Reverse alternate k characters in a string

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

eegksfgroeeks


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