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Left Rotation of a String

Reverse a String – Complete Tutorial

Last Updated : 04 Jun, 2025

Try it on GfG Practice

Given a string s, the task is to reverse the string. Reversing a string means rearranging the characters such that the first character becomes the last, the second character becomes second last and so on.

Examples:

Input: s = "GeeksforGeeks"
Output: "skeeGrofskeeG"
Explanation : The first character G moves to last position, the second character e moves to second-last and so on.
Input: s = "abdcfe"
Output: "efcdba"
Explanation: The first character a moves to last position, the second character b moves to second-last and so on.

Table of Content

Using backward traversal – O(n) Time and O(n) Space

The idea is to start at the last character of the string and move backward, appending each character to a new string res. This new string res will contain the characters of the original string in reverse order.

C++

// C++ program to reverse a string using backward traversal

#include <iostream>
#include <string>
using namespace std;

string reverseString(string& s) {
    string res;
  
  	// Traverse on s in backward direction
  	// and add each charecter to a new string
    for (int i = s.size() - 1; i >= 0; i--) {
        res += s[i];
    }
    return res;
}

int main() {
    string s = "abdcfe";
    string res = reverseString(s);
    cout << res;
    return 0;
}

C

// C program to reverse a string using backward traversal

#include <stdio.h>
#include <string.h>

char *reverseString(char *s) {
    int n = strlen(s);
    char *res = (char *)malloc((n + 1) * sizeof(char));

    int j = 0;
    
    // Traverse on s in backward direction
    // and add each character to a new string
    for (int i = n - 1; i >= 0; i--) {
        res[j] = s[i];
        j++;
    }

    // Null-terminate the result string
    res[n] = '\0';
    return res;
}

int main() {
    char s[] = "abdcfe";
    char *res = reverseString(s);
    printf("%s", res);
    return 0;
}

Java

// Java program to reverse a string using backward traversal

class GfG {
    static String reverseString(String s) {
        StringBuilder res = new StringBuilder();
  
        // Traverse on s in backward direction
        // and add each character to a new string
        for (int i = s.length() - 1; i >= 0; i--) {
            res.append(s.charAt(i));
        }
        return res.toString();
    }

    public static void main(String[] args) {
        String s = "abdcfe";
        String res = reverseString(s);
        System.out.print(res);
    }
}

Python

# Python program to reverse a string using backward traversal

def reverseString(s):
    res = []
  
    # Traverse on s in backward direction
    # and add each character to the list
    for i in range(len(s) - 1, -1, -1):
        res.append(s[i])

    # Convert list back to string
    return ''.join(res)

if __name__ == "__main__":
    s = "abdcfe"
    print(reverseString(s))

C#

// C# program to reverse a string using backward traversal

using System;
using System.Text;

class GfG {
    static string reverseString(string s) {
        StringBuilder res = new StringBuilder();
  
        // Traverse on s in backward direction
        // and add each character to a new string
        for (int i = s.Length - 1; i >= 0; i--) {
            res.Append(s[i]);
        }
      
        // Convert StringBuilder to string
        return res.ToString(); 
    }

    static void Main(string[] args) {
        string s = "abdcfe";
        string res = reverseString(s);
        Console.WriteLine(res);
    }
}

JavaScript

// JavaScript program to reverse a string using backward traversal

function reverseString(s) {
    let res = [];
  
    // Traverse on s in backward direction
    // and add each character to the array
    for (let i = s.length - 1; i >= 0; i--) {
        res.push(s[i]);
    }
    return res.join('');
}

const s = "abdcfe";
console.log(reverseString(s));

Output

efcdba

Time Complexity: O(n) for backward traversal
Auxiliary Space: O(n) for storing the reversed string.

Using Two Pointers - O(n) Time and O(1) Space

The idea is to maintain two pointers: left and right, such that left points to the beginning of the string and right points to the end of the string.
While left pointer is less than the right pointer, swap the characters at these two positions. After each swap, increment the left pointer and decrement the right pointer to move towards the center of the string. This will swap all the characters in the first half with their corresponding character in the second half.

