10 are:"<<std::endl;
    
    // Method 2: Start at 2 and increment by 2
    for (int i = 1; i <= 10; i++) {
        std::cout << i << " ";
    }
    
    std::cout << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program starts by reading the limit N from the user. The for loop is structured with three parts:

1. Initialization: int i = 1 sets the counter to the starting value.
2. Condition: i <= N ensures the loop continues as long as the counter is less than or equal to the target number.
3. Update: ++i increments the counter by 1 after each iteration. Inside the loop, std::cout << i << " "; prints the current value of the counter, effectively printing every number from 1 to N.

#include <iostream>

int main() {
    std::cout << "Even numbers between 1 and 20 are:\n";
    
    // Method 2: Start at 2 and increment by 2
    for (int i = 2; i <= 20; i += 2) {
        std::cout << i << " ";
    }
    
    std::cout << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

• This solution uses the more efficient approach (Method 2).
• The for loop is initialized with int i = 2, as 2 is the first even number in the range. The condition i <= 20 ensures we don’t exceed the limit.
• Crucially, the update statement is i += 2, which means the counter skips all odd numbers, jumping directly from one even number to the next (e.g., 2,4,6,8,…). This reduces the total number of loop iterations by half compared to checking every number.

#include <iostream>

int main() {
    int N = 10;
    long long sum = 0; // Use long long for the sum to prevent overflow for large N
    int i = 1;
    
    // The 'while' loop runs as long as the counter 'i' is less than or equal to N
    while (i <= N) {
        sum += i; // Add the current number to the running sum
        i++;      // Increment the counter
    }
    
    std::cout << "The sum of the first " << N << " natural numbers is: " << sum << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

A while loop is used here. We initialize sum to 0 and the counter i to 1. The loop condition i <= N checks if there are still numbers to be added. Inside the loop:

1. sum += i; adds the current value of i to the total sum.
2. i++; increments the counter. This is crucial for two reasons: it moves to the next number, and it ensures the loop condition will eventually become false, preventing an infinite loop. Once i exceeds N, the loop terminates, and the final sum is printed.

#include <iostream>

int main() {
    int N = 10;
    long long evenSum = 0;
    long long oddSum = 0;
    
    // Loop from 1 to N
    for (int i = 1; i <= N; ++i) {
        // Check if the current number 'i' is even or odd
        if (i % 2 == 0) {
            evenSum += i;
        } else {
            oddSum += i;
        }
    }
    
    std::cout << "Results up to " << N << ":" << std::endl;
    std::cout << "Sum of even numbers: " << evenSum << std::endl;
    std::cout << "Sum of odd numbers: " << oddSum << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program initializes two accumulator variables, evenSum and oddSum, to 0. A for loop iterates through every number i from 1 to N. Inside the loop, the expression i % 2 == 0 is checked.

• If the remainder of i divided by 2 is 0, the number is even, and it’s added to evenSum.
• Otherwise (in the else block), the number is odd, and it’s added to oddSum. This single loop efficiently classifies and sums all the numbers in the given range.

#include <iostream>

int main() {
    int n = 5;
    long long factorial = 1; // Start with 1 as 0! = 1 and it's the multiplication identity
    
    // Handle the special case for 0!
    if (n == 0) {
        // Factorial is already 1, nothing more to do
    } else {
        // Loop from 1 up to n
        for (int i = 1; i <= n; ++i) {
            factorial *= i; // Equivalent to: factorial = factorial * i;
        }
    }
    
    std::cout << n << "! (Factorial of " << n << ") is: " << factorial << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program handles input validation and the edge case where n=0 (where 0!=1). For any n>0, the for loop executes. The variable factorial is initialized to 1. The loop runs from i=1 to i=n.

• In the first iteration, factorial becomes 1×1=1.
• In the second, factorial becomes 1×2=2.
• In the third, factorial becomes 2×3=6. …and so on, accumulating the product until the loop finishes. Using long long for the factorial variable helps accommodate the rapid growth of factorials (e.g., 13! exceeds the capacity of a standard 32-bit int).

