Given a string str and an integer k, the task is to reverse alternate k characters of the given string. If characters present are less than k, leave them as it is.
Examples:
Input: str = “geeksforgeeks”, k = 3
Output: eegksfgroeeks
Input: str = “abcde”, k = 2
Output: bacde
Approach: The idea is to first reverse k characters, then jump onto the next k characters by adding 2 * k to the index and so on until the complete string is modified.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the string after // reversing the alternate k characters string revAlternateK(string s, int k, int len)
{ for ( int i = 0; i < s.size();) {
// If there are less than k characters
// starting from the current position
if (i + k > len)
break ;
// Reverse first k characters
reverse(s.begin() + i, s.begin() + i + k);
// Skip the next k characters
i += 2 * k;
}
return s;
} // Driver code int main()
{ string s = "geeksforgeeks" ;
int len = s.length();
int k = 3;
cout << revAlternateK(s, k, len);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to return the string after // reversing the alternate k characters static String revAlternateK(String s,
int k, int len)
{ for ( int i = 0 ; i < s.length();)
{
// If there are less than k characters
// starting from the current position
if (i + k > len)
break ;
// Reverse first k characters
s = s.substring( 0 , i) + new String( new StringBuilder(
s.substring(i, i + k)).reverse()) +
s.substring(i + k);
// Skip the next k characters
i += 2 * k;
}
return s;
} // Driver code public static void main(String[] args)
{ String s = "geeksforgeeks" ;
int len = s.length();
int k = 3 ;
System.out.println(revAlternateK(s, k, len));
} } // This code is contributed by 29AjayKumar |
# Python3 implementation of the approach # Function to return the string after # reversing the alternate k characters def revAlternateK(s, k, Len ):
i = 0
while (i < len (s)):
# If there are less than k characters
# starting from the current position
if (i + k > Len ):
break
# Reverse first k characters
ss = s[i:i + k]
s = s[:i] + ss[:: - 1 ] + s[i + k:]
# Skip the next k characters
i + = 2 * k
return s;
# Driver code s = "geeksforgeeks"
Len = len (s)
k = 3
print (revAlternateK(s, k, Len ))
# This code is contributed by mohit kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the string after // reversing the alternate k characters static String revAlternateK(String s,
int k, int len)
{ for ( int i = 0; i < s.Length;)
{
// If there are less than k characters
// starting from the current position
if (i + k > len)
break ;
// Reverse first k characters
s = s.Substring(0, i) +
reverse(s.Substring(i, k).ToCharArray(), 0, k - 1) +
s.Substring(i + k);
// Skip the next k characters
i += 2 * k;
}
return s;
} static String reverse( char []str, int start, int end)
{ // Temporary variable to store character
char temp;
while (start <= end)
{
// Swapping the first and last character
temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
return String.Join( "" , str);
} // Driver code public static void Main(String[] args)
{ String s = "geeksforgeeks" ;
int len = s.Length;
int k = 3;
Console.WriteLine(revAlternateK(s, k, len));
} } // This code contributed by Rajput-Ji |
<script> // Javascript implementation of the approach // Function to return the string after // reversing the alternate k characters function revAlternateK(s,k,len)
{ for (let i = 0; i < s.length;)
{
// If there are less than k characters
// starting from the current position
if (i + k > len)
break ;
// Reverse first k characters
s = s.substring(0, i) + s.substring(i, i + k).split( "" ).reverse().join( "" ) +
s.substring(i + k);
// Skip the next k characters
i += 2 * k;
}
return s;
} // Driver code let s = "geeksforgeeks" ;
let len = s.length; let k = 3; document.write(revAlternateK(s, k, len)); // This code is contributed by patel2127 </script> |
eegksfgroeeks
Time Complexity: O(N) where N is length of given string “s”
Auxiliary Space: O(1)
Approach 2:
The idea to solve this problem is to traverse the string, and while traversing reverse first K characters, then skip next K characters, then again reverse next K characters and so on until the complete string is modified.
