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Print reverse of a string using recursion

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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

                    

C

// C program to reverse a string using recursion
# include <stdio.h>
 
/* Function to print reverse of the passed string */
void reverse(char *str)
{
   if (*str)
   {
       reverse(str+1);
       printf("%c", *str);
   }
}
 
/* Driver program to test above function */
int main()
{
   char a[] = "Geeks for Geeks";
   reverse(a);
   return 0;
}

                    

Java

// Java program to reverse a string using recursion
 
class StringReverse
{
    /* Function to print reverse of the passed string */
    void reverse(String str)
    {
        if ((str==null)||(str.length() <= 1))
           System.out.println(str);
        else
        {
            System.out.print(str.charAt(str.length()-1));
            reverse(str.substring(0,str.length()-1));
        }
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        String str = "Geeks for Geeks";
        StringReverse obj = new StringReverse();
        obj.reverse(str);
    }   
}

                    

Python

# Python program to reverse a string using recursion
 
# Function to print reverse of the passed string
def reverse(string):
    if len(string) == 0:
        return
     
    temp = string[0]
    reverse(string[1:])
    print(temp, end='')
 
# Driver program to test above function
string = "Geeks for Geeks"
reverse(string)
 
# A single line statement to reverse string in python
# string[::-1]
 
# This code is contributed by Bhavya Jain

                    

C#

// C# program to reverse
// a string using recursion
using System;
 
class GFG
{
    // Function to print reverse
    // of the passed string
    static void reverse(String str)
    {
        if ((str == null) || (str.Length <= 1))
        Console.Write(str);
     
        else
        {
            Console.Write(str[str.Length-1]);
            reverse(str.Substring(0,(str.Length-1)));
        }
    }
     
    // Driver Code
    public static void Main()
    {
        String str = "Geeks for Geeks";
        reverse(str);
    }
}
 
// This code is contributed by Sam007

                    

Javascript

<script>
        // JavaScript Program for the above approach
 
 
        /* Function to print reverse of the passed string */
        function reverse(str, len) {
            if (len == str.length) {
                return;
            }
            reverse(str, len + 1);
 
            document.write(str[len]);
 
        }
 
        /* Driver program to test above function */
 
        let a = "Geeks for Geeks";
 
        reverse(a, 0);
 
    // This code is contributed by Potta Lokesh
    </script>

                    

PHP

<?php
// PHP program to reverse
// a string using recursion
 
// Function to print reverse
// of the passed string
function reverse($str)
{
    if (($str == null) ||
        (strlen($str) <= 1))
    echo ($str);
 
    else
    {
        echo ($str[strlen($str) - 1]);
        reverse(substr($str, 0,
               (strlen($str) - 1)));
    }
}
 
// Driver Code
$str = "Geeks for Geeks";
reverse($str);
 
// This code is contributed by
// Manish Shaw(manishshaw1)
?>

                    

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) where n is the length of string
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;
}
 
// This is code is contributed by anuragayu

                    

C

// C program to reverse a string using recursion
 
#include <stdio.h>
 
/* 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
    printf("%c", 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;
}
 
// This is code is contributed by anuragayu

                    

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  /* Function to print reverse of the passed string */
  static 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
    System.out.print(temp); // printing each stored character while recurring back
  }
 
  public static void main(String[] args)
  {
    char a[] = "Geeks for Geeks".toCharArray();
    int n = a.length;
    reverse(a, 0, n);
  }
}
 
// This code is contributed by aadityaburujwale.

                    

Python3

# Python3 program to reverse a string using recursion
def reverse(string, index, n):
    if index == n:   # return if we reached at last index or at the end of the string
        return
    temp = string[index]   # storing each character starting from index 0 in function call stack;
    reverse(string, index+1, n)  # calling recursive function by increasing index everytime
    print(temp, end="")   # printing each stored character while recurring back
 
# Driver code
string = "Geeks for Geeks"
n = len(string)
reverse(string, 0, n)
 
# This code is contributed by Potta Lokesh

                    

C#

// Include namespace system
using System;
 
public class GFG
{
  // Function to print reverse of the passed string
  public static 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;
    }
    var temp = str[index];
 
    // storing each character starting from index 0 in function call stack;
    GFG.reverse(str, index + 1, n);
 
    // calling recursive function by increasing index everytime
    Console.Write(temp);
  }
  public static void Main(String[] args)
  {
    char[] a = "Geeks for Geeks".ToCharArray();
    var n = a.Length;
    GFG.reverse(a, 0, n);
  }
}
 
// This code is contributed by aadityaburujwale.

                    

Javascript

// JavaScript program to reverse a string using recursion
 
function reverse(string, index, n) {
if (index === n) { // return if we reached at last index or at the end of the string
return;
}
var temp = string[index]; // storing each character starting from index 0 in function call stack;
reverse(string, index+1, n); // calling recursive function by increasing index everytime
console.log(temp); // printing each stored character while recurring back
}
 
// Driver code
var string = "Geeks for Geeks";
var n = string.length;
reverse(string, 0, n);
 
// This code is contributed by pradeepkumarppk2003

                    

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.

C++

#include <iostream>
using namespace std;
 
// Function to reverse a string using recursion
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;
}
 
// This code is contributed by akshitaguprzj3

                    

Java

import java.util.*;
 
public class Main {
    // 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()));
    }
}

                    

Python3

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__":
    input_str = "Geeks for geeks"
    print(reverse(input_str, len(input_str)))

                    

C#

using System;
 
class Program
{
    // Function to reverse a string using recursion
    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

// Javascript program to reverse string using recursion
 
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))
  
 // This code is contributed by rithikpathak123

                    

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.

Explanation: This problem can be expressed 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”



 

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Anurag Mishra and Aarti_Rathi .
 



Last Updated : 16 Nov, 2023
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