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++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output

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++

// 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

// 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

# 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#

// 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

// 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.

Output

eegksfgroeeks

Time Complexity: O(N)
Auxiliary Space: O(N)