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Print reverse of a string using recursion

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  • Difficulty Level : Easy
  • Last Updated : 18 Nov, 2022
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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


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)
?>


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>


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;
}
 
// 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.


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.


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 you want to share more information about the topic discussed above.
 

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


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