#include <iostream>

int main() {
    int number = 2;    
    std::cout << "Multiplication Table for " << number << " (1 to 10):" <<std::endl;
    
    // Loop from 1 to 10
    for (int i = 1; i <= 10; ++i) {
        int product = number * i;
        
        // Print the result in the format: 5 x 1 = 5
        std::cout << number << " x " << i << " = " << product << std::endl;
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program reads the number for which the table is needed. The for loop controls the multiplier i, which ranges from 1 to 10. In each iteration:

1. The product is calculated by multiplying the user’s number by the current loop multiplier i.
2. std::cout prints the full line in the standard multiplication table format. The loop ensures that the calculation and output are performed exactly 10 times, covering the full range requested.

#include <iostream>

int main() {
    long long number = 7568;
    int count = 0;
    
    // Handle the special case where the input is 0
    if (number == 0) {
        count = 1;
    } else {
        // Use the absolute value to handle negative numbers (e.g., -123 has 3 digits)
        long long tempNumber = std::abs(number);
        
        // Loop continues until the number is reduced to 0
        while (tempNumber > 0) {
            tempNumber /= 10; // Remove the last digit
            count++;          // Increment the digit count
        }
    }
    
    std::cout << "The number " << number << " has " << count << " digits." << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The solution uses a while loop to repeatedly remove the last digit of the number until the number becomes 0.

1. The initial check handles the input 0, which has 1 digit.
2. For non-zero numbers, the absolute value is used to correctly count digits for negative inputs (e.g., −7568).
3. The condition while (tempNumber > 0) keeps the loop running.
4. The update tempNumber /= 10 is integer division; for 7568, it becomes 756, then 75, then 7, then 0.
5. The counter count is incremented in each step, representing the digit that was just “removed.”

#include <iostream>

int main() {
    int originalNumber = 7568;
    long long reversedNumber = 0; // Use long long for safety
    
    int tempNumber = originalNumber; // Work with a copy
    
    // Loop until all digits are processed
    while (tempNumber != 0) {
        int digit = tempNumber % 10;         // Get the last digit
        reversedNumber = reversedNumber * 10 + digit; // Build the reversed number
        tempNumber /= 10;                    // Remove the last digit
    }
    
    std::cout << "The reverse of " << originalNumber << " is: " << reversedNumber << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This C++ program takes an integer input from the user and prints its reverse.

1. It first asks the user to enter an integer and validates the input.
2. It then copies that number into a temporary variable tempNumber.
3. Inside a while loop, it repeatedly extracts the last digit using % 10, adds it to reversedNumber (after shifting existing digits left by multiplying by 10), and removes the last digit from tempNumber using /= 10.
4. The loop continues until all digits are processed.
5. Finally, it displays the reversed number.
6. If input = 1234, then output = 4321.

#include <iostream>

int main() {
    int n = 121;
    long long reversedN = 0;
    
    int originalN = n; // Store the original number for comparison
    int tempN = n;     // Use a temporary variable for manipulation
    
    // Handle negative numbers (e.g., -121 is usually not considered a palindrome)
    if (n < 0) {
        tempN = -n;
        originalN = -n;
    }
    
    // Loop to reverse the number
    while (tempN != 0) {
        int digit = tempN % 10;
        reversedN = reversedN * 10 + digit;
        tempN /= 10;
    }
    
    // Compare the original and reversed numbers
    if (reversedN == originalN) {
        std::cout << originalN << " is a **palindrome**." << std::endl;
    } else {
        std::cout << originalN << " is **not** a palindrome." << std::endl;
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

• This solution first copies the input n into originalN (which must not be changed) and tempN (which the loop modifies).
• The while loop is identical to the one used in the reversal problem (Exercise 8). After the loop completes, reversedN holds the reversed version of the input.
• The program then uses a simple if statement to compare reversedN with the invariant originalN. If they are equal, the number is a palindrome (e.g., 121 reversed is 121).