Illustration:

C++

// C++ program to reverse a string using two pointers

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

string reverseString(string &s) {
    int left = 0, right = s.length() - 1;

    // Swap characters from both ends till we reach
    // the middle of the string
    while (left < right) {
        swap(s[left], s[right]);
        left++;
        right--;
    }
  
    return s;
}

int main() {
    string s = "abdcfe";
    cout << reverseString(s);
    return 0;
}

C

// C program to reverse a string using two pointers

#include <stdio.h>
#include <string.h>

char* reverseString(char *s) {
    int left = 0, right = strlen(s) - 1;

    // Swap characters from both ends till we reach
    // the middle of the string
    while (left < right) {
        char temp = s[left];
        s[left] = s[right];
        s[right] = temp;
        left++;
        right--;
    }
  
    return s;
}

int main() {
    char s[] = "abdcfe"; 
    printf("%s", reverseString(s)); 
    return 0;
}

Java

// Java program to reverse a string using two pointers

class GfG {
    static String reverseString(String s) {
        int left = 0, right = s.length() - 1;
      
        // Use StringBuilder for mutability
        StringBuilder res = new StringBuilder(s);
      
        // Swap characters from both ends till we reach
        // the middle of the string
        while (left < right) {
            char temp = res.charAt(left);
            res.setCharAt(left, res.charAt(right));
            res.setCharAt(right, temp);
            left++;
            right--;
        }
  
        // Convert StringBuilder back to string
        return res.toString();
    }

    public static void main(String[] args) {
        String s = "abdcfe";
        System.out.println(reverseString(s));
    }
}

Python

# Python program to reverse a string using two pointers

# Function to reverse a string using two pointers

def reverseString(s):
    left = 0
    right = len(s) - 1
    
    # Convert string to a list for mutability
    s = list(s)  
    
    # Swap characters from both ends till we reach
    # the middle of the string
    while left < right:
        s[left], s[right] = s[right], s[left]
        left += 1
        right -= 1
    
    # Convert list back to string
    return ''.join(s)  

if __name__ == "__main__":
    s = "abdcfe"
    print(reverseString(s))

C#

// C# program to reverse a string using two pointers

using System;
using System.Text;

class GfG {
    static string reverseString(string s) {
      
        // Use StringBuilder for mutability
        StringBuilder res = new StringBuilder(s); 
        int left = 0, right = res.Length - 1;

        // Swap characters from both ends till we reach
        // the middle of the string
        while (left < right) {
            char temp = res[left];
            res[left] = res[right];
            res[right] = temp;
            left++;
            right--;
        }
  
        // Convert StringBuilder back to string
        return res.ToString(); 
    }

    static void Main(string[] args) {
        string s = "abdcfe";
        Console.WriteLine(reverseString(s));
    }
}

JavaScript

// JavaScript program to reverse a string using two pointers

function reverseString(s) {
    let left = 0, right = s.length - 1;

    // Convert string to array for mutability
    s = s.split(''); 
    
    // Swap characters from both ends till we reach
    // the middle of the string
    while (left < right) {
        [s[left], s[right]] = [s[right], s[left]];
        left++;
        right--;
    }
  
    return s.join('');
}

const s = "abdcfe";
console.log(reverseString(s));

Output

efcdba

Time Complexity: O(n)
Auxiliary Space: O(1)

Using Recursion - O(n) Time and O(n) Space

The idea is to use recursion and define a recursive function that takes a string as input and reverses it. Inside the recursive function,
Swap the first and last element.
Recursively call the function with the remaining substring.

C++

// C++ Program to reverse an array using Recursion

#include <iostream>
#include <vector>
using namespace std;

// recursive function to reverse a string from l to r
void reverseStringRec(string &s, int l, int r) {
  
    // If the substring is empty, return
    if(l >= r)
        return;
  
    swap(s[l], s[r]);
  	
    // Recur for the remaining string
    reverseStringRec(s, l + 1, r - 1);
}

// function to reverse a string
string reverseString(string &s) {
    int n = s.length();
    reverseStringRec(s, 0, n - 1);
  	return s;
}

int main() {
    string s = "abdcfe";
  	cout << reverseString(s) << endl;
    return 0;
}

C

// C program to reverse a string using Recursion

#include <stdio.h>
#include <string.h>

// Recursive function to reverse a string from l to r
void reverseStringRec(char *s, int l, int r) {
    if (l >= r)
        return;

    // Swap the characters at the ends
    char temp = s[l];
    s[l] = s[r];
    s[r] = temp;

    // Recur for the remaining string
    reverseStringRec(s, l + 1, r - 1);
}

char* reverseString(char *s) {
    int n = strlen(s);
    reverseStringRec(s, 0, n - 1);
  	return s;
}

int main() {
    char s[] = "abdcfe";
  
    printf("%s\n", reverseString(s));
    return 0;
}

Java

// Java program to reverse a string using Recursion

class GfG {
  
    // Recursive function to reverse a string from l to r
    static void reverseStringRec(char[] s, int l, int r) {
        if (l >= r)
            return;

