Write a recursive function to print reverse of a given string.
Program:
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.
C++
#include <bits/stdc++.h>
using namespace std;
void reverse(string str)
{
if(str.size() == 0)
{
return;
}
reverse(str.substr(1));
cout << str[0];
}
int main()
{
string a = "Geeks for Geeks";
reverse(a);
return 0;
}
|
C
# include <stdio.h>
void reverse(char *str)
{
if (*str)
{
reverse(str+1);
printf("%c", *str);
}
}
int main()
{
char a[] = "Geeks for Geeks";
reverse(a);
return 0;
}
|
Java
class StringReverse
{
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));
}
}
public static void main(String[] args)
{
String str = "Geeks for Geeks";
StringReverse obj = new StringReverse();
obj.reverse(str);
}
}
|
Python
def reverse(string):
if len(string) == 0:
return
temp = string[0]
reverse(string[1:])
print(temp, end='')
string = "Geeks for Geeks"
reverse(string)
|
C#
using System;
class GFG
{
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)));
}
}
public static void Main()
{
String str = "Geeks for Geeks";
reverse(str);
}
}
|
PHP
<?php
function reverse($str)
{
if (($str == null) ||
(strlen($str) <= 1))
echo ($str);
else
{
echo ($str[strlen($str) - 1]);
reverse(substr($str, 0,
(strlen($str) - 1)));
}
}
$str = "Geeks for Geeks";
reverse($str);
?>
|
Javascript
<script>
function reverse(str, len) {
if (len == str.length) {
return;
}
reverse(str, len + 1);
document.write(str[len]);
}
let a = "Geeks for Geeks";
reverse(a, 0);
</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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
