Multiplication (Recursive) #

Given two integers m & n, calculate and return their multiplication using recursion.
You can only use subtraction and addition for your calculation.
No other operators are allowed.
Input format : m and n (in different lines)

Sample Input

3
5

Sample Output

15
Code 
int multiplyNumbers(int m, int n)
{
    if(m==0|| n==0)// either of them could becime 0 depending on the first
        return 0;
    return n + multiplyNumbers(m-1, n);
}

Count Zeros #

Given an integer n, count and return the number of zeros that are present in the given integer using recursion.

Input Format : Integer n
Output Format : No. of 0s

Sample Input

10204

Sample Output

2
Code 
int countZeros(int n)
{
    if(n==0)
        return 0;
    if(n%10==0)
        return 1 + countZeros(n/10);
    else
        return 0 + countZeros(n/10);
}

Geometric Sum #

Given k, find the geometric sum i.e. 1 + 1/2 + 1/4 + 1/8 + ... + 1/(2^k)
using recursion. Return the answer.

Sample Input

3

Sample Output

1.875
Code 
double geometricSum(int k)
{
    if(k==0)
        return 1;
    return 1.0 + geometricSum(k-1)/2.0;
}

Check Palindrome (recursive) #

Check if a given String S is palindrome or not (using recursion). Return true or false.
Input Format : String S
Output Format : true or false

Sample Input 1

racecar

Sample Output 1

true

Sample Input 2

ninja

Sample Output 2

false
Code 
bool helper(char* start, char* end)
{
    if(end<=start) // if equal, then true. A letter is same as itself, lame
        return true; // when they pass
    if(*start==*end)
        return helper(start+1, end-1);
    else
        return false;
}

bool checkPalindrome(char input[])
{
    char* end = input;
    while(*end!=0)
        end++; // end at the null
    return helper(input, end-1); // we need to seek from the back
}

//Way better a solution then the CN solution. Elegance guarenteed

Sum of Digits (recursive) #

Write a recursive function that returns the sum of the digits of a given integer.

Sample Input

12345

Sample Output

15
Code 
int sumOfDigits(int n)
{
    if(n==0)  // or n<=9 is better, 1 step less
        return 0;
    return n%10 + sumOfDigits(n/10);
}