Program to find the Encrypted word

Given a string, the given string is an encrypted word, the task is to decrypt the given string to get the original word.

Examples:

Input: str = "abcd"
Output: bdee
Explanation:
a -> a + 1 -> b
b -> b + 2 -> d
c -> c + 2 -> e
d -> d + 1 -> e

Input: str = "xyz"
Output: yaa
Explanation:
x -> x + 1 -> y
y -> y + 2 -> a
z -> z + 1 -> a

Approach:

  • Let the length of the string be n.
  • then the encrypted string will be:
  • Print the string after finding the scrypted word.

Below is the implementation of the above approach:

C++

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// C++ program to implement 
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the encrypted string
void findWord(string c, int n)
{
    int co = 0, i;
  
    // to store the encrypted string
    string s(n, ' ');
  
    for (i = 0; i < n; i++) {
        if (i < n / 2)
            co++;
        else
            co = n - i;
  
        // after 'z', it should go to a.
        if (c[i] + co <= 122)
            s[i] = (char)((int)c[i] + co);
        else
            s[i] = (char)((int)c[i] + co - 26);
    }
    cout << s;
}
  
// Driver code
int main()
{
    string s = "abcd";
    findWord(s, s.length());
    return 0;
}

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Java

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// Java program to implement the above approach
import java.util.*;
import java.io.*;
  
class GFG
{
  
// Static function declared to find
// the encrypted string
public static void findWord(String c, int n)
{
    int co = 0, i;
  
    // Character array to store
    //the encrypted string
    char s[] = new char[n];
      
    for (i = 0; i < n ; i++)
    {
        if (i < n / 2)
            co++;
        else
            co = n - i;
  
        // after 'z', it should go to a.
        if ((c.charAt(i) + co) <= 122)
            s[i] = (char)((int)c.charAt(i) + co);
        else
            s[i] = (char)((int)c.charAt(i) + co - 26);
    }
      
    // storing the character array in the string.
    String str = Arrays.toString(s);
    System.out.println(str);
}
  
// Driver code
public static void main(String args[])
{
    String s = "abcd";
    findWord(s, s.length());
}
}
  
// This code is contributed by Animesh_Gupta

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Python3

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# Python3 program to implement 
# the above approach
  
# Function to find the encrypted string
def findWord(c, n):
    co = 0
      
    # to store the encrypted string
    s = [0] * n
    for i in range(n):
        if (i < n / 2):
            co += 1
        else:
            co = n - i
          
        # after 'z', it should go to a.
        if (ord(c[i]) + co <= 122):
            s[i] = chr(ord(c[i]) + co)
        else:
            s[i] = chr(ord(c[i]) + co - 26)
    print(*s, sep = "")
  
# Driver code
s = "abcd"
findWord(s, len(s))
  
# This code is contributed by SHUBHAMSINGH10

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C#

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// C# program to implement the above approach
using System;
  
class GFG
{
  
// Static function declared to find
// the encrypted string
public static void findWord(String c, int n)
{
    int co = 0, i;
  
    // Character array to store
    // the encrypted string
    char []s = new char[n];
      
    for (i = 0; i < n ; i++)
    {
        if (i < n / 2)
            co++;
        else
            co = n - i;
  
        // after 'z', it should go to a.
        if ((c[i] + co) <= 122)
            s[i] = (char)((int)c[i] + co);
        else
            s[i] = (char)((int)c[i] + co - 26);
    }
      
    // storing the character array in the string.
    String str = String.Join("",s);
    Console.WriteLine(str);
}
  
// Driver code
public static void Main(String []args)
{
    String s = "abcd";
    findWord(s, s.Length);
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

bdee

Time Complexity: O(N)

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