#include <iostream>

int main() {
    int number = 1234;
    int sumOfDigits = 0;
    
    // Use the absolute value of the number for the calculation
    int tempNumber = std::abs(number);
    
    // Loop until the temporary number is reduced to 0
    while (tempNumber > 0) {
        int digit = tempNumber % 10; // Extract the last digit
        sumOfDigits += digit;        // Add the digit to the sum
        tempNumber /= 10;            // Remove the last digit
    }
    
    std::cout << "The sum of the digits of " << number << " is: " << sumOfDigits << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This problem uses the same digit-extraction pattern as the reversal and digit-counting problems. The while loop continuously breaks down the number.

1. tempNumber % 10 isolates the rightmost digit.
2. sumOfDigits += digit adds this digit to the total.
3. tempNumber /= 10 shifts the remaining digits one position right, effectively dropping the digit that was just processed. The loop terminates when tempNumber becomes 0. For example, with 456: the digits 6, 5, and 4 are extracted sequentially, and 6+5+4=15. The use of std::abs() ensures the logic works correctly for negative numbers, as the sum of digits is usually considered the sum of the absolute values of the digits

#include <iostream>

int main() {
    int base = 2, exponent = 4;
    long long result = 1; // Use long long for the result as power values grow fast
    
    // Any number raised to the power of 0 is 1 (except 0^0, which is undefined/1)
    if (exponent == 0) {
        // result remains 1, no loop needed
    } else {
        // Loop 'exponent' number of times
        for (int i = 0; i < exponent; ++i) {
            result *= base; // result = result * base;
        }
    }
    
    std::cout << base << " raised to the power of " << exponent << " is: " << result << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

1. It takes two inputs: base and exponent.
2. It validates both inputs — the exponent must be non-negative.
3. It initializes result to 1.
4. If the exponent is 0, the result remains 1 (since any number to the power 0 is 1).
5. Otherwise, it multiplies result by base repeatedly in a for loop that runs exponent times.
6. Finally, it prints the calculated power.

#include <iostream>

// Helper function for manual power calculation (as per the spirit of Ex 11)
long long calculatePower(int base, int exponent) {
    long long res = 1;
    for (int i = 0; i < exponent; ++i) {
        res *= base;
    }
    return res;
}

int main() {
    int originalN = 153;
    long long sumOfPowers = 0;
    int digitCount = 0;
    
    int tempN = originalN; 
    
    // 1. Loop to count the number of digits (D)
    while (tempN > 0) {
        tempN /= 10;
        digitCount++;
    }
    
    tempN = originalN; // Reset tempN to original value
    
    // 2. Loop to calculate the sum of each digit raised to the power D
    while (tempN > 0) {
        int digit = tempN % 10;
        sumOfPowers += calculatePower(digit, digitCount);
        tempN /= 10;
    }
    
    // 3. Comparison
    if (sumOfPowers == originalN) {
        std::cout << originalN << " is an Armstrong number." << std::endl;
    } else {
        std::cout << originalN << " is not an Armstrong number." << std::endl;
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program is solved using two consecutive loops.

1) The first while loop determines D, the total number of digits in the number.
2) The second while loop then iterates over the digits again:

• It extracts the digit using the modulo operator.
• It uses the helper function calculatePower(digit, digitCount) (which uses an internal for loop) to raise the digit to the power D.
• It adds this result to sumOfPowers. Finally, the result is compared to the originalN. If they match, the number is Armstrong.

#include <iostream>

int main() {
    int N = 8;
    long long n1 = 0, n2 = 1, nextTerm;
    
    std::cout << "Fibonacci Series up to " << N << " terms:\n";
    
    if (N >= 1) {
        std::cout << n1; // Print 0
    }
    if (N >= 2) {
        std::cout << " " << n2; // Print 1
    }
    
    // Loop starting from the 3rd term (i=3) up to N
    for (int i = 3; i <= N; ++i) {
        nextTerm = n1 + n2; // Calculate the next term
        std::cout << " " << nextTerm;
        
        // Update terms for the next iteration
        n1 = n2;
        n2 = nextTerm;
    }
    
    std::cout << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program handles the edge cases N=1 and N=2 separately since they are fixed starting values. For N≥3, the for loop takes over. The crucial part is the variable update:

• nextTerm = n1 + n2: The new term is calculated from the two previous terms (n1 and n2).
• n1 = n2: The second-to-last term (n1) is updated to become the previous term (n2).
• n2 = nextTerm: The previous term (n2) is updated to become the newly calculated term (nextTerm). This shifting process effectively moves the calculation forward, always maintaining the last two terms needed to find the next number in the series.

