Decode the string encoded with the given algorithm - GeeksforGeeks (2024)

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Decode the string encoded with the given algorithm - GeeksforGeeks (1)

Last Updated : 31 May, 2021

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Given a decoded string str which was decoded with the following encoding algorithm:
Write down the middle character of the string then delete it and repeat the process until there are no characters left. For example, “abba” will be encoded as “bbaa”.
Note that the middle character is the first character of the two middle characters when the length of the string is even.
Examples:

Input: "ofrsgkeeeekgs"Output: geeksforgeeksInput: str = "bbaa"Output: abba

Approach: It can be observed that while decoding the string, the first letter of the encoded string becomes the median of the decoded string. So first, write the very first character of the encoded string and remove it from the encoded string then start adding the first character of the encoded string first to the left and then to the right of the decoded string and do this task repeatedly till the encoded string becomes empty.
For example:

Encoded String Decoded Stringofrsgkeeeekgs ofrsgkeeeekgs forsgkeeeekgs forsgkeeeekgs sforgkeeeekgs sforgkeeeekgs ksforgeeeekgs ksforgeeeekgs eksforgeeekgs eksforgeeekgs eeksforgeekgs eeksforgeekgs geeksgorgeeks geeksforgeeks

Below is the implementation of the above approach:

C++

// C++ implementation of the approach

#include <bits/stdc++.h>

using namespace std;

// Function to decode and print

// the original string

void decodeStr(string str, int len)

{

// To store the decoded string

char c[len] = "";

int med, pos = 1, k;

// Getting the mid element

if (len % 2 == 1)

med = len / 2;

else

med = len / 2 - 1;

// Storing the first element of the

// string at the median position

c[med] = str[0];

// If the length is even then store

// the second element also

if (len % 2 == 0)

c[med + 1] = str[1];

// k represents the number of characters

// that are already stored in the c[]

if (len & 1)

k = 1;

else

k = 2;

for (int i = k; i < len; i += 2) {

c[med - pos] = str[i];

// If string length is odd

if (len % 2 == 1)

c[med + pos] = str[i + 1];

// If it is even

else

c[med + pos + 1] = str[i + 1];

pos++;

}

// Print the decoded string

for (int i = 0; i < len; i++)

cout << c[i];

}

// Driver code

int main()

{

string str = "ofrsgkeeeekgs";

int len = str.length();

decodeStr(str, len);

return 0;

}

 
Java
 
// Java implementation of the approach
class GFG{
// Function to decode and print
// the original String
static void decodeStr(String str, int len)
{
// To store the decoded String
char []c = new char[len];
int med, pos = 1, k;
// Getting the mid element
if (len % 2 == 1)
med = len / 2;
else
See Also
Python encode() and decode() Functions - AskPythonMy Guide to Encoding/Decoding Text Data with PythonStata | Australia Indonesia New ZealandTextEncoder: encode() method - Web APIs | MDN
med = len / 2 - 1;
// Storing the first element of the
// String at the median position
c[med] = str.charAt(0);
// If the length is even then store
// the second element also
if (len % 2 == 0)
c[med + 1] = str.charAt(1);
// k represents the number of characters
// that are already stored in the c[]
if (len % 2 == 1)
k = 1;
else
k = 2;
for(int i = k; i < len; i += 2)
{
c[med - pos] = str.charAt(i);
// If String length is odd
if (len % 2 == 1)
c[med + pos] = str.charAt(i + 1);
// If it is even
else
c[med + pos + 1] = str.charAt(i + 1);
pos++;
}
// Print the decoded String
for (int i = 0; i < len; i++)
System.out.print(c[i]);
}
// Driver code
public static void main(String[] args)
{
String str = "ofrsgkeeeekgs";
int len = str.length();
decodeStr(str, len);
}
}
// This code is contributed by 29AjayKumar
 
 
Python3
 
# Python3 implementation of the 
# above approach
# Function to decode and print 
# the original string 
def decodeStr(str, len):
# To store the decoded string
c = ["" for i in range(len)]
pos = 1
# Getting the mid element 
if(len % 2 == 1):
med = int(len / 2)
else:
med = int(len / 2 - 1)
# Storing the first element 
# of the string at the 
# median position
c[med] = str[0]
# If the length is even 
# then store the second 
# element also
if(len % 2 == 0):
c[med + 1] = str[1]
# k represents the number 
# of characters that are 
# already stored in the c[]
if(len & 1):
k = 1
else:
k = 2
for i in range(k, len, 2):
c[med - pos] = str[i]
# If string length is odd
if(len % 2 == 1):
c[med + pos] = str[i + 1]
# If it is even
else:
c[med + pos + 1] = str[i + 1]
pos += 1
# Print the decoded string
print(*c, sep = "")
# Driver code
str = "ofrsgkeeeekgs"
len = len(str)
decodeStr(str, len)
# This code is contributed by avanitrachhadiya2155
 
