Program to find the Encrypted word Last Updated : 02 Feb, 2023 Comments Improve Suggest changes Like Article Like Report 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: bdeeExplanation:a -> a + 1 -> bb -> b + 2 -> dc -> c + 2 -> ed -> d + 1 -> e Input: str = "xyz"Output: yaaExplanation:x -> x + 1 -> yy -> y + 2 -> az -> 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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