#include <iostream>
#include <algorithm> // Required for std::min

int main() {
    int A = 5, B = 10, gcd = 1;

    int minimum = std::min(A, B);
    
    // Loop from 1 up to the smallest of A or B
    for (int i = 1; i <= minimum; ++i) {
        // Check if i divides both A and B perfectly
        if (A % i == 0 && B % i == 0) {
            gcd = i; // Store this factor. The last one stored will be the GCD.
        }
    }
    
    std::cout << "The GCD (HCF) of " << A << " and " << B << " is: " << gcd << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

• The program finds the smaller of the two numbers using std::min() to set the limit for the for loop, as the GCD cannot be larger than the smallest number.
• The loop iterates through every number i from 1 up to this limit.
• The condition if (A % i == 0 && B % i == 0) checks if i is a common factor of both A and B. When a common factor is found, the gcd variable is updated.
• Because the loop progresses in increasing order, the last common factor stored in gcd upon loop termination will necessarily be the Greatest Common Divisor.

#include <iostream>
#include <algorithm> // Required for std::max

int main() {
    int A = 45, B = 10, larger;
   
    // The LCM must be at least as large as the larger number
    larger = std::max(A, B);
    
    // Loop starting from the larger number
    int i = larger;
    while (true) { 
        // Check if 'i' is divisible by both A and B
        if (i % A == 0 && i % B == 0) {
            std::cout << "The LCM of " << A << " and " << B << " is: " << i << std::endl;
            break; // Exit the loop immediately once the first multiple is found
        }
        i++; // Check the next integer
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program efficiently searches for the LCM by starting the search at max(A,B) (since the LCM cannot be smaller than the largest number). A while(true) loop is used to iterate through candidate numbers i. In each iteration:

• The condition i % A == 0 && i % B == 0 checks if i is a common multiple of A and B.
• The first time this condition is met, i is the smallest number that satisfies the condition, thus it is the Least Common Multiple (LCM).
• The break statement immediately terminates the loop once the LCM is found. If the condition is false, i is incremented, and the next candidate number is checked.

#include <iostream>
#include <string>
#include <cctype> // Required for tolower() and isalpha()

int main() {
    std::string text = "PYnative";
    int vowelCount = 0;
    int consonantCount = 0;

    // Loop through each character in the string
    for (char c : text) { // C++11 range-based for loop
        // 1. Convert to lowercase for case-insensitive checking
        char lowerC = std::tolower(c);
        
        // 2. Check if the character is an alphabet
        if (std::isalpha(lowerC)) {
            // 3. Check if it's a vowel
            if (lowerC == 'a' || lowerC == 'e' || lowerC == 'i' || lowerC == 'o' || lowerC == 'u') {
                vowelCount++;
            } else {
                // 4. If it's an alphabet but not a vowel, it's a consonant
                consonantCount++;
            }
        }
    }
    
    std::cout << "Analysis of the string: "<< text << std::endl;
    std::cout << "Total Vowels: " << vowelCount << std::endl;
    std::cout << "Total Consonants: " << consonantCount << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The solution uses a range-based for loop (a concise C++ feature) to iterate over every char in the input text.

• std::tolower(c) ensures that comparisons are case-insensitive (e.g., ‘A’ is treated the same as ‘a’).
• std::isalpha(lowerC) filters out non-alphabetic characters (numbers, spaces, punctuation). Only letters proceed to the next step.
• The if condition explicitly checks for the five vowel characters.
• The else block catches all characters that passed the isalpha check but failed the vowel check, classifying them as consonants.