 
C#
 
// C# implementation of the approach
using System;
class GFG{
// Function to decode and print
// the original String
static void decodeStr(String str, int len)
{
// To store the decoded String
char []c = new char[len];
int med, pos = 1, k;
// Getting the mid element
if (len % 2 == 1)
med = len / 2;
else
med = len / 2 - 1;
// Storing the first element of the
// String at the median position
c[med] = str[0];
// If the length is even then store
// the second element also
if (len % 2 == 0)
c[med + 1] = str[1];
// k represents the number of characters
// that are already stored in the c[]
if (len % 2 == 1)
k = 1;
else
k = 2;
for(int i = k; i < len; i += 2)
{
c[med - pos] = str[i];
// If String length is odd
if (len % 2 == 1)
c[med + pos] = str[i + 1];
See Also
ENCODE function
// If it is even
else
c[med + pos + 1] = str[i + 1];
pos++;
}
// Print the decoded String
for(int i = 0; i < len; i++)
Console.Write(c[i]);
}
// Driver code
public static void Main(String[] args)
{
String str = "ofrsgkeeeekgs";
int len = str.Length;
decodeStr(str, len);
}
}
// This code is contributed by sapnasingh4991
 
 
Javascript
 
<script>
// JavaScript implementation of the approach
// Function to decode and print
// the original string
function decodeStr(str, len)
{
// To store the decoded string
var c = Array(len).fill("");
var med, pos = 1, k;
// Getting the mid element
if (len % 2 == 1)
med = parseInt(len / 2);
else
med = parseInt(len / 2) - 1;
// Storing the first element of the
// string at the median position
c[med] = str[0];
// If the length is even then store
// the second element also
if (len % 2 == 0)
c[med + 1] = str[1];
// k represents the number of characters
// that are already stored in the c[]
if (len & 1)
k = 1;
else
k = 2;
for (var i = k; i < len; i += 2) {
c[med - pos] = str[i];
// If string length is odd
if (len % 2 == 1)
c[med + pos] = str[i + 1];
// If it is even
else
c[med + pos + 1] = str[i + 1];
pos++;
}
// Print the decoded string
for (var i = 0; i < len; i++)
{
document.write(c[i]);
}
}
// Driver code
var str = "ofrsgkeeeekgs";
var len = str.length;
decodeStr(str, len);
</script>
 
 
Output:
geeksforgeeks
The Complexity: O(n)

 
 
 
   
 
 
 
 
 
 
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 `; }, error:function(data) { jQuery('.spinner-loading-overlay:eq(0)').remove(); jQuery('#suggestion-modal-alert').html("Something went wrong."); jQuery('#suggestion-modal-alert').show(); error_msg = true; } }); } else{ jQuery('.spinner-loading-overlay:eq(0)').remove(); jQuery('#suggestion-modal-alert').html("Minimum 5 Words and Maximum Character limit is 2000."); jQuery('#suggestion-modal-alert').show(); jQuery('#suggestion-section-textarea').focus(); error_msg = true; } } else{ jQuery('.spinner-loading-overlay:eq(0)').remove(); jQuery('#suggestion-modal-alert').html("Enter atleast four words !"); jQuery('#suggestion-modal-alert').show(); jQuery('#suggestion-section-textarea').focus(); error_msg = true; } if(error_msg){ setTimeout(() => { jQuery('#suggestion-section-textarea').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; } // load the captcha script and set the token loadScript('https://www.google.com/recaptcha/api.js?render=6LdMFNUZAAAAAIuRtzg0piOT-qXCbDF-iQiUi9KY',[], function() { 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(); $('.improvement-reason-modal').hide(); $('.create-improvement-redirection-to-write').attr('href',writeUrl + 'improve-post/' + `${result.id}` + '/', '_blank'); $('.create-improvement-redirection-to-write')[0].click(); }, error:function(e) { $('.spinner-loading-overlay:eq(0)').remove(); var result = e.responseJSON; if(result.detail.non_field_errors.length){ $('.improve-modal--improve-content .improve-modal--improve-content-modified').text(`${result.detail.non_field_errors}.`); 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"); $('.improvement-reason-modal').hide(); } }, }); }); 
 
 
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; }  

                                                        Decode the string encoded with the given algorithm - GeeksforGeeks (2024)
                        

                            
                        

                    
                                                                                
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