        // Swap the characters at the ends
        char temp = s[l];
        s[l] = s[r];
        s[r] = temp;

        // Recur for the remaining string
        reverseStringRec(s, l + 1, r - 1);
    }

    // Function to reverse a string
    static String reverseString(String s) {
        char[] arr = s.toCharArray();
        reverseStringRec(arr, 0, arr.length - 1);
        return new String(arr);
    }
	
    public static void main(String[] args) {
        String s = "abdcfe";
        System.out.println(reverseString(s));
    }
}

Python

# Python program to reverse a string using Recursion

# Recursive Function to reverse a string
def reverseStringRec(arr, l, r):
    if l >= r:
        return

    # Swap the characters at the ends
    arr[l], arr[r] = arr[r], arr[l]

    # Recur for the remaining string
    reverseStringRec(arr, l + 1, r - 1)

def reverseString(s):
  
  	# Convert string to list of characters
    arr = list(s)  
    reverseStringRec(arr, 0, len(arr) - 1)
    
    # Convert list back to string
    return ''.join(arr)  

if __name__ == "__main__":
    s = "abdcfe"
    print(reverseString(s))

C#

// C# program to reverse a string using Recursion

using System;
using System.Text;

class GfG {
  
    // recursive function to reverse a string from l to r
    static void reverseStringRec(StringBuilder s, int l, int r) {
        if (l >= r)
            return;

        char temp = s[l];
        s[l] = s[r];
        s[r] = temp;

        // Recur for the remaining string
        reverseStringRec(s, l + 1, r - 1);
    }

    // function to reverse a string
    static string reverseString(string input) {
        StringBuilder s = new StringBuilder(input);
        int n = s.Length;
        reverseStringRec(s, 0, n - 1);
        return s.ToString();
    }

    static void Main() {
        string s = "abdcfe";
        Console.WriteLine(reverseString(s));
    }
}

JavaScript

// JavaScript program to reverse a string using Recursion

// Recursive Function to reverse a string
function reverseStringRec(res, l, r) {
    if (l >= r) return;

    // Swap the characters at the ends
    [res[l], res[r]] = [res[r], res[l]];

    // Recur for the remaining string
    reverseStringRec(res, l + 1, r - 1);
}

function reverseString(s) {

	// Convert string to array of characters
    let res = s.split('');  
    reverseStringRec(res, 0, res.length - 1);
    
    // Convert array back to string
    return res.join('');  
}

let s = "abdcfe";
console.log(reverseString(s));

Output

efcdba

Time Complexity: O(n) where n is length of string
Auxiliary Space: O(n)

Using Stack - O(n) Time and O(n) Space

The idea is to use stack for reversing a string because Stack follows Last In First Out (LIFO) principle. This means the last character you add is the first one you'll take out. So, when we push all the characters of a string into the stack, the last character becomes the first one to pop.

Illustration:

C++

// C++ program to reverse a string using stack

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

string reverseString(string &s) {
    stack<char> st;
  
    // Push the charcters into stack
    for (int i = 0; i < s.size(); i++) {
        st.push(s[i]);
    }

    // Pop the characters of stack into the original string
    for (int i = 0; i < s.size(); i++) {
        s[i] = st.top();
        st.pop();
    }
  
  	return s;
}

int main() {
    string s = "abdcfe";
    cout << reverseString(s);
    return 0;
}

Java

// Java program to reverse a string using stack
import java.util.*;

class GfG {
    static String reverseString(String s) {
        Stack<Character> st = new Stack<>();
      
        // Push the characters into stack
        for (int i = 0; i < s.length(); i++)
            st.push(s.charAt(i));

        StringBuilder res = new StringBuilder();
      
        // Pop the characters of stack into the original string
        for (int i = 0; i < s.length(); i++)
            res.append(st.pop());
        
        return res.toString();
    }

    public static void main(String[] args) {
        String s = "abdcfe";
        System.out.println(reverseString(s));
    }
}

Python

# Python program to reverse a string using stack
def reverseString(s):
    stack = []
    
    # Push the characters into stack
    for char in s:
        stack.append(char)

    # Prepare a list to hold the reversed characters
    rev = [''] * len(s)

    # Pop the characters from stack into the reversed list
    for i in range(len(s)):
        rev[i] = stack.pop()

    # Join the list to form the reversed string
    return ''.join(rev)

if __name__ == "__main__":
    s = "abdcfe"
    print(reverseString(s))