#include <iostream>

int main() {
    int arr[5] = {10, 20, 30, 40, 50};
    long long sum = 0;
    
    // 2. Loop to calculate the sum (using range-based for loop)
    for (int element : arr) {
        sum += element;
    }
    std::cout << "\nThe sum of all elements in the array is: " << sum << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

• The for loop uses a range-based loop (for (int element : arr)). This construct iterates over the elements of the array directly, assigning the value of the current element to the variable element in each step.
• sum += element; accumulates the total. This approach is cleaner and less error-prone than managing an index variable manually for summation.

#include <iostream>
#include <vector>
#include <limits> // Required for std::numeric_limits

int main() {
    int size;
    
    std::cout << "Enter the number of elements: ";
    if (!(std::cin >> size) || size <= 0) return 1;
    
    std::vector<int> arr(size);
    
    // Input loop omitted for brevity, assuming arr is filled.
    // For demonstration, let's assume the array is filled:
    std::cout << "Enter " << size << " integers:\n";
    for (int i = 0; i < size; ++i) {
        std::cout << "Element " << i + 1 << ": ";
        if (!(std::cin >> arr[i])) return 1;
    }
    
    // Check for empty array case
    if (size == 0) {
        std::cout << "Array is empty." << std::endl;
        return 0;
    }
    
    // Initialize max and min to the first element's value
    int maxElement = arr[0];
    int minElement = arr[0];
    
    // Loop through the array starting from the second element (index 1)
    for (int i = 1; i < size; ++i) {
        if (arr[i] > maxElement) {
            maxElement = arr[i]; // New maximum found
        }
        
        if (arr[i] < minElement) {
            minElement = arr[i]; // New minimum found
        }
    }
    
    std::cout << "\nMaximum element is: " << maxElement << std::endl;
    std::cout << "Minimum element is: " << minElement << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The solution uses a std::vector for dynamic array handling.

After the array is populated, the maxElement and minElement variables are initialized to the value of the first element (arr[0]). This guarantees that the initial values are actual numbers present in the array, making the comparison accurate regardless of the potential range of the array values.

The for loop then starts from the second element (index 1). In each step:

• if (arr[i] > maxElement) updates the maximum if the current element is larger.
• if (arr[i] < minElement) updates the minimum if the current element is smaller. By the time the loop finishes iterating through all elements, maxElement and minElement will hold the correct extreme values.

#include <iostream>
#include <cmath> // Required for sqrt()

int main() {
    int N = 11;
    bool isPrime = true;
      
    // Handle special cases: 0, 1, and 2
    if (N <= 1) {
        isPrime = false;
    } else if (N == 2) {
        isPrime = true;
    } else if (N % 2 == 0) {
        // Optimization: all other even numbers are not prime
        isPrime = false;
    } else {
        // Optimization: loop only up to the square root of N
        // We only check odd divisors (i=3, i+=2)
        for (int i = 3; i * i <= N; i += 2) {
            if (N % i == 0) {
                isPrime = false;
                break; // Found a divisor, no need to check further
            }
        }
    }
    
    if (isPrime) {
        std::cout << N << " is a prime number." << std::endl;
    } else {
        std::cout << N << " is not a prime number." << std::endl;
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The program uses a boolean flag isPrime, initialized to true.

• Edge cases N≤1 and N=2 are handled first.
• An initial check for divisibility by 2 quickly handles all even numbers greater than 2.
• The main for loop iterates using the ​square root of N optimization. The condition i * i <= N is mathematically equivalent to i≤ square root of N​ but avoids the sqrt() function call in every iteration.
• The step i += 2 ensures we only check odd divisors (since even divisors were already covered).
• If N % i == 0 is ever true, a divisor other than 1 and N is found, so isPrime is set to false, and the break statement stops the loop immediately, maximizing efficiency.

