Decode a median string to the original string

Given a string s written in median form, change it back to the original string. The median letter in a string is the letter that is in the middle of the string. If the string’s length is even, the median letter is the left of the two middle letters. The given string is formed by writing down the median letter of the word, then deleting it and repeating the process until there are no letters left. 

Examples:

Input: eekgs
Output: geeks 
Explanation: in the original string “geeks” 
can be written in median form by picking up 
e first then, again e, then k then g and at
the end s. As these are the median when the
median letter is picked and deleted. 

Input: abc 
Output: bac 
Explanation: median of bac is a, then median
of bc is b, then median of c is c.  

To find the answer we can iterate through the given encoded string from left to right and add each letter in the answer string, one letter to the beginning, next letter to the end, next letter to begin, and so on. If n is even then the first letter must be added to the beginning and the second letter to the end. In the other case, the first letter to the end, second to the beginning. We need to make it until we do not add all letters from the given string. 

Note: For strings with even length, when we add the first character to begin and the second character to the end then the remaining string will always be of even length. The same is true for strings with odd lengths.

Given below is the implementation of the above approach

C++

// C++ program to decode a median string
// to the original string

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

// function to calculate the median back string
string decodeMedianString(string s)
{
    // length of string
    int l = s.length();

    // initialize a blank string
    string s1 = "";

    // Flag to check if length is even or odd
    bool isEven = (l % 2 == 0)? true : false;

    // traverse from first to last
    for (int i = 0; i < l; i += 2) {

        // if len is even then add first character
        // to beginning of new string and second
        // character to end
        if (isEven) {
            s1 = s[i] + s1;
            s1 += s[i + 1];
        } else {

            // if current length is odd and is
            // greater than 1
            if (l - i > 1) {

                // add first character to end and
                // second character to beginning
                s1 += s[i];
                s1 = s[i + 1] + s1;
            } else {

                // if length is 1, add character
                // to end
                s1 += s[i];
            }
        }
    }

    return s1;
}

// driver program
int main()
{
    string s = "eekgs";
    cout << decodeMedianString(s);
    return 0;
}

Java

// java program to decode a median
// string to the original string

public class GFG {

    // function to calculate the
    // median back string
    static String decodeMedianString(String s)
    {

        // length of string
        int l = s.length();

        // initialize a blank string
        String s1 = "";

        // Flag to check if length is
        // even or odd
        boolean isEven = (l % 2 == 0) ?
                          true : false;

        // traverse from first to last
        for (int i = 0; i < l; i += 2)
        {

            // if len is even then add
            // first character to
            // beginning of new string
            // and second character to
            // end
            if (isEven) {
                s1 = s.charAt(i) + s1;
                s1 += s.charAt(i+1);
            }
            else {

                // if current length is
                // odd and is greater
                // than 1
                if (l - i > 1) {

                    // add first character
                    // to end and second
                    // character to
                    // beginning
                    s1 += s.charAt(i);
                    s1 = s.charAt(i+1) + s1;
                }
                else {

                    // if length is 1,
                    // add character
                    // to end
                    s1 += s.charAt(i);
                }
            }
        }

        return s1;
    }

    // Driver code
    public static void main(String args[])
    {
        String s = "eekgs";

        System.out.println(
                    decodeMedianString(s));
    }
}

// This code is contributed by Sam007.

Python3

# Python3 program to decode a median
# string to the original string

# function to calculate the median
# back string
def decodeMedianString(s):

    # length of string
    l = len(s)

    # initialize a blank string
    s1 = ""

    # Flag to check if length is
    # even or odd
    if(l % 2 == 0):
        isEven = True
    else:
        isEven = False

    # traverse from first to last
    for i in range(0, l, 2):

        # if len is even then add first
        # character to beginning of new
        # string and second character to end
        if (isEven):
            s1 = s[i] + s1
            s1 += s[i + 1]
        else :

            # if current length is odd and
            # is greater than 1
            if (l - i > 1):

                # add first character to end and
                # second character to beginning
                s1 += s[i]
                s1 = s[i + 1] + s1
            else:

                # if length is 1, add character
                # to end
                s1 += s[i]

    return s1

# Driver Code
if __name__ == '__main__':
    s = "eekgs"
    print(decodeMedianString(s))

# This code is contributed by
# Sanjit_Prasad

C#

// C# program to decode a median
// string to the original string
using System;

class GFG {

    // function to calculate the
    // median back string
    static string decodeMedianString(string s)
    {

        // length of string
        int l = s.Length;

        // initialize a blank string
        string s1 = "";

        // Flag to check if length is
        // even or odd
        bool isEven = (l % 2 == 0) ?
                         true : false;

        // traverse from first to last
        for (int i = 0; i < l; i += 2)
        {

            // if len is even then add
            // first character to
            // beginning of new string
            // and second character to
            // end
            if (isEven) {
                s1 = s[i] + s1;
                s1 += s[i + 1];
            }
            else {

                // if current length is
                // odd and is greater
                // than 1
                if (l - i > 1) {

                    // add first character
                    // to end and second
                    // character to
                    // beginning
                    s1 += s[i];
                    s1 = s[i + 1] + s1;
                }
                else {

                    // if length is 1,
                    // add character
                    // to end
                    s1 += s[i];
                }
            }
        }

        return s1;
    }

    // Driver code
    public static void Main ()
    {
        string s = "eekgs";
        Console.WriteLine(
               decodeMedianString(s));
    }
}

// This code is contributed by Sam007.

Javascript

<script>
// javascript program to decode a median
// string to the original string



    // function to calculate the
    // median back string
    function decodeMedianString( s) {

        // length of string
        var l = s.length;

        // initialize a blank string
        var s1 = "";

        // Flag to check if length is
        // even or odd
        var isEven = (l % 2 == 0) ? true : false;

        // traverse from first to last
        for (i = 0; i < l; i += 2) {

            // if len is even then add
            // first character to
            // beginning of new string
            // and second character to
            // end
            if (isEven) {
                s1 = s.charAt(i) + s1;
                s1 += s.charAt(i + 1);
            } else {

                // if current length is
                // odd and is greater
                // than 1
                if (l - i > 1) {

                    // add first character
                    // to end and second
                    // character to
                    // beginning
                    s1 += s.charAt(i);
                    s1 = s.charAt(i + 1) + s1;
                } else {

                    // if length is 1,
                    // add character
                    // to end
                    s1 += s.charAt(i);
                }
            }
        }

        return s1;
    }

    // Driver code

        var s = "eekgs";

        document.write(decodeMedianString(s));

// This code contributed by Rajput-Ji
</script>

Output

geeks

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.

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