Encrypt the given string with the following operations
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:
C++
// C++ implementation of the above approach:#include <bits/stdc++.h>#define MAX 26using namespace std;// Function to return the encrypted stringstring encryptStr(string str, int n, int x){ // Reduce x because rotation of // length 26 is unnecessary x = x % MAX; // Calculate the frequency of characters int freq[MAX] = {0}; for(int i = 0; i < n; i++) { freq[str[i] - 'a']++; } for(int i = 0; i < n; i++) { // If the frequency of current character // is even then increment it by x if (freq[str[i] - 'a'] % 2 == 0) { int pos = (str[i] - 'a' + x) % MAX; str[i] = (char)(pos + 'a'); } // Else decrement it by x else { int pos = (str[i] - 'a' - x); if (pos < 0) { pos += MAX; } str[i] = (char)(pos + 'a'); } } // Return the count return str;}// Driver codeint main(){ string s = "abcda"; int n = s.size(); int x = 3; cout << encryptStr(s, n, x) << endl; return 0;}// This code is contributed by avanitrachhadiya2155 |
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)); }} |
Python3
# Python3 implementation of the above approach:MAX = 26# Function to return the encrypted stringdef 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 codes = "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 |
Javascript
<script>// Javascript implementation of the above approach:var MAX = 26;// Function to return the encrypted stringfunction encryptStr(str, n, x){ // Reduce x because rotation of // length 26 is unnecessary x = x % MAX; // Calculate the frequency of characters var freq = Array(MAX).fill(0); for(var i = 0; i < n; i++) { freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++; } for(var i = 0; i < n; i++) { // If the frequency of current character // is even then increment it by x if (freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] % 2 == 0) { var pos = (str[i].charCodeAt(0) - 'a'.charCodeAt(0) + x) % MAX; str[i] = String.fromCharCode(pos + 'a'.charCodeAt(0)); } // Else decrement it by x else { var pos = (str[i].charCodeAt(0) - 'a'.charCodeAt(0) - x); if (pos < 0) { pos += MAX; } str[i] = String.fromCharCode(pos + 'a'.charCodeAt(0)); } } // Return the count return str.join('');}// Driver codevar s = "abcda";var n = s.length;var x = 3;document.write( encryptStr(s.split(''), n, x));</script> |
Output:
dyzad
Time Complexity: O(N)
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