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++
- Java
- Python3
- C#
- Javascript
C++
// 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
Python3
C#
Javascript
geeksforgeeks
The Complexity: O(n)