Follow the steps to solve the problem:
- Traverse till the end of original string
- Traverse backwards from i+k to i and store the characters in resultant string
- Update i to i+k
- Traverse from i to i + k now, and store the characters in resultant string
- Return the original string
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the string after // reversing the alternate K characters string revAlternateK(string s, int k, int n)
{ string ans = "" ;
int i = 0, j = 0;
// Traverse till the end
// of original string
while (i < n) {
// Traverse backwards from i+k to i
// and store the characters
// in resultant string
for (j = min(i + k, n) - 1; j >= i; j--)
ans += s[j];
i = min(i + k, n);
// Traverse from i to i+k and store
// the characters in resultant string
for (j = i; j < min(i + k, n); j++)
ans += s[j];
i = j;
}
// Return ans
return ans;
} // Driver code int main()
{ string str = "geeksforgeeks" ;
int N = str.length();
int K = 3;
cout << revAlternateK(str, K, N);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG {
// Function to return the string after
// reversing the alternate K characters
static String revAlternate(String s, int k, int n)
{
String ans = "" ;
int i = 0 , j = 0 ;
// Traverse till the end
// of original string
while (i < n) {
// Traverse backwards from i+k to i
// and store the characters
// in resultant string
for (j = Math.min(i + k, n) - 1 ; j >= i; j--) {
ans += s.charAt(j);
}
i = Math.min(i + k, n);
// Traverse from i to i+k and store
// the characters in resultant string
for (j = i; j < Math.min(i + k, n); j++) {
ans += s.charAt(j);
}
i = j;
}
// Return ans
return ans;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
int N = str.length();
int K = 3 ;
System.out.print(revAlternate(str, K, N));
}
} // This code is contributed by lokeshmvs21. |
# Python3 implementation of the approach # Function to return the string after # reversing the alternate K characters def revAlternateK(s, k, n):
ans = ""
i = 0
j = 0
# Traverse till the end
# of original string
while (i < n):
# Traverse backwards from i+k to i
# and store the characters
# in resultant string
for j in range ( min (i + k,n) - 1 ,i - 1 , - 1 ):
ans + = s[j]
i = min (i + k, n)
# Traverse from i to i+k and store
# the characters in resultant string
for j in range (i, min (i + k,n)):
ans + = s[j]
i = j + 1
# Return ans
return ans
# Driver code str = "geeksforgeeks"
N = len ( str )
K = 3
print (revAlternateK( str , K, N))
# This code is contributed by akashish__ |
// C# implementation of the approach using System;
public class GFG {
// Function to return the string after
// reversing the alternate K characters
static string revAlternate( string s, int k, int n)
{
string ans = "" ;
int i = 0, j = 0;
// Traverse till the end
// of original string
while (i < n) {
// Traverse backwards from i+k to i
// and store the characters
// in resultant string
for (j = Math.Min(i + k, n) - 1; j >= i; j--) {
ans += s[j];
}
i = Math.Min(i + k, n);
// Traverse from i to i+k and store
// the characters in resultant string
for (j = i; j < Math.Min(i + k, n); j++) {
ans += s[j];
}
i = j;
}
// Return ans
return ans;
}
static public void Main()
{
// Code
string str = "geeksforgeeks" ;
int N = str.Length;
int K = 3;
Console.Write(revAlternate(str, K, N));
}
} // This code is contributed by lokesh. |
// JavaScript implementation of the approach // Function to return the string after // reversing the alternate K characters function revAlternate(s, k, n){
let ans = "" ;
let i = 0, j = 0;
// Traverse till the end
// of original string
while (i < n) {
// Traverse backwards from i+k to i
// and store the characters
// in resultant string
for (j = Math.min(i + k, n) - 1; j >= i; j--) {
ans += s[j];
}
i = Math.min(i + k, n);
// Traverse from i to i+k and store
// the characters in resultant string
for (j = i; j < Math.min(i + k, n); j++) {
ans += s[j];
}
i = j;
}
// Return ans
return ans;
} let str = "geeksforgeeks" ;
let N = str.length; let K = 3; console.log(revAlternate(str, K, N)); // This code is contributed by lokeshmvs21. |
eegksfgroeeks
Time Complexity: O(N)
Auxiliary Space: O(N)