LastIndexOfNumber.py

# Given an array of length N and an integer x, you need to find and return the last index of integer x present in the array. Return -1 if it is not present in the array.
# Last index means - if x is present multiple times in the array, return the index at which x comes last in the array.
# You should start traversing your array from 0, not from (N - 1).
# Do this recursively. Indexing in the array starts from 0.

# Sample Input :
# 4
# 9 8 10 8
# 8
# Sample Output :
# 3

from math import *
from collections import *
from sys import *
from os import *

## Read input as specified in the question.
## Print output as specified in the question.

def lastindex(arr, x):
    l=len(arr)
    if l==0:
        return -1
    
    smallerlist = arr[1:]
    smallerlistoutput = lastindex(smallerlist, x)

    if smallerlistoutput != -1:
        return smallerlistoutput + 1
    else:
        if arr[0] == x:
            return 0
        else:
            return -1

from sys import setrecursionlimit
setrecursionlimit(11000)
n=int(input())
arr=list(int(i) for i in input().strip().split(' '))
x=int(input())
print(lastindex(arr, x))