Reverse alternate k characters in a string
Last Updated :
16 Dec, 2022
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++
#include <bits/stdc++.h>
using namespace std;
string revAlternateK(string s, int k, int len)
{
for ( int i = 0; i < s.size();) {
if (i + k > len)
break ;
reverse(s.begin() + i, s.begin() + i + k);
i += 2 * k;
}
return s;
}
int main()
{
string s = "geeksforgeeks" ;
int len = s.length();
int k = 3;
cout << revAlternateK(s, k, len);
return 0;
}
|
Java
class GFG
{
static String revAlternateK(String s,
int k, int len)
{
for ( int i = 0 ; i < s.length();)
{
if (i + k > len)
break ;
s = s.substring( 0 , i) + new String( new StringBuilder(
s.substring(i, i + k)).reverse()) +
s.substring(i + k);
i += 2 * k;
}
return s;
}
public static void main(String[] args)
{
String s = "geeksforgeeks" ;
int len = s.length();
int k = 3 ;
System.out.println(revAlternateK(s, k, len));
}
}
|
Python3
def revAlternateK(s, k, Len ):
i = 0
while (i < len (s)):
if (i + k > Len ):
break
ss = s[i:i + k]
s = s[:i] + ss[:: - 1 ] + s[i + k:]
i + = 2 * k
return s;
s = "geeksforgeeks"
Len = len (s)
k = 3
print (revAlternateK(s, k, Len ))
|
C#
using System;
class GFG
{
static String revAlternateK(String s,
int k, int len)
{
for ( int i = 0; i < s.Length;)
{
if (i + k > len)
break ;
s = s.Substring(0, i) +
reverse(s.Substring(i, k).ToCharArray(), 0, k - 1) +
s.Substring(i + k);
i += 2 * k;
}
return s;
}
static String reverse( char []str, int start, int end)
{
char temp;
while (start <= end)
{
temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
return String.Join( "" , str);
}
public static void Main(String[] args)
{
String s = "geeksforgeeks" ;
int len = s.Length;
int k = 3;
Console.WriteLine(revAlternateK(s, k, len));
}
}
|
Javascript
<script>
function revAlternateK(s,k,len)
{
for (let i = 0; i < s.length;)
{
if (i + k > len)
break ;
s = s.substring(0, i) + s.substring(i, i + k).split( "" ).reverse().join( "" ) +
s.substring(i + k);
i += 2 * k;
}
return s;
}
let s = "geeksforgeeks" ;
let len = s.length;
let k = 3;
document.write(revAlternateK(s, k, len));
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
string revAlternateK(string s, int k, int n)
{
string ans = "" ;
int i = 0, j = 0;
while (i < n) {
for (j = min(i + k, n) - 1; j >= i; j--)
ans += s[j];
i = min(i + k, n);
for (j = i; j < min(i + k, n); j++)
ans += s[j];
i = j;
}
return ans;
}
int main()
{
string str = "geeksforgeeks" ;
int N = str.length();
int K = 3;
cout << revAlternateK(str, K, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
static String revAlternate(String s, int k, int n)
{
String ans = "" ;
int i = 0 , j = 0 ;
while (i < n) {
for (j = Math.min(i + k, n) - 1 ; j >= i; j--) {
ans += s.charAt(j);
}
i = Math.min(i + k, n);
for (j = i; j < Math.min(i + k, n); j++) {
ans += s.charAt(j);
}
i = j;
}
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));
}
}
|
Python3
def revAlternateK(s, k, n):
ans = ""
i = 0
j = 0
while (i < n):
for j in range ( min (i + k,n) - 1 ,i - 1 , - 1 ):
ans + = s[j]
i = min (i + k, n)
for j in range (i, min (i + k,n)):
ans + = s[j]
i = j + 1
return ans
str = "geeksforgeeks"
N = len ( str )
K = 3
print (revAlternateK( str , K, N))
|
C#
using System;
public class GFG {
static string revAlternate( string s, int k, int n)
{
string ans = "" ;
int i = 0, j = 0;
while (i < n) {
for (j = Math.Min(i + k, n) - 1; j >= i; j--) {
ans += s[j];
}
i = Math.Min(i + k, n);
for (j = i; j < Math.Min(i + k, n); j++) {
ans += s[j];
}
i = j;
}
return ans;
}
static public void Main()
{
string str = "geeksforgeeks" ;
int N = str.Length;
int K = 3;
Console.Write(revAlternate(str, K, N));
}
}
|
Javascript
function revAlternate(s, k, n){
let ans = "" ;
let i = 0, j = 0;
while (i < n) {
for (j = Math.min(i + k, n) - 1; j >= i; j--) {
ans += s[j];
}
i = Math.min(i + k, n);
for (j = i; j < Math.min(i + k, n); j++) {
ans += s[j];
}
i = j;
}
return ans;
}
let str = "geeksforgeeks" ;
let N = str.length;
let K = 3;
console.log(revAlternate(str, K, N));
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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