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 = "geeksforgeeks"
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++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to change every character
// of the string to another character
void 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 Code
int main()
{
string str = "nayan";
changeString(str);
return 0;
}
// Jsvs code for the above approach
import 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.
# Python code for the above approach
# Function to change every character
# of the string to another character
def 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 Code
str = "cycleofalphabet";
changeString(str);
# This code is contributed by Saurabh Jaiswal
// C# code for the above approach
using 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.
<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)