Invertir k caracteres alternativos en una string

Dada una string str y un entero k , la tarea es invertir k caracteres alternativos de la string dada. Si los caracteres presentes son menores que k, déjelos como están.
Ejemplos: 
 

Entrada: str = “geeksforgeeks”, k = 3 
Salida: eegksfgroeeks
Entrada: str = “abcde”, k = 2 
Salida: bacde 
 

Enfoque: la idea es primero invertir k caracteres, luego saltar a los siguientes k caracteres agregando 2 * k al índice y así sucesivamente hasta que se modifique la string completa.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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

Enfoque 2:  

La idea para resolver este problema es atravesar la string y, mientras se recorre, invertir los primeros K caracteres, luego omitir los siguientes K caracteres, luego nuevamente invertir los siguientes K caracteres y así sucesivamente hasta que se modifique la string completa.

Siga los pasos para resolver el problema:

• Atraviesa hasta el final de la string original
  • Recorre hacia atrás desde i+k hasta i y almacena los caracteres en la string resultante
  • Actualizar i a i+k
  • Recorra de i a i + k ahora, y almacene los caracteres en la string resultante
• Devolver la string original

A continuación se muestra la implementación del enfoque anterior:

// 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;
}

eegksfgroeeks

Complejidad temporal: O(N)
Espacio auxiliar: O(N)