Invertir una array hasta una posición dada

Dada una array arr[] y una posición en la array, k. Escriba un nombre de función inversa (a[], k) tal que invierta el subarreglo arr[0..k-1]. El espacio adicional utilizado debe ser O(1) y la complejidad de tiempo debe ser O(k). 
Ejemplo: 

Input:
arr[] = {1, 2, 3, 4, 5, 6}
    k = 4

Output:
arr[] = {4, 3, 2, 1, 5, 6} 

Le recomendamos encarecidamente que minimice su navegador y que pruebe esto usted mismo primero.

A continuación se muestra la implementación de la misma. 

// C++ program to reverse a subarray arr[0..k-1]
#include <bits/stdc++.h>
using namespace std;
 
// Reverse subarray a[0..k-1]
void reverse(int a[], int n, int k)
{
    if (k > n)
    {
        cout << "Invalid k";
        return;
    }
 
    // One by one reverse first and last elements of a[0..k-1]
    for (int i = 0; i < k/2; i++)
        swap(a[i], a[k-i-1]);
}
 
// Driver program
int main()
{
    int a[] = {1, 2, 3, 4, 5, 6};
    int n = sizeof(a) / sizeof(int), k = 4;
 
    reverse(a, n, k);
 
    for (int i = 0; i < n; ++i)
        printf("%d ", a[i]);
 
    return 0;
}

// java program to reverse a
// subarray arr[0..k-1]
 
public class GFG {
 
    // Reverse subarray a[0..k-1]
    static void reverse(int []a, int n, int k)
    {
        if (k > n)
        {
            System.out.println( "Invalid k");
            return;
        }
     
        // One by one reverse first
        // and last elements of a[0..k-1]
        for (int i = 0; i < k / 2; i++)
        {
            int tempswap = a[i];
                a[i] = a[k - i - 1];
                a[k - i - 1] = tempswap;            
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        int []a = {1, 2, 3, 4, 5, 6};
        int n = a.length, k = 4;
        reverse(a, n, k);
        for (int i = 0; i < n; ++i)
            System.out.print(a[i] + " ");
    }
}
 
// This code is contributed by Sam007.

# python program to reverse a subarray
# arr[0..k-1]
from __future__ import print_function
 
# Reverse subarray a[0..k-1]
def reverse(a, n, k):
     
    if (k > n):
        print( "Invalid k")
        return
     
    # One by one reverse first and
    # last elements of a[0..k-1]
    for i in range(0, (int)(k/2)):
        temp = a[i]
        a[i] = a[k-i-1]
        a[k-i-1] = temp
         
# Driver program
a = [1, 2, 3, 4, 5, 6]
n = len(a)
k = 4
 
reverse(a, n, k);
 
for i in range(0, n):
    print(a[i], end=" ")
     
# This code is contributed by Sam007.

// C# program to reverse a
// subarray arr[0..k-1]
using System;
 
class GFG {
     
static void SwapNum(ref int x, ref int y)
{
    int tempswap = x;
    x = y;
    y = tempswap;            
}
     
// Reverse subarray a[0..k-1]
static void reverse(int []a, int n,
                             int k)
{
    if (k > n)
    {
        Console.Write( "Invalid k");
        return;
    }
 
    // One by one reverse first
    // and last elements of a[0..k-1]
    for (int i = 0; i < k / 2; i++)
        SwapNum(ref a[i], ref a[k - i - 1]);
         
}
 
// Driver Code
public static void Main()
{
    int []a = {1, 2, 3, 4, 5, 6};
    int n = a.Length, k = 4;
 
    reverse(a, n, k);
 
    for (int i = 0; i < n; ++i)
        Console.Write(a[i] + " ");
}
}
 
// This code is contributed by Sam007

<script>
 
// Javascript program to reverse
// a subarray arr[0..k-1]
 
// Reverse subarray a[0..k-1]
function reverse( a, n, k)
{
    if (k > n)
    {
        document.write("Invalid k");
        return;
    }
 
    // One by one reverse first
    // and last elements of a[0..k-1]
    for (let i = 0; i < Math.floor(k/2); i++)
    {
      let temp = a[i] ;
      a[i] = a[k-i-1] ;
      a[k-i-1] = temp ;
    }
     
}
    // driver code
     
    let a = [1, 2, 3, 4, 5, 6];
    let n = a.length, k = 4;
 
    reverse(a, n, k);
 
    for (let i = 0; i < n; ++i)
        document.write(a[i] + " ");
 
</script>

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Complejidad del tiempo: O(k)

Método 2 (usando STL):

En este método, utilizaremos una función STL de C++ incorporada denominada inversa. Esta función completa la tarea de invertir K elementos de la array en tiempo O(K) y tampoco utiliza espacio adicional.

La implementación de este método se encuentra a continuación.

// C++ program to reverse the first K
// elements using in-built function
#include <bits/stdc++.h>
using namespace std;
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
    int k = 4;
 
    // STL function to reverse element
    // from 0 index to K-1 index.
    reverse(arr, arr + k);
    // printing the array after reversing
    // first K elements.
    for (int i = 0; i < 8; i++) {
        cout << arr[i] << " ";
    }
    return 0;
}
 
// this code is contributed by Machhaliya Muhammad

# Python3 program to reverse the first K
# elements using in-built function
arr = [ 1, 2, 3, 4, 5, 6, 7, 8 ]
k = 4
 
# Using list slicing to reverse the array
# from 0 index to K-1 index.
arr[:k] = arr[:k][::-1]
 
# printing the array after reversing
# first K elements.
print(*arr)
 
# This code is contributed by phasing17

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