Print reverse of a string using recursion

Last Updated : 06 Mar, 2025

Given a string, the task is to print the given string in reverse order using recursion.

Examples:

Input: s = "Geeks for Geeks"
Output: "skeeG rof skeeG"
Explanation: After reversing the input string we get "skeeG rof skeeG".
Input: s = "Reverse a string Using Recursion"
Output: "noisruceR gnisU gnirts a esreveR"
Explanation: After reversing the input string we get "noisruceR gnisU gnirts a esreveR".

[Approach - 1] - Make a Recursive Call and Then Process the First Char

The idea for this approach is to make a recursive call for the substring starting from the second character and then print the first character.

C++

#include <bits/stdc++.h>
using namespace std;

string reverse(string str) 
{ 
    if(str.size() == 0)
        return str;
    return reverse(str.substr(1)) + str[0];
}

int main() 
{ 
    string str = "Geeks for Geeks"; 
    cout << reverse(str); 
    return 0; 
}

#include <stdio.h>

void reverse(const char *str) 
{
    if (*str == '\0') 
        return;
    reverse(str + 1);
    putchar(*str);
}

int main() 
{
    char str[] = "Geeks for Geeks";
    reverse(str);
    return 0;
}

Java

public class Main {
    public static String reverse(String str) {
        if (str.isEmpty())
            return str;
        return reverse(str.substring(1)) + str.charAt(0);
    }

    public static void main(String[] args) {
        String str = "Geeks for Geeks";
        System.out.println(reverse(str));
    }
}

Python

# Function to reverse a string

def reverse(str):
    if len(str) == 0:
        return str
    return reverse(str[1:]) + str[0]

# Main function
if __name__ == '__main__':
    str = 'Geeks for Geeks'
    print(reverse(str))

using System;

class Program {
    static string Reverse(string str) {
        if (str.Length == 0)
            return str;
        return Reverse(str.Substring(1)) + str[0];
    }

    static void Main() {
        string str = "Geeks for Geeks";
        Console.WriteLine(Reverse(str));
    }
}

JavaScript

// Function to reverse a string

function reverse(str) {
    if (str.length === 0)
        return str;
    return reverse(str.substring(1)) + str[0];
}

// Main function
const str = 'Geeks for Geeks';
console.log(reverse(str));

Output

skeeG rof skeeG

Time Complexity: O(n)
Auxiliary Space: O(n)

[Approach - 2] - Process the Last Char and Then Make Recursive Call

This idea is to break the problem in terms of smaller instance of same subproblem.
str= "Geeks for geeks"
reverse string of str = last character of str + reverse string of remaining str = "s" + reverse string of "Geeks for geek" = "skeeg rof skeeG"

C++

#include <bits/stdc++.h>
using namespace std;

string reverse(string str, int len)
{
    if (len < 1)
    {
        return "";
    }
    // Base case
    if (len == 1)
    {
        return string(1, str[0]);
    }
    return str[len - 1] + reverse(str, len - 1);
}

int main()
{
    string str = "Geeks for geeks";
    cout << reverse(str, str.length()) << endl;
    return 0;
}

Java

import java.util.*;

public class GfG {
    // Function to reverse a string using recursion
    public static String reverse(String str, int len)
    {
        if (len < 1) {
            return "";
        }

        // Base case
        if (len == 1) {
            return String.valueOf(str.charAt(0));
        }

        return str.charAt(len - 1) + reverse(str, len - 1);
    }

    public static void main(String[] args)
    {
        String str = "Geeks for geeks";
        System.out.println(reverse(str, str.length()));
    }
}

Python

def reverse(string, length):
    if length < 1:
        return ""

    # Base case
    if length == 1:
        return string[0]

    return string[length - 1] + reverse(string, length - 1)


if __name__ == "__main__":
    str = "Geeks for geeks"
    print(reverse(str, len(str)))

using System;

class GfG {
    static string Reverse(string str, int len)
    {
        // Base case: If the string length is less than 1,
        // return an empty string
        if (len < 1) {
            return "";
        }
        // Base case: If the string has only one character,
        // return that character as a string
        if (len == 1) {
            return str[0].ToString();
        }
        // Recursive case: Concatenate the last character of
        // the string with the reversed substring
        return str[len - 1] + Reverse(str, len - 1);
    }

    static void Main()
    {
        string str = "Geeks for geeks";
        // Call the Reverse function and print the reversed
        // string
        Console.WriteLine(Reverse(str, str.Length));
    }
}

JavaScript

function reverse(str, len)
{
    if (len < 1) {
        return
    }
    // base case
    if (len === 1) {
        return str[0]
    }

    return str[len - 1] + reverse(str, len - 1)
}

// Driver code

let str = "Geeks for geeks"
console.log(reverse(str, str.length))

Output

skeeg rof skeeG

Time Complexity: O(n)
Auxiliary Space: O(n)

[Approach - 3] - Optimized- Process from Both Ends

The idea is to begin with both corners, swap the corner characters and then make recursive call for the remaining string.

C++

#include <bits/stdc++.h>
using namespace std;

void reverse(string &str, int start, int end) 
{
    if (start >= end) 
        return;
    swap(str[start], str[end]);
    reverse(str, start + 1, end - 1);
}

int main() 
{ 
    string str = "Geeks for Geeks"; 
    reverse(str, 0, str.size() - 1);
    cout << str;  
    return 0; 
}

Java

// Java program to reverse a string
class ReverseString {
    static void reverse(StringBuilder str, int start, int end) {
        if (start >= end) 
            return;
        char temp = str.charAt(start);
        str.setCharAt(start, str.charAt(end));
        str.setCharAt(end, temp);
        reverse(str, start + 1, end - 1);
    }

    public static void main(String[] args) {
        StringBuilder str = new StringBuilder("Geeks for Geeks");
        reverse(str, 0, str.length() - 1);
        System.out.println(str);
    }
}

Python

# Python program to reverse a string
def reverse(str, start, end):
    if start >= end:
        return
    str[start], str[end] = str[end], str[start]  # Swap characters
    reverse(str, start + 1, end - 1)

if __name__ == '__main__':
    str = list("Geeks for Geeks")
    reverse(str, 0, len(str) - 1)
    print(''.join(str))

JavaScript

// JavaScript implementation
function reverse(str, start, end) {
    if (start >= end) 
        return;
    str = str.split('');
    [str[start], str[end]] = [str[end], str[start]];
    str = str.join('');
    reverse(str, start + 1, end - 1);
    return str;
}

let str = "Geeks for Geeks";
str = reverse(str, 0, str.length - 1);
console.log(str);

Time Complexity: O(n)
Auxiliary Space: O(n)

Print reverse of a string using recursion

Anurag Mishra and Aarti_Rathi

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Practice Tags :

Print reverse of a string using recursion

[Approach - 1] - Make a Recursive Call and Then Process the First Char

[Approach - 2] - Process the Last Char and Then Make Recursive Call

[Approach - 3] - Optimized- Process from Both Ends

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