C#

// C# program to reverse a string using stack

using System;
using System.Collections.Generic;
using System.Text;

class GfG {
    static string reverseString(string s) {
        Stack<char> st = new Stack<char>();

        // Push the characters into stack
        for (int i = 0; i < s.Length; i++)
            st.Push(s[i]);

        // Create a StringBuilder to hold the reversed string
        StringBuilder res = new StringBuilder();

        // Pop the characters from stack into the StringBuilder
        while (st.Count > 0)
            res.Append(st.Pop());

        // Convert the StringBuilder back to a string
        return res.ToString();
    }

    static void Main() {
        string s = "abdcfe";
        Console.WriteLine(reverseString(s));
    }
}

JavaScript

// JavaScript program to reverse a string using stack

function reverseString(s) {

    // To store the characters of the original string.
    const stack = [];

    // Push the characters into the stack.
    for (let i = 0; i < s.length; i++) {
        stack.push(s[i]);
    }

    // Create an array to hold the reversed characters
    const res = new Array(s.length);
	
    // Pop the characters of the stack and store in the array
    for (let i = 0; i < s.length; i++) {
        res[i] = stack.pop();
    }

    // Join the array to form the reversed string
    return res.join('');
}

let str = "abdcfe";
console.log(reverseString(str));

Output

efcdba

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

Using Inbuilt methods - O(n) Time and O(1) Space

The idea is to use built-in reverse method to reverse the string. If built-in method for string reversal does not exist, then convert string to array or list and use their built-in method for reverse. Then convert it back to string.

C++

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

string reverseString(string &s) {
    reverse(s.begin(), s.end());
  	return s;
}

int main() {
    string s = "abdcfe"; 
    cout << reverseString(s) ;
    return 0;
}

Java

// Java program to reverse a string using StringBuffer class

import java.io.*;
import java.util.*;

class GFG {
    static String stringReverse(String s) {
    	StringBuilder res = new StringBuilder(s);
        res.reverse();
      	return res.toString();
    }
  	
    public static void main(String[] args) {
        String s = "abdcfe";
        System.out.println(stringReverse(s));
    }
}

Python

# Function to reverse a string
def reverseString(s):
  	
    # Reverse the string using slicing
    return s[::-1]

if __name__ == "__main__":
    str = "abdcfe"
    print(reverseString(str))

C#

// C# program to reverse a string
using System;

class GfG {
    static string reverseString(string s) {
      	
        // Reverse the string using the built-in method
        char[] arr = s.ToCharArray();
        Array.Reverse(arr);
        return new string(arr);
    }

    static void Main() {
        string s = "abdcfe"; 
        Console.WriteLine(reverseString(s));
    }
}

JavaScript

// Function to reverse a string
function reverseString(s) {
	
    // convert to array to make it mutable
	s = s.split('');
    
    // Reverse the string using the built-in method
    s = s.reverse();
    return s.join('');
}

let str = "abdcfe"; 
console.log(reverseString(str));

Output

efcdba

Time Complexity: O(n)
Auxiliary Space: O(1) in C++ and python and O(n) in Java, C# and JavaScript (extra space is used to store in array or list or StringBuilder for reversal).

