Given two strings S1 and S2 as input. Your task is to concatenate two strings and then reverse the string. Finally print the reversed string.
Example 1:
Input: S1 = "Geeks" , S2 = "forGeeks"
Output: "skeeGrofskeeG" 
Explanation: Concatenating S1 and S2 to 
get "GeeksforGeeks" then reversing it to 
"skeeGrofskeeG".
Example 2:
Input: S1 = "Practice" , S2 = "Geeks"
Output: "skeeGecitcarP"
Explanation: Concatenating S1 and S2 to
get "PracticeGeeks" then reversing it to
"skeegecitcarP".
Your Task:  
You dont need to read input or print anything. Complete the function conRevstr() which takes S1 and S2 as input parameter and returns the sting final_s after concatenating and reversing.
Expected Time Complexity: O(|S1| + |S2|)
Expected Auxiliary Space: O(|S1| + |S2|)
Constraints:
1<= |S1| <=1000
1<= |S2| <=1000

Source: geeksforgeeks | Difficulty: EASY
Tags: String algorithms

Data structures

Starter code: #User function Template for python3
class Solution:
    def conRevstr (ob, S1, S2):
        # code here 

Solutions:
class Solution:

	def conRevstr(ob, S1, S2):
		addedString = S1 + S2
		reversedString = addedString[::-1]
		return reversedString


class Solution:

	def conRevstr(ob, S1, S2):
		a = S1 + S2
		b = a[::-1]
		return b


class Solution:

	def conRevstr(ob, S1, S2):
		return (S1 + S2)[::-1]


class Solution:

	def conRevstr(ob, S1, S2):
		c = S1 + S2
		b = c[::-1]
		return b


class Solution:

	def conRevstr(ob, S1, S2):
		S = S1 + S2
		s = S[::-1]
		return s


class Solution:

	def conRevstr(ob, S1, S2):
		str = S1 + S2
		word = str[::-1]
		return word


class Solution:

	def conRevstr(ob, S1, S2):
		s = S1 + S2
		st = s[::-1]
		return st


class Solution:

	def conRevstr(ob, S1, S2):
		S = S1 + S2
		REV = ''
		for L in S:
			REV = L + REV
		return REV


class Solution:

	def conRevstr(ob, S1, S2):
		m = S1 + S2
		k = m[::-1]
		return k


class Solution:

	def conRevstr(ob, S1, S2):
		s11 = S1[::-1]
		s22 = S2[::-1]
		f = s22 + s11
		return f


class Solution:

	def conRevstr(ob, S1, S2):
		S3 = S1 + S2
		k = S3[::-1]
		return k


class Solution:

	def conRevstr(ob, S1, S2):
		String1 = S1 + S2
		string2 = String1[::-1]
		return string2


class Solution:

	def conRevstr(ob, S1, S2):
		ob.s1 = S1
		ob.s2 = S2
		k = ob.s1[::-1]
		l = ob.s2[::-1]
		j = l + k
		return j


class Solution:

	def conRevstr(ob, S1, S2):
		q = S1 + S2
		m = q[::-1]
		return m


class Solution:

	def conRevstr(ob, S1, S2):
		S1 = S1 + S2
		S2 = ''
		for i in range(len(S1) - 1, -1, -1):
			S2 = S2 + S1[i]
		return S2


class Solution:

	def conRevstr(ob, S1, S2):
		s1 = S1[::-1]
		s2 = S2[::-1]
		return s2 + s1


class Solution:

	def conRevstr(ob, S1, S2):
		s = S1 + S2
		ans = s[::-1]
		return ans


class Solution:

	def conRevstr(ob, S1, S2):
		s = S1 + S2
		res = ''
		for i in s:
			res = i + res
		return res


class Solution:

	def conRevstr(ob, S1, S2):
		res_ = S1 + S2
		result = res_[::-1]
		return result


class Solution:

	def conRevstr(ob, S1, S2):
		D = S1[-1::-1]
		S = S2[-1::-1]
		return S + D


class Solution:

	def conRevstr(ob, S1, S2):
		st = ''
		sum_st = ''
		sum_st = S1 + S2
		for i in sum_st:
			st = i + st
		return st


class Solution:

	def conRevstr(ob, S1, S2):
		S = S1 + S2
		SR = S[::-1]
		return SR


class Solution:

	def conRevstr(ob, S1, S2):
		S3 = S1 + S2
		reverse = S3[::-1]
		return reverse