Given a string str and a positive integer N, the task is to reverse N characters and skip N characters until the end of the string to generate the encrypted message.
Examples:
Input: str = “ihTs suohld ebeas!y”, K = 3
Output: This should be easy!
Explanation:
Reverse “ihT” -> “Thi”
“s” remains the same
“uoh” -> “hou”, and so on.
Input: str = “!ysae eb dluohs sihT”, K = 30
Output: This should be easy!
Explanation:
Since 30 is larger than the length of the given string(= 20), just print the reverse of the string.
Approach: Follow the steps below to solve the problem:
- Traverse the given string.
- Increment the iterator by 2 * N.
- Reverse N characters step by step.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int decryptString(string s, unsigned int N)
{
for (unsigned int i = 0; i < s.size();
i += 2 * N) {
auto end = s.begin() + i + N;
if (i + N > s.size())
end = s.end();
reverse(s.begin() + i, end);
}
cout << s << endl;
}
int main()
{
string s = "ihTs suohld ebeas!y" ;
unsigned int N = 3;
decryptString(s, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
static String reverse_(String s , int start , int end)
{
String str= "" ;
for ( int i = 0 ;i<start;i++)
str += s.charAt(i);
for ( int i = end- 1 ;i>=start;i--)
str += s.charAt(i);
for ( int i = end; i<s.length();i++)
str += s.charAt(i);
return str;
}
static void decryptString(String s, int N)
{
for ( int i = 0 ; i < s.length(); i += 2 *N) {
int end = i + N;
if (i + N > s.length())
end = s.length();
s = reverse_(s, i, end);
}
System.out.println(s);
}
public static void main(String[] args)
{
String s = "ihTs suohld ebeas!y" ;
int N = 3 ;
decryptString(s, N);
}
}
|
C#
using System;
class GFG
{
static string Reverse_( string s, int start, int end)
{
string str = "" ;
for ( int i = 0; i < start; i++)
str += s[i];
for ( int i = end - 1; i >= start; i--)
str += s[i];
for ( int i = end; i < s.Length; i++)
str += s[i];
return str;
}
static void DecryptString( string s, int N)
{
for ( int i = 0; i < s.Length; i += 2 * N)
{
int end = i + N;
if (i + N > s.Length)
end = s.Length;
s = Reverse_(s, i, end);
}
Console.WriteLine(s);
}
public static void Main( string [] args)
{
string s = "ihTs suohld ebeas!y" ;
int N = 3;
DecryptString(s, N);
}
}
|
Python3
def decryptString(s, N):
for i in range ( 0 , len (s), 2 * N):
if (i + N < len (s)):
end = s[i + N]
if (i + N > len (s)):
end = s[ - 1 ]
if (i = = 0 ):
s = s[i + N - 1 :: - 1 ] + s[i + N:]
else :
s = s[:i] + s[i + N - 1 :i - 1 : - 1 ] + s[i + N:]
print (s)
if __name__ = = "__main__" :
s = "ihTs suohld ebeas!y"
N = 3
decryptString(s, N)
|
Javascript
function reverse_(s , start , end)
{
end -= 1;
while (start<end)
{
var temp = s[start];
s[start] = s[end];
s[end] = temp;
start++;
end--;
}
return s;
}
function decryptString( s, N)
{
for ( var i = 0; i < s.length;i += 2 * N) {
var end = i + N;
if (i + N > s.length)
end = s.length;
s = reverse_(s, i, end);
}
console.log(s);
}
var s = "ihTs suohld ebeas!y" ;
var N = 3;
decryptString(s, N);
|
Output:
This should be easy!
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!
Last Updated :
23 Jan, 2023
Like Article
Save Article