Left Rotation of a String

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'); $('.spinner-loading-overlay').show(); jQuery.ajax({ url: writeApiUrl + 'create-improvement-post/?v=1', type: "POST", contentType: 'application/json; charset=utf-8', dataType: 'json', xhrFields: { withCredentials: true }, data: JSON.stringify({ gfg_id: post_id }), success:function(result) { $('.spinner-loading-overlay:eq(0)').remove(); $('.improve-modal--overlay').hide(); $('.unlocked-status--improve-modal-content').css("display","none"); $('.create-improvement-redirection-to-write').attr('href',writeUrl + 'improve-post/' + `${result.id}` + '/', '_blank'); $('.create-improvement-redirection-to-write')[0].click(); }, error:function(e) { showErrorMessage(e.responseJSON,e.status) }, }); } else { if(loginData && !loginData.isLoggedIn) { $('.improve-modal--overlay').hide(); if ($('.header-main__wrapper').find('.header-main__signup.login-modal-btn').length) { $('.header-main__wrapper').find('.header-main__signup.login-modal-btn').click(); } return; } } }); $('.left-arrow-icon_wrapper').on('click',function(){ if($('.improve-modal--suggestion').is(":visible")) $('.improve-modal--suggestion').hide(); else{ } $('.improve-modal--improvement').show(); }); const showErrorMessage = (result,statusCode) => { if(!result) return; $('.spinner-loading-overlay:eq(0)').remove(); if(statusCode == 403) { $('.improve-modal--improve-content.error-message').html(result.message); jQuery('.improve-modal--overlay').show(); jQuery('.improve-modal--improvement').show(); $('.locked-status--impove-modal').css("display","block"); $('.unlocked-status--improve-modal-content').css("display","none"); $('.improve-modal--improvement').attr("status","locked"); return; } } function suggestionCall() { var editorValue = suggestEditorInstance.getValue(); var suggest_val = $(".ContentEditable__root").find("[data-lexical-text='true']").map(function() { return $(this).text().trim(); }).get().join(' '); suggest_val = suggest_val.replace(/\s+/g, ' ').trim(); var array_String= suggest_val.split(" ") //array of words var gCaptchaToken = $("#g-recaptcha-response-suggestion-form").val(); var error_msg = false; if(suggest_val != "" && array_String.length >=4){ if(editorValue.length { jQuery('.ContentEditable__root').focus(); jQuery('#suggestion-modal-alert').hide(); }, 3000); } } document.querySelector('.suggest-bottom-btn').addEventListener('click', function(){ jQuery('body').append('

'); jQuery('.spinner-loading-overlay').show(); if(loginData && loginData.isLoggedIn) { suggestionCall(); return; } // script for grecaptcha loaded in loginmodal.html and call function to set the token setGoogleRecaptcha(); }); $('.improvement-bottom-btn.create-improvement-btn').click(function() { //create improvement button is clicked $('body').append('

'); $('.spinner-loading-overlay').show(); // send this option via create-improvement-post api jQuery.ajax({ url: writeApiUrl + 'create-improvement-post/?v=1', type: "POST", contentType: 'application/json; charset=utf-8', dataType: 'json', xhrFields: { withCredentials: true }, data: JSON.stringify({ gfg_id: post_id }), success:function(result) { $('.spinner-loading-overlay:eq(0)').remove(); $('.improve-modal--overlay').hide(); $('.create-improvement-redirection-to-write').attr('href',writeUrl + 'improve-post/' + `${result.id}` + '/', '_blank'); $('.create-improvement-redirection-to-write')[0].click(); }, error:function(e) { showErrorMessage(e.responseJSON,e.status); }, }); });

Continue without supporting

`; $('body').append(adBlockerModal); $('body').addClass('body-for-ad-blocker'); const modal = document.getElementById("adBlockerModal"); modal.style.display = "block"; } function handleAdBlockerClick(type){ if(type == 'disabled'){ window.location.reload(); } else if(type == 'info'){ document.getElementById("ad-blocker-div").style.display = "none"; document.getElementById("ad-blocker-info-div").style.display = "flex"; handleAdBlockerIconClick(0); } } var lastSelected= null; //Mapping of name and video URL with the index. const adBlockerVideoMap = [ ['Ad Block Plus','https://media.geeksforgeeks.org/auth-dashboard-uploads/abp-blocker-min.mp4'], ['Ad Block','https://media.geeksforgeeks.org/auth-dashboard-uploads/Ad-block-min.mp4'], ['uBlock Origin','https://media.geeksforgeeks.org/auth-dashboard-uploads/ub-blocke-min.mp4'], ['uBlock','https://media.geeksforgeeks.org/auth-dashboard-uploads/U-blocker-min.mp4'], ] function handleAdBlockerIconClick(currSelected){ const videocontainer = document.getElementById('ad-blocker-info-div-gif'); const videosource = document.getElementById('ad-blocker-info-div-gif-src'); if(lastSelected != null){ document.getElementById("ad-blocker-info-div-icons-"+lastSelected).style.backgroundColor = "white"; document.getElementById("ad-blocker-info-div-icons-"+lastSelected).style.borderColor = "#D6D6D6"; } document.getElementById("ad-blocker-info-div-icons-"+currSelected).style.backgroundColor = "#D9D9D9"; document.getElementById("ad-blocker-info-div-icons-"+currSelected).style.borderColor = "#848484"; document.getElementById('ad-blocker-info-div-name-span').innerHTML = adBlockerVideoMap[currSelected][0] videocontainer.pause(); videosource.setAttribute('src', adBlockerVideoMap[currSelected][1]); videocontainer.load(); videocontainer.play(); lastSelected = currSelected; }