Given a string s, the task is to encrypt the string in the following way:
- If the frequency of current character is even, then increment current character by x.
- If the frequency of current character is odd, then decrement current character by x.
Note: All the operations are circular that is adding 1 to ‘z’ will give ‘a’ and subtracting 1 from ‘a’ will give ‘z’
Examples:
Input :s=”abcda”, x=3
Output :dyzad
‘a’ appear 2 times in the string, hence incrementing ‘a’ by 3 becomes ‘d’
‘b’ appear 1 times in the string, hence decrementing ‘b’ by 3 becomes ‘y’
‘c’ appear 1 times in the string, hence decrementing ‘c’ by 3 becomes ‘z’
‘d’ appear 1 times in the string, hence decrementing ‘d’ by 3 becomes ‘a’
‘a’ appear 2 times in the string, hence incrementing ‘a’ by 3 becomes ‘d’
Hence the string becomes “dyzad”Input :s=”aabbc”, x=5
Output :ffggx
Approach:
- Create a frequency array to store the frequency of each character .
- Traverse the given string and for every character if its frequency is even, then increment it by x else if the frequency is odd decrement it by x.
Below is the implementation of the above approach:
Java
// Java implementation of the above approach: public class GFG { static final int MAX = 26; // Function to return the encrypted string static String encryptStr(String str, int n, int x) { // Reduce x because rotation of // length 26 is unnecessary x = x % MAX; char arr[] = str.toCharArray(); // calculate the frequency of characters int freq[] = new int[MAX]; for (int i = 0; i < n; i++) freq[arr[i] - 'a']++; for (int i = 0; i < n; i++) { // If the frequency of current character // is even then increment it by x if (freq[arr[i] - 'a'] % 2 == 0) { int pos = (arr[i] - 'a' + x) % MAX; arr[i] = (char)(pos + 'a'); } // Else decrement it by x else { int pos = (arr[i] - 'a' - x); if (pos < 0) pos += MAX; arr[i] = (char)(pos + 'a'); } } // Return the count return String.valueOf(arr); } // Driver code public static void main(String[] args) { String s = "abcda"; int n = s.length(); int x = 3; System.out.println(encryptStr(s, n, x)); } } |
Python
# Python3 implementation of the above approach: MAX = 26 # Function to return the encrypted strring def encryptstrr(strr, n, x): # Reduce x because rotation of # length 26 is unnecessary x = x % MAX arr = list(strr) # calculate the frequency of characters freq = [0]*MAX for i in range(n): freq[ord(arr[i]) - ord('a')] += 1 for i in range(n): # If the frequency of current character # is even then increment it by x if (freq[ord(arr[i]) - ord('a')] % 2 == 0): pos = (ord(arr[i]) - ord('a') + x) % MAX arr[i] = chr(pos + ord('a')) # Else decrement it by x else: pos = (ord(arr[i]) - ord('a') - x) if (pos < 0): pos += MAX arr[i] = chr(pos + ord('a')) # Return the count return "".join(arr) # Driver code s = "abcda"n = len(s) x = 3print(encryptstrr(s, n, x)) # This code is contributed by # shubhamsingh10 |
C#
// C# implementation of the above approach: using System; class GFG { static int MAX = 26; // Function to return the encrypted string public static char[] encryptStr(String str, int n, int x) { // Reduce x because rotation of // length 26 is unnecessary x = x % MAX; char[] arr = str.ToCharArray(); // calculate the frequency of characters int[] freq = new int[MAX]; for (int i = 0; i < n; i++) freq[arr[i] - 'a']++; for (int i = 0; i < n; i++) { // If the frequency of current character // is even then increment it by x if (freq[arr[i] - 'a'] % 2 == 0) { int pos = (arr[i] - 'a' + x) % MAX; arr[i] = (char)(pos + 'a'); } // Else decrement it by x else { int pos = (arr[i] - 'a' - x); if (pos < 0) pos += MAX; arr[i] = (char)(pos + 'a'); } } // Return the count return arr; } // Driver code public static void Main(String[] args) { String s = "abcda"; int n = s.Length; int x = 3; Console.WriteLine(encryptStr(s, n, x)); } } // This code is contributed by // sanjeev2552 |
dyzad
Time Complexity: O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Encrypt a string into the Rovarspraket (The Robber Language)
- Encrypt string with product of number of vowels and consonants in substring of size k
- Encrypt a string by repeating i-th character i times
- Program to Encrypt a String using ! and @
- Construct a binary string following the given constraints
- Restore original String from given Encrypted String by the given operations
- Find the final co-ordinates reached by following a sequence of directions
- Minimum number of given operations required to convert a string to another string
- Minimum given operations required to convert a given binary string to all 1's
- Minimum reduce operations to convert a given string into a palindrome
- Final string after performing given operations
- Maximum number of given operations to remove the entire string
- Minimum operations required to make the string satisfy the given condition
- Minimum number of operations required to obtain a given Binary String
- Minimum operations required to convert all characters of a String to a given Character
- Find value after N operations to remove N characters of string S with given constraints
- Minimum operations to transform given string to another by moving characters to front or end
- Kth character from the Nth string obtained by the given operations
- Given a string and an integer k, find the kth sub-string when all the sub-strings are sorted according to the given condition
- Count of operations to make a binary string"ab" free
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
