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 approach#include <iostream>using namespace std;// Function to decrypt and// print the new stringvoid 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 Codeint 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 approachclass GFG{// Function to decrypt and// print the new Stringstatic 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 Codepublic 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 approach# Function to decrypt and# print the new stringdef 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 CodeN = 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 approachusing System;class GFG{// Function to decrypt and// print the new Stringstatic 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 Codepublic 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// print the new Stringfunction 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; // Print 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(N)
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