C++ Program To Print Reverse of a String Using Recursion

Write a recursive function to print the reverse of a given string. 
Code:

C++

// C++ program to reverse a string using recursion
#include <bits/stdc++.h>
using namespace std;

/* Function to print reverse of the passed string */
void reverse(string str)
{
    if(str.size() == 0)
    {
        return;
    }
    reverse(str.substr(1));
    cout << str[0];
}

/* Driver program to test above function */
int main()
{
    string a = "Geeks for Geeks";
    reverse(a);
    return 0;
}

// This is code is contributed by rathbhupendra

Output:

skeeG rof skeeG

Explanation: Recursive function (reverse) takes string pointer (str) as input and calls itself with next location to passed pointer (str+1). Recursion continues this way when the pointer reaches ‘\0’, all functions accumulated in stack print char at passed location (str) and return one by one.

Time Complexity: O(n^2) as substr() method has a time complexity of O(k) where k is the size of the returned string. So for every recursive call, we are reducing the size of the string by one, which leads to a series like (k-1)+(k-2)+…+1 = k*(k-1)/2 = O(k^2) = O(n^2)
See Reverse a string for other methods to reverse string.
Auxiliary Space: O(n)

Efficient Approach: 

We can store each character in recursive stack and then can print while coming back as shown in the below code: 

C++

// C++ program to reverse a string using recursion
#include <bits/stdc++.h>
using namespace std;

/* Function to print reverse of the passed string */
void reverse(char *str, int index, int n)
{
    if(index == n)      // return if we reached at last index or at the end of the string
    {
        return;
    }
    char temp = str[index];    // storing each character starting from index 0 in function call stack;
    reverse(str, index+1, n);  // calling recursive function by increasing index everytime
    cout << temp;              // printing each stored character while recurring back
}

/* Driver program to test above function */
int main()
{
    char a[] = "Geeks for Geeks";
    int n = sizeof(a) / sizeof(a[0]);
    reverse(a, 0, n);
    return 0;
}

Output:

skeeG rof skeeG

Time Complexity: O(n) where n is size of the string

Auxiliary Space: O(n) where n is the size of string, which will be used in the form of function call stack of recursion.

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.

Please suggest if someone has a better solution that is more efficient in terms of space and time.