Given a string and a number N, we need to mirror the characters from N-th position up to the length of the string in the alphabetical order. In mirror operation, we change ‘a’ to ‘z’, ‘b’ to ‘y’, and so on.
Examples:
Input : N = 3
paradox
Output : paizwlc
We mirror characters from position 3 to end.
Input : N = 6
pneumonia
Output : pnefnlmrz
Below are different characters and their mirrors.
Mirroring the alphabetical order means that a corresponds to z, b corresponds to y. Which means that first character becomes the last and so on. Now, to achieve this we maintain a string(or a character array) which contains the English alphabets in lower case. Now from the pivot point up to the length, we can look up the reverse alphabetical order of a character by using its ASCII value as an index. Using the above technique, we transform the given string in the required one.
Implementation:
C++
#include <iostream>
#include <string>
using namespace std;
string compute(string str, int n)
{
string reverseAlphabet = "zyxwvutsrqponmlkjihgfedcba";
int l = str.length();
for (int i = n; i < l; i++)
str[i] = reverseAlphabet[str[i] - 'a'];
return str;
}
int main()
{
string str = "pneumonia";
int n = 4;
string answer = compute(str, n - 1);
cout << answer;
return 0;
}
|
Java
import java.io.*;
class GeeksforGeeks {
static String compute(String str, int n)
{
String reverseAlphabet = "zyxwvutsrqponmlkjihgfedcba";
int l = str.length();
String answer = "";
for (int i = 0; i < n; i++)
answer = answer + str.charAt(i);
for (int i = n; i < l; i++)
answer = answer + reverseAlphabet.charAt(str.charAt(i) - 'a');
return answer;
}
public static void main(String args[])
{
String str = "pneumonia";
int n = 4;
System.out.print(compute(str, n - 1));
}
}
|
Python3
def compute(st, n):
reverseAlphabet = "zyxwvutsrqponmlkjihgfedcba"
l = len(st)
answer = ""
for i in range(0, n):
answer = answer + st[i];
for i in range(n, l):
answer = (answer +
reverseAlphabet[ord(st[i]) - ord('a')]);
return answer;
st = "pneumonia"
n = 4
answer = compute(st, n - 1)
print(answer)
|
C#
using System;
class GFG {
static String compute(string str, int n)
{
string reverseAlphabet =
"zyxwvutsrqponmlkjihgfedcba";
int l = str.Length;
String answer = "";
for (int i = 0; i < n; i++)
answer = answer + str[i];
for (int i = n; i < l; i++)
answer = answer +
reverseAlphabet[str[i]- 'a'];
return answer;
}
public static void Main()
{
string str = "pneumonia";
int n = 4;
Console.Write(compute(str, n - 1));
}
}
|
php
<?php
function compute($str, $n)
{
$reverseAlphabet =
"zyxwvutsrqponmlkjihgfedcba";
$l = strlen($str);
$answer = "";
for ($i = 0; $i < $n; $i++)
$answer = $answer.$str[$i];
for ($i = $n; $i < $l; $i++){
$answer = $answer.$reverseAlphabet[
ord($str[$i])- ord('a')];
}
return $answer;
}
$str = "pneumonia";
$n = 4;
$answer = compute($str, $n - 1);
echo $answer;
?>
|
Javascript
<script>
function compute(str, n)
{
let reverseAlphabet =
"zyxwvutsrqponmlkjihgfedcba";
let l = str.length;
let answer = "";
for (let i = 0; i < n; i++)
answer = answer + str[i];
for (let i = n; i < l; i++)
answer = answer + reverseAlphabet[str[i].charCodeAt()- 'a'.charCodeAt()];
return answer;
}
let str = "pneumonia";
let n = 4;
document.write(compute(str, n - 1));
</script>
|
Output:
pnefnlmrz
Time Complexity: O(n), where n represents the size of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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