#include <iostream>
#include <cmath>

int main() {
    std::cout << "Prime numbers between 1 and 100 are:\n";
    
    // Outer loop: iterate through every number N from 2 to 100
    for (int N = 2; N <= 100; ++N) {
        bool isPrime = true; // Assume current number is prime initially
        
        // Inner loop: check for divisibility (optimization: check up to sqrt(N))
        // Start check from 2 up to sqrt(N)
        for (int i = 2; i * i <= N; ++i) {
            if (N % i == 0) {
                isPrime = false; // Found a divisor, so N is not prime
                break;           // No need to check other divisors
            }
        }
        
        // If the flag is still true after the inner loop, the number is prime
        if (isPrime) {
            std::cout << N << " ";
        }
    }
    
    std::cout << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

• The Outer Loop (for (int N = 2; N <= 100; ++N)) picks each number N to be tested.
• The isPrime flag is reset to true for each new number N.
• The Inner Loop (for (int i = 2; i * i <= N; ++i)) tests N for divisibility by i. If a factor is found, isPrime becomes false, and the inner loop breaks.
• After the inner loop completes, the outer if condition checks the final status of isPrime. Only if it remains true (meaning no divisors were found) is the number N printed. This process is repeated for every number up to 100.

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i)
    for (int i = 0; i < N; ++i) {
        
        // Inner loop: controls columns (j)
        for (int j = 0; j < N; ++j) {
            std::cout << "* "; // Print star and a space
        }
        
        std::cout << "\n"; // Move to the next line after completing a row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

For an N=4 square, the outer loop runs 4 times (for rows). In each iteration of the outer loop, the inner loop runs 4 times, printing one ‘*‘ for each column. The core structure of nested loops for 2D patterns is:

for (rows) {
    for (columns) {
        // Print character
    }
    // Print newline
}Code language: C++ (cpp)

This ensures that the correct number of characters are printed across the row before moving down to print the next row.

#include <iostream>

int main() {
    int N;
    
    std::cout << "Enter the height (N) for the triangle: ";
    if (!(std::cin >> N) || N <= 0) return 1;
    
    // Outer loop: controls rows (i) from 1 to N
    for (int i = 1; i <= N; ++i) {
        
        // Inner loop: controls columns (j). It runs 'i' times.
        for (int j = 1; j <= i; ++j) {
            std::cout << "*"; // No space needed here for tighter look
        }
        
        std::cout << std::endl; // Newline for the next row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

• The outer loop counter i not only tracks the current row number but also dictates the number of columns to be printed in that row.
• When i=1 (Row 1), the inner loop condition j <= 1 runs once, printing one ‘*‘.
• When i=4 (Row 4), the inner loop condition j <= 4 runs four times, printing four *.
• By making the inner loop’s termination condition dependent on the outer loop’s counter, the desired triangular shape is generated.

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i) from N down to 1
    for (int i = N; i >= 1; --i) {
        
        // Inner loop: controls columns (j). It runs 'i' times.
        for (int j = 1; j <= i; ++j) {
            std::cout << "*"; 
        }
        
        std::cout << "\n"; // Newline for the next row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The structure is similar to the standard right triangle, but the outer loop initialization and update are reversed:

• Initialization: int i = N (starts with the longest row).
• Condition: i >= 1 (continues until the shortest row is printed).
• Update: --i (decreases the row length by one in each iteration)

Since the inner loop’s condition (j <= i) still depends on the outer loop counter i, the number of stars printed starts at N and decreases by 1 in each subsequent row.

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i)
    for (int i = 1; i <= N; ++i) {
        
        // 1. Inner loop for leading spaces
        // The number of spaces is (N - i)
        for (int j = 1; j <= N - i; ++j) {
            std::cout << " ";
        }
        
        // 2. Inner loop for stars
        // The number of stars is (2 * i - 1)
        for (int k = 1; k <= 2 * i - 1; ++k) {
            std::cout << "*";
        }
        
        std::cout << std::endl; // Newline for the next row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This pattern needs two inner loops within the outer row loop.

Space Loop: As the row number i increases, the number of spaces needed decreases. The formula N−i correctly calculates this:

• Row 1 (i=1): N−1 spaces.
• Row N (i=N): 0 spaces.

Star Loop: The number of stars in a standard pyramid increases by 2 in each row. The formula 2×i−1 correctly calculates this:

• Row 1 (i=1): 2(1)−1=1 star.
• Row 2 (i=2): 2(2)−1=3 stars.

