Program to find the Encrypted word

0 0 3 minutes read

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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

// Function to find the encrypted string
function findWord(c, n) {
    let co = 0, i;
 
    // to store the encrypted string
    let s = new Array(n).fill(' ');
 
    for (i = 0; i < n; i++) {
        if (i < n / 2)
            co++;
        else
            co = n - i;
 
        // after 'z', it should go to a.
        if (c.charCodeAt(i) + co <= 122)
            s[i] = String.fromCharCode(c.charCodeAt(i) + co);
        else
            s[i] = String.fromCharCode(c.charCodeAt(i) + co - 26);
    }
    console.log(s.join(''));
}
 
// Driver code
let s = "abcd";
findWord(s, s.length);

Output:

bdee

Time Complexity: O(N)
Auxiliary Space: O(N), The extra space is used to store the result.

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