Kanchan Rai

Day 7: 100 Days of DSA

Recursion

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Kanchan Rai
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3 min read

Welcome to Day 2 of my 100 Days of DSA challenge! Today, I solved five problems focused on recursion. Here's a quicCk look at the questions I tackled and how I approached them 🤖✨ Check out my GitHub repository for all my solutions and progress. Let’s keep learning!

1. To find factorial of a number using recursion.

We can make a function called factorial, which will be called again and again until the value of n reaches 1. The base condition for this is n==0, here we stop finding the result.

Code:

// factorial of a number using recursion

#include <iostream>
using namespace std;

int factorial(int n) {
    if (n == 0) {
        return 1;
    }
    return n * factorial(n - 1);
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;
    cout << "Factorial of " << n << " is " << factorial(n) << endl;
    return 0;
}

Output:

2. Solving the tower of hanoi problem for three disks

This is a very famous problem in recursion, where we have to shift the disks from rod1 to rod3, given the rules:

  1. only one disk can be moved at a time.

  2. a disk can be placed on a larger disk only.

Code:

// Tower of hanoi problem for 3 disks

#include <iostream>
using namespace std;

void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
    if (n == 1) {
        cout << "Move disk 1 from rod " << from_rod << " to rod " << to_rod << endl;
        return;
    }
    towerOfHanoi(n - 1, from_rod, aux_rod, to_rod);
    cout << "Move disk " << n << " from rod " << from_rod << " to rod " << to_rod << endl;
    towerOfHanoi(n - 1, aux_rod, to_rod, from_rod);
}

int main() {
    int n = 3; // Number of disks
    towerOfHanoi(n, 'A', 'C', 'B'); // A, B and C are names of rods
    return 0;
}

Output:

3. Display all sequences of a string

Code:

// prnt allsubsequences of a string

#include <iostream>
using namespace std;

void printSubsequences(string input, string output) {
    if (input.empty()) {
        cout << output << endl;
        return;
    }
    printSubsequences(input.substr(1), output + input[0]);
    printSubsequences(input.substr(1), output);
}

int main() {
    string input = "abc";
    string output = "";
    printSubsequences(input, output);
    return 0;
}

Output:

4. To find the nth fibonacci number

The Fibonacci sequence is defined as:
F(n)=F(n−1)+F(n−2)with base cases F(0)=0 and F(1)=1. A recursive function captures this directly by expressing F(n)F(n) in terms of smaller subproblems.

Code:

// To find the nth Fibonacci number, the program uses a recursive function.

#include <iostream>
using namespace std;

int fibonacci(int n) {
    if (n <= 1) {
        return n;
    }
    return fibonacci(n - 1) + fibonacci(n - 2);
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;
    cout << "Fibonacci of " << n << " is " << fibonacci(n) << endl;
    return 0;
}

Code:

5. To find the sum of digitd in a number

To find the this we use a recursive function in which we find the last digit by using the modulo operator by 10 and keep on adding it.

// To find the sum of digits of a number using recursion

#include <iostream>
using namespace std;

int sumOfDigits(int n) {
    if (n == 0) {
        return 0;
    }
    return n % 10 + sumOfDigits(n / 10);
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;
    cout << "Sum of digits of " << n << " is " << sumOfDigits(n) << endl;
    return 0;
}

Output:

Happy Coding!

Recursion
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