These two inner loops work together to align the pattern centrally.

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i) from N down to 1
    for (int i = N; i >= 1; --i) {
        
        // 1. Inner loop for leading spaces
        // The number of spaces increases: N - i is constant, but N - i is the number of rows already processed from the bottom
        for (int j = 1; j <= N - i; ++j) {
            std::cout << " ";
        }
        
        // 2. Inner loop for stars
        // The number of stars decreases: (2 * i - 1)
        for (int k = 1; k <= 2 * i - 1; ++k) {
            std::cout << "*";
        }
        
        std::cout << std::endl; // Newline
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The solution adapts the logic from the standard pyramid by reversing the outer loop’s iteration direction: i goes from N down to 1.

Space Loop: The number of spaces is determined by how many rows have been printed so far from the top, which is equivalent to N−i.

• Row 1 (i=N): N−N=0 spaces.
• Row 2 (i=N−1): N−(N−1)=1 space.

Star Loop: The formula 2×i−1 still determines the number of stars, but because i is decreasing, the stars decrease in odd steps:

• Row 1 (i=N=3): 2(3)−1=5 stars.
• Row 2 (i=2): 2(2)−1=3 stars.

#include <iostream>
#include <iomanip> // Required for std::setw

int main() {
    std::cout << "--- 1 to 10 Multiplication Chart ---\n\n";
    
    // Outer loop: controls the row multiplier (i)
    for (int i = 1; i <= 10; ++i) {
        
        // Inner loop: controls the column multiplier (j)
        for (int j = 1; j <= 10; ++j) {
            int product = i * j;
            
            // Use std::setw(4) to ensure each number takes 4 characters, 
            // guaranteeing clean column alignment.
            std::cout << std::setw(4) << product; 
        }
        
        std::cout << std::endl; // Newline for the next multiplication row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This is the standard application of nested loops for grid/table generation.

• The Outer Loop determines which row we are printing (e.g., the “5 times” row when i=5).
• The Inner Loop iterates through the multipliers j=1 through 10.
• The line int product = i * j; calculates the result.
• std::setw(4) is used to format the output. It forces the printed number to occupy a minimum field width of 4 characters, aligning the table neatly regardless of whether the product is single-digit (5) or three-digit (100).

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i)
    for (int i = 1; i <= N; ++i) {
        
        // Inner loop: controls columns (j). Runs 'i' times.
        // Print the column counter 'j' in each iteration
        for (int j = 1; j <= i; ++j) {
            std::cout << j << " "; // Print the column number 'j'
        }
        
        std::cout << std::endl; // Newline for the next row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This pattern combines the structure of the right triangle pattern (Exercise 22) with the logic of printing a dynamic value.

• The Outer Loop defines the row i.
• The Inner Loop runs j from 1 to i.
• By printing the inner loop counter j (std::cout << j << " ";), the value printed automatically resets to 1 at the beginning of every new row and increments up to the limit i defined by the outer loop. This results in the sequence 1, then 1,2, then 1,2,3, and so on.

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i) and the number to be printed
    for (int i = 1; i <= N; ++i) {
        
        // Inner loop: controls columns (j). Runs 'i' times.
        for (int j = 1; j <= i; ++j) {
            // Print the row number 'i'
            std::cout << i << " "; 
        }
        
        std::cout << std::endl; // Newline for the next row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This pattern builds on the right triangle structure.

• The Outer Loop counter i determines both the value to be printed and the length of the row.
• The Inner Loop runs j from 1 up to i, correctly generating the triangular shape.
• The key is that we print i inside the inner loop. When i=3, the inner loop runs three times, and in all three iterations, it prints 3, resulting in ‘3 3 3‘.

#include <iostream>

int main() {
    int N = 4;
    
    // Outer loop: controls rows (i) from N down to 1
    for (int i = N; i >= 1; --i) {
        
        // Inner loop: controls columns (j). It starts at N and decreases to 'i'.
        for (int j = N; j >= i; --j) {
            std::cout << j << " "; // Print the current column number 'j'
        }
        
        std::cout << std::endl; // Newline for the next row
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This pattern uses both the outer and inner loops in a decreasing fashion.

