Power #

Write a program to find x to the power n (i.e. x^n). Take x and n from the user. You need to return the answer.
Do this recursively.

Input format : Two integers x and n (separated by space)
Output Format : x^n (i.e. x raise to the power n)
Constraints :
1 <= x <= 30
0 <= n <= 30

Sample Input 1

3 4

Sample Output 1

Sample Input 2

2 5

Sample Output 2

Code

int power(int x, int n)
{
    if(n==0)
        return 1;
    return x * power(x, n - 1);
}

// can be optimized using odd even value of the exponent

Print Numbers #

Given the code to print number from 1 to n in increasing order recursively. But it contains few bugs, that you need to rectify such that all the test cases should pass.
Input Format : Integer n
Output Format : Numbers from 1 to n (separated by space)
Constraints :
1 <= n <= 10000

Sample Input

Sample Output

1 2 3 4 5 6

Code

void print(int n)
{
    if(n == 1)
    {
        cout << n << " ";
        return;
    }

    print(n - 1);        // induction hypothesis and the next step in one go.
    cout << n << " ";
}

Number of Digits #

Given the code to find out and return the number of digits present in a number recursively. But it contains few bugs, that you need to rectify such that all the test cases should pass.
Input Format : Integer n
Output Format : Count of digits
Constraints :
1 <= n <= 10^6

Sample Input

Sample Output

Code

int count(int n){
    if(n < 10)
    {
        return 1;
    }

    int smallAns = count(n / 10);
    return smallAns + 1;
}

Sum of Array #

Given an array of length N, you need to find and return the sum of all elements of the array.
Do this recursively.
Input Format :
Line 1 : An Integer N i.e. size of array
Line 2 : N integers which are elements of the array, separated by spaces
Output Format :
Sum
Constraints :
1 <= N <= 10^3

Sample Input

3
9 8 9

Sample Output 1

Code

int sum(int input[], int n)
{
    if(n==0)
        return 0;
    return input[0] + sum(input + 1, n - 1);
}

Check Number #

Given an array of length N and an integer x, you need to find if x is present in the array or not. Return true or false.
Do this recursively.
Input Format :
Line 1 : An Integer N i.e. size of array
Line 2 : N integers which are elements of the array, separated by spaces
Line 3 : Integer x
Output Format :
true or false
Constraints :
1 <= N <= 10^3

Sample Input

3
9 8 10
8

Sample Output

true

Code

bool checkNumber(int input[], int size, int x)
{
    if(size==0)
        return false; // base case
    // hypothesis -> assumed to be true

    // return true if the current one is the key or that remaining has it
    if(input[0]==x || checkNumber(input+1, size-1, x)==true)
        return true;
    return false;
}

First Index of Number #

Given an array of length N and an integer x, you need to find and return the first index of integer x present in the array. Return -1 if it is not present in the array.
First index means, the index of first occurrence of x in the input array.
Do this recursively. Indexing in the array starts from 0.

Input Format :
Line 1 : An Integer N i.e. size of array
Line 2 : N integers which are elements of the array, separated by spaces
Line 3 : Integer x
Output Format :
first index or -1
Constraints :
1 <= N <= 10^3

Sample Input

4
9 8 10 8
8

Sample Output

Code

int firstIndex(int input[], int size, int x)
{
    if(size==0)
        return -1; // if at all we reach an end

    if(input[0]==x)
        return 0;

    // if this is don't find it

    int k = firstIndex(input+1, size-1, x);
    // hav to take this variable as I have to compare the value I get, and 'pass' it.

    if(k==-1)
        return -1;
    return k+1;     // for add operation, we used k as the dummy variable
}

All Indices of Number #

Given an array of length N and an integer x, you need to find all the indexes where x is present in the input array. Save all the indexes in an array (in increasing order).
Do this recursively. Indexing in the array starts from 0.
Input Format :
Line 1 : An Integer N i.e. size of array
Line 2 : N integers which are elements of the array, separated by spaces
Line 3 : Integer x
Output Format :
indexes where x is present in the array (separated by space)
Constraints :
1 <= N <= 10^3

Sample Input

5
9 8 10 8 8
8

Sample Output

1 3 4

Code

int allIndexes(int input[], int size, int x, int output[])
{
    static int index = -1;
    if(size==0)
        return 0;
    if(input[++index]==x)
    {
        *output = index;
        return 1 + allIndexes(input, size-1, x, output+1);
    }
    return 0 + allIndexes(input, size-1, x, output);
}
 // done easy. Just needed a static variable or use a helper function with a remenant of the static variable