Find First index of array using recursion

         First Index (recursive)

Problem Description:​ ​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.

Sample Input:
 ​5 (Size of array)
1 4 -1 3 4 (The array elements)
4 (x)

Sample Output:
 ​1

How to approach?

Doing this iteratively is trivial. All we need to do is to start a loop from the 0​th index to the end of the array. If we encounter ‘x’, then we return the index we’re currently on.

If the whole loop is finished and we didn’t return any value, that means ‘x’ is not present on the array, so we return -1.
We can do the same process recursively. Let’s create a function that returns an integer
and takes in the following parameters:
1. The array itself
2. x
3. A start index (initially 0)

Now, we will check whether the value element at index ‘start’ is equal to ‘x’. If yes, then
we straight-away return ‘start’, otherwise we recurse on our function passing the following arguments: the array, x and ​start + 1

What will be the base case? The most trivial input would be an array of size 0.

In that  case, we won’t have to do any work and we can straisize 0 won’t contain any elements. Also, when we reach the end of an array, that signifies that the array that we now have to search is technically empty, so we can
return -1 even in that case.

Therefore, our base case will look something like this:
                 if(start == arr.length()):
                         return -1
U
→➡️The pseudo-code for this approach is shown on the next page.

function solve(arr, x, start):
if(start = arr.length()):
return -1
if(arr[start] = x):
return start
return solve(arr, x, start + 1)
function firstIndex(arr, x):
return solve(arr, x, 0)

First Index of Number - Question

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.

CODE:

def firstIndex(arr, x, si):
    l=len(arr) 
    if si==l:
        return -1
    if arr[si]==x:
        return si
    
    smalleroutput= firstIndex(arr, x, si+1) 
    return smalleroutput
    # Please add your code here

pass

# Main
from sys import setrecursionlimit
setrecursionlimit(11000)
n=int(input())
arr=list(int(i) for i in input().strip().split(' '))
x=int(input())
print(firstIndex(arr, x, 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 :
1