1) The Outer Loop counter i runs backwards (from N to 1), defining the ending number for each row.
2) The Inner Loop counter j controls the numbers printed in the row. It is initialized to N and decrements (--j).
3) The Inner Loop Condition j >= i is key.

• When i=1 (first row), the loop runs as long as j≥1, printing 5,4,3,2,1.
• When i=4 (fourth row), the loop runs as long as j≥4, printing 5,4.
• When i=5 (last row), the loop runs as long as j≥5, printing 5.
• By printing the decreasing inner counter j, we achieve the desired pattern.

#include <iostream>
#include <string>

int main() {
    const std::string CORRECT_PASSWORD = "Pass1212";
    std::string userAttempt;
    int attempts = 0;
    const int MAX_ATTEMPTS = 3;
    bool isAuthenticated = false;
    
    std::cout << "--- Password Lock Simulation ---\n";

    do {
        attempts++; // Increment attempt counter first
        std::cout << "Attempt " << attempts << "/" << MAX_ATTEMPTS << ". Enter password: ";
        
        // Read the user's input, allowing spaces
        std::getline(std::cin, userAttempt); 
        
        if (userAttempt == CORRECT_PASSWORD) {
            isAuthenticated = true;
            break; // Exit loop immediately on success
        } 
        
        if (attempts < MAX_ATTEMPTS) {
            std::cout << "Wrong Password. Access denied. Try again.\n";
        }
        
    } while (attempts < MAX_ATTEMPTS); // Loop condition checks if max attempts haven't been reached
    
    // Final result output
    if (isAuthenticated) {
        std::cout << "\nAccess GRANTED. Welcome!\n";
    } else {
        std::cout << "\nMaximum attempts reached. Account locked.\n";
    }
    
    return 0;
}Code language: C++ (cpp)

Explanation:

The do-while loop is ideal here because the password check must happen at least once.

• The attempts counter tracks the usage. It is incremented at the start of the do block.
• The std::getline function is used to read the full line of input, ensuring the password can contain spaces if required.
• If the password matches, the isAuthenticated flag is set, and break immediately exits the loop.
• The while condition (attempts < MAX_ATTEMPTS) only allows the loop to repeat if the maximum limit of 3 attempts hasn’t been hit.
• After the third failed attempt, attempts becomes 3, the condition fails, and the loop terminates, leading to the “Account locked” message.

#include <iostream>
#include <string>
#include <vector>
#include <sstream> // Required for stringstream

int main() {
    std::string sentence = "This is a test";
    std::vector<std::string> words;
    
    // Check if the input is empty
    if (sentence.empty()) {
        std::cout << "No words entered." << std::endl;
        return 0;
    }
    
    // 1. Use stringstream to break the sentence into words
    std::stringstream ss(sentence);
    std::string word;
    
    // Loop (implicitly a while loop) to read each word into the vector
    while (ss >> word) {
        words.push_back(word);
    }
    
    std::cout << "\nSentence with words reversed:\n";
    
    // 2. Loop through the vector backwards
    // Start at the last index (size - 1) and stop at index 0
    for (int i = words.size() - 1; i >= 0; --i) {
        std::cout << words[i];
        
        // Print a space after every word except the last one
        if (i > 0) {
            std::cout << " ";
        }
    }
    
    std::cout << std::endl;
    
    return 0;
}Code language: C++ (cpp)

Explanation:

This solution uses two distinct control flow mechanisms:

• Word Extraction (Implicit Loop): The statement while (ss >> word) uses the overloaded stream extraction operator (>>). This automatically tokenizes the string ss by whitespace, extracting words one by one and storing them sequentially in the words vector. The while loop terminates when ss reaches the end of the string.
• Reverse Output (Explicit Loop): A standard for loop is used to iterate over the words vector. By setting the initialization to words.size() - 1, the condition to i >= 0, and the update to –i, the loop iterates from the last word to the first, achieving the reverse order output.