# Decode the string encoded with the given algorithm

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: geeksforgeeks

Input: 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 String
ofrsgkeeeekgs           o
frsgkeeeekgs            fo
rsgkeeeekgs             for
sgkeeeekgs              sfor
gkeeeekgs               sforg
keeeekgs                ksforg
eeeekgs                 ksforge
eeekgs                  eksforge
eekgs                   eksforgee
ekgs                    eeksforgee
kgs                     eeksforgeek
gs                      geeksgorgeek
s                       geeksforgeeks

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include 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         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];                 // 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



Output:

geeksforgeeks

The Complexity: O(n)

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