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String formed with middle character of every right substring followed by left sequentially

  • Last Updated : 09 Jun, 2021

Given a string str of length N, the task is to decrypt it using a given set of decryption rules and print the decrypted string. 
The decryption rules are as follows: 
 

  • Start with the middle character of the string str and print it.
  • Repetitively traverse the right substring and print its middle character.
  • Repeat the same procedure for the left substring too.

Examples: 
 

Input: N = 4, str = "abcd" 
Output: bcda
Explanation:
              abcd  ------ b
              / \
             a   cd ------ c
            /     \
           a       d ----- d
          /
         a --------------- a
Hence, the final string is "bcda".

Input: N = 6, str = "gyuitp"
Output: utpigy
Explanation:
            gyuitp ------- u
             / \
           gy  itp ------- t
          /    /\
         gy   i  p ------ p
         /   /
        gy  i ----------- i
        /
       gy --------------- g
       \
        y  -------------- y
Hence, the final string is "utpigy".

 

Approach: 
The main idea is to use recursion. Keep on dividing the whole string into left and right substrings and print the middle element of every such substring, until the string is left with a single character and can not be further divided.
Detailed steps for this approach are as follows: 
 

  • Initialize start = 0, end = N -1, denoting the first and last character of the string.
  • Print the character at the middle of the string, that is mid = (start + end) / 2.
  • Recursively traverse its right substring (start = mid +1, end) followed by its left substring (start, mid – 1).
  • Repeat the above steps for each substring traversed. Continue until the entire string is traversed and print the given string.

Below is the implementation of the above approach: 
 



C++




// C++ implementation of
// the above appraoch
 
#include <iostream>
using namespace std;
 
// Function to decrypt and
// print the new string
void decrypt(string Str,
             int Start, int End)
{
    // If the whole string
    // has been traversed
    if (Start > End) {
        return;
    }
 
    // To calculate middle
    // index of the string
    int mid = (Start + End) >> 1;
 
    // Print the character
    // at middle index
    cout << Str[mid];
 
    // Recursively call
    // for right-substring
    decrypt(Str, mid + 1, End);
 
    // Recursive call
    // for left-substring
    decrypt(Str, Start, mid - 1);
}
 
// Driver Code
int main()
{
 
    int N = 4;
    string Str = "abcd";
    decrypt(Str, 0, N - 1);
    cout << "\n";
 
    N = 6;
    Str = "gyuitp";
    decrypt(Str, 0, N - 1);
 
    return 0;
}

Java




// Java implementation of
// the above appraoch
class GFG{
 
// Function to decrypt and
// print the new String
static void decrypt(String Str,
            int Start, int End)
{
    // If the whole String
    // has been traversed
    if (Start > End)
    {
        return;
    }
 
    // To calculate middle
    // index of the String
    int mid = (Start + End) >> 1;
 
    // Print the character
    // at middle index
    System.out.print(Str.charAt(mid));
 
    // Recursively call
    // for right-subString
    decrypt(Str, mid + 1, End);
 
    // Recursive call
    // for left-subString
    decrypt(Str, Start, mid - 1);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 4;
    String Str = "abcd";
    decrypt(Str, 0, N - 1);
    System.out.print("\n");
 
    N = 6;
    Str = "gyuitp";
    decrypt(Str, 0, N - 1);
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 implementation of
# the above appraoch
 
# Function to decrypt and
# print the new string
def decrypt(Str, Start, End):
 
    # If the whole string
    # has been traversed
    if (Start > End):
        return;
     
    # To calculate middle
    # index of the string
    mid = (Start + End) >> 1;
 
    # Print the character
    # at middle index
    print(Str[mid], end = "");
 
    # Recursively call
    # for right-substring
    decrypt(Str, mid + 1, End);
 
    # Recursive call
    # for left-substring
    decrypt(Str, Start, mid - 1);
 
# Driver Code
N = 4;
Str = "abcd";
decrypt(Str, 0, N - 1);
print();
 
N = 6;
Str = "gyuitp";
decrypt(Str, 0, N - 1);
 
# This code is contributed by Code_Mech

C#




// C# implementation of
// the above appraoch
using System;
class GFG{
 
// Function to decrypt and
// print the new String
static void decrypt(String Str,
            int Start, int End)
{
    // If the whole String
    // has been traversed
    if (Start > End)
    {
        return;
    }
 
    // To calculate middle
    // index of the String
    int mid = (Start + End) >> 1;
 
    // Print the character
    // at middle index
    Console.Write(Str[mid]);
 
    // Recursively call
    // for right-subString
    decrypt(Str, mid + 1, End);
 
    // Recursive call
    // for left-subString
    decrypt(Str, Start, mid - 1);
}
 
// Driver Code
public static void Main()
{
    int N = 4;
    String Str = "abcd";
    decrypt(Str, 0, N - 1);
    Console.Write("\n");
 
    N = 6;
    Str = "gyuitp";
    decrypt(Str, 0, N - 1);
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to decrypt and
// prlet the new String
function decrypt(Str, Start, End)
{
 
    // If the whole String
    // has been traversed
    if (Start > End)
    {
        return;
    }
 
    // To calculate middle
    // index of the String
    let mid = (Start + End) >> 1;
 
    // Prlet the character
    // at middle index
    document.write(Str[mid]);
 
    // Recursively call
    // for right-subString
    decrypt(Str, mid + 1, End);
 
    // Recursive call
    // for left-subString
    decrypt(Str, Start, mid - 1);
}
 
 
// Driver Code
    let N = 4;
    let Str = "abcd";
    decrypt(Str, 0, N - 1);
    document.write("<br/>");
 
    N = 6;
    Str = "gyuitp";
    decrypt(Str, 0, N - 1);
 
// This code is contributed by sanjoy_62.
</script>
Output: 
bcda
utpigy

 

Time complexity: O(N) 
Auxiliary Space: O(1) 
 

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