Given string str of size N consisting of lowercase English alphabets, the task is to encode the given string as follows:
- change every character of that string to another character
- the distance between the changed character and the current character is the same as the distance between the current character and ‘a’.
- Also, assume that the character’s array forms a cycle, i.e. after ‘z’ the cycle starts again from ‘a’.
Examples:
Input: str = “neveropen”
Output: “miiukkcimiiuk”
Explanation:
g changed to m (distance between g & a is 6, distance between m & g is 6)
e changed to i (distance between e & a is 4, distance between i & e is 4)
and same for other characters as well.Input: str = “cycleofalphabet”
Output: “ewewickaweoacim”
Approach: This problem can be solved using the following steps:
- Run a loop from i=0 to i<N and traverse each character of the string. For each character str[i]:
- Find the distance between str[i] and ‘a’, i.e. dist=str[i]-‘a’.
- Now, if (dist+(str[i]-‘a’)) > 26, this means that ‘z’ is exceeded, so
- Otherwise, change str[i] to str[i]+dist.
- Print the string as the answer to this problem.
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to change every character// of the string to another charactervoid changeString(string str){ for (auto& x : str) { int dist = x - 'a'; // If 'z' is exceeded if (dist + (x - 'a') >= 26) { dist = (dist + (x - 'a')) % 26; x = 'a' + dist; } // If 'z' is not exceeded else { x = x + dist; } } cout << str << endl;}// Driver Codeint main(){ string str = "nayan"; changeString(str); return 0;} |
Java
// Jsvs code for the above approachimport java.util.*;class GFG { // Function to change every character // of the string to another character static void changeString(String str) { char[] ch = str.toCharArray(); for (int i = 0; i < str.length(); i++) { int dist = ch[i] - 'a'; // If 'z' is exceeded if (dist + (ch[i] - 'a') >= 26) { dist = (dist + (ch[i] - 'a')) % 26; ch[i] = (char)('a' + dist); } // If 'z' is not exceeded else { ch[i] = (char)(ch[i] + dist); } } String s = new String(ch); System.out.println(s); } // Driver Code public static void main(String[] args) { String str = "cycleofalphabet"; changeString(str); }}// This code is contributed by ukasp. |
Python3
# Python code for the above approach # Function to change every character# of the string to another characterdef changeString(str): str = list(str) for x in range(len(str)): dist = ord(str[x]) - ord('a') # If 'z' is exceeded if (dist + (ord(str[x]) - ord('a')) >= 26): dist = (dist + (ord(str[x]) - ord('a'))) % 26; str[x] = chr(ord('a') + dist); # If 'z' is not exceeded else: str[x] = chr(ord(str[x]) + dist); str = "".join(str) print(str)# Driver Codestr = "cycleofalphabet";changeString(str);# This code is contributed by Saurabh Jaiswal |
C#
// C# code for the above approachusing System;using System.Collections;class GFG{ // Function to change every character // of the string to another character static void changeString(string str) { char[] ch = str.ToCharArray(); for(int i = 0; i < str.Length; i++) { int dist = ch[i] - 'a'; // If 'z' is exceeded if (dist + (ch[i] - 'a') >= 26) { dist = (dist + (ch[i] - 'a')) % 26; ch[i] = (char)('a' + dist); } // If 'z' is not exceeded else { ch[i] = (char)(ch[i] + dist); } } string s = new string(ch); Console.WriteLine(s); } // Driver Code public static void Main() { string str = "cycleofalphabet"; changeString(str); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to change every character // of the string to another character function changeString(str) { str = str.split('') for (let x = 0; x < str.length; x++) { let dist = str[x].charCodeAt(0) - 'a'.charCodeAt(0); // If 'z' is exceeded if (dist + (str[x].charCodeAt(0) - 'a'.charCodeAt(0)) >= 26) { dist = (dist + (str[x].charCodeAt(0) - 'a'.charCodeAt(0))) % 26; str[x] = String.fromCharCode('a'.charCodeAt(0) + dist); } // If 'z' is not exceeded else { str[x] = String.fromCharCode(str[x].charCodeAt(0) + dist); } } str = str.join('') document.write(str + "<br>") } // Driver Code let str = "cycleofalphabet"; changeString(str); // This code is contributed by Potta Lokesh </script> |
Output
ewewickaweoacim
Time Complexity: O(N)
Auxiliary Space: O(1)
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