Decode the string encoded with the given algorithm

31 May, 2021
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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to decode and print// the original stringvoid 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 codeint main(){    string str = "ofrsgkeeeekgs";    int len = str.length();     decodeStr(str, len);     return 0;}
// Java implementation of the approachclass GFG{ // Function to decode and print// the original Stringstatic 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 codepublic static void main(String[] args){    String str = "ofrsgkeeeekgs";    int len = str.length();     decodeStr(str, len);}} // This code is contributed by 29AjayKumar
# Python3 implementation of the# above approach # Function to decode and print# the original stringdef 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 codestr = "ofrsgkeeeekgs"len = len(str)decodeStr(str, len) # This code is contributed by avanitrachhadiya2155
// C# implementation of the approachusing System; class GFG{ // Function to decode and print// the original Stringstatic 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 codepublic static void Main(String[] args){    String str = "ofrsgkeeeekgs";    int len = str.Length;     decodeStr(str, len);}} // This code is contributed by sapnasingh4991
<script>  // JavaScript implementation of the approach // Function to decode and print// the original stringfunction 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 codevar str = "ofrsgkeeeekgs";var len = str.length;decodeStr(str, len); </script>
geeksforgeeks
 
The Complexity: O(n) 
29AjayKumar
sapnasingh4991
avanitrachhadiya2155
rutvik_56
encoding-decoding
Strings
Strings
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