Dada una array arr[] que consta de 2* N elementos en forma de { a 1 , a 2 , …, a N , b 1 , b 2 , …, b N } , la tarea es barajar la array a {a 1 , b 1 , a 2 , b 2 , …, a n , b 1 } sin usar espacio adicional.
Ejemplos :
Entrada: arr[] = { 1, 3, 5, 2, 4, 6 }
Salida: 1 2 3 4 5 6
Explicación:
La salida contiene los elementos en forma de { a 1 , b 1 , a 2 , b 2 , a 3 , b 3 }.Entrada: arr[] = {1, 2, 3, -1, -2, -3, }
Salida: 1 -1 2 -2 3 -3
Enfoque basado en divide y vencerás: si el tamaño de una array es una potencia de 2 , entonces sigue el artículo Baraja 2n enteros en formato {a1, b1, a2, b2, a3, b3, ……, an, bn} usando divide y conquistar la técnica.
Complejidad de tiempo: O(N * log(N))
Espacio auxiliar: O(1)
Enfoque alternativo: el enfoque anterior funcionará para todos los valores posibles de N dividiendo recursivamente la array de manera que la longitud de ambas mitades sea uniforme. Siga los pasos a continuación para resolver el problema:
- Defina una función recursiva, digamos shuffle(start, end) .
- Si la longitud de la array es divisible por 4 , calcule el punto medio de la array, digamos mid = start + (end – start + 1) / 2.
- De lo contrario, mid = start + (end – start + 1) / 2 – 1.
- Calcule los puntos medios de ambos subarreglos, digamos mid1 = start + (mid – start)/2 , y mid2 = mid + (end – mid + 1) / 2.
- Invierta los subarreglos en los rangos [mid1, mid2], [mid1, mid-1] y [mid, mid2 – 1].
- Realice llamadas recursivas para los subarreglos [start, mid – 1] y [mid, end] , es decir, shuffle(start, mid – 1) y shuffle(mid, end) respectivamente.
- Finalmente, imprima la array .
Ilustración:
Considere una array arr[] = {a1, a2, a3, b1, b2, b3}:
- Divida la array en dos mitades, ambas de igual longitud, es decir, a1, a2 : a3, b1, b2, b3.
- Invierta la mitad de la primera mitad a la mitad de la segunda mitad, es decir, a1, b1 : a3, a2, b2, b3.
- Ahora, invierta la mitad de la primera mitad a la mitad del subarreglo [0, 5], a1, b1 : a3, a2, b2, b3.
- Ahora invierta la mitad del subarreglo [0, 5] a la mitad de la segunda mitad, a1, b1 : a2, a3, b2, b3.
- Llame recursivamente a las arrays {a1, b1} y {a2, a3, b2, b3} .
- Ahora la array {a2, a3, b2, b3} se modifica a {a2, b2, a3, b3} después de aplicar las operaciones anteriores.
- Ahora el arr[] se modifica a {a1, b1, a2, b2, a3, b3} .
A continuación se muestra la implementación del enfoque anterior:
C
// C program for the above approach
#include <stdio.h>
// Function to reverse the array from the
// position 'start' to position 'end'
void reverse(int arr[], int start, int end)
{
// Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for (int i = 0; i < mid; i++) {
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return;
}
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
void shuffleArrayUtil(int arr[], int start, int end)
{
int i;
// Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4) {
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
}
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
void shuffleArray(int arr[], int N,
int start, int end)
{
// Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for (int i = 0; i < N; i++)
printf("%d ", arr[i]);
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
return 0;
}
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to reverse the array from the
// position 'start' to position 'end'
void reverse(int arr[], int start, int end)
{
// Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for (int i = 0; i < mid; i++) {
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return;
}
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
void shuffleArrayUtil(int arr[], int start, int end)
{
int i;
// Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4) {
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
}
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
void shuffleArray(int arr[], int N, int start, int end)
{
// Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for (int i = 0; i < N; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
return 0;
}
// This code is contributed by jainlovely450.
Java
// Java program for the above approach
class GFG{
// Function to reverse the array from the
// position 'start' to position 'end'
static void reverse(int arr[], int start, int end)
{
// Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for (int i = 0; i < mid; i++)
{
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return;
}
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
static void shuffleArrayUtil(int arr[], int start, int end)
{
int i;
// Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4 > 0)
{
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
}
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
static void shuffleArray(int arr[], int N,
int start, int end)
{
// Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for (int i = 0; i < N; i++)
System.out.printf("%d ", arr[i]);
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = arr.length;
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for the above approach
# Function to reverse the array from the
# position 'start' to position 'end'
def reverse(arr, start, end):
# Stores mid of start and end
mid = (end - start + 1) // 2
# Traverse the array in
# the range [start, end]
for i in range(mid):
# Stores arr[start + i]
temp = arr[start + i]
# Update arr[start + i]
arr[start + i] = arr[end - i]
# Update arr[end - i]
arr[end - i] = temp
return arr
# Utility function to shuffle the given array
# in the of form {a1, b1, a2, b2, ....an, bn}
def shuffleArrayUtil(arr, start, end):
i = 0
# Stores the length of the array
l = end - start + 1
# If length of the array is 2
if (l == 2):
return
# Stores mid of the { start, end }
mid = start + l // 2
# Divide array into two
# halves of even length
if (l % 4):
# Update mid
mid -= 1
# Calculate the mid-points of
# both halves of the array
mid1 = start + (mid - start) // 2
mid2 = mid + (end + 1 - mid) // 2
# Reverse the subarray made
# from mid1 to mid2
arr = reverse(arr, mid1, mid2 - 1)
# Reverse the subarray made
# from mid1 to mid
arr = reverse(arr, mid1, mid - 1)
# Reverse the subarray made
# from mid to mid2
arr = reverse(arr, mid, mid2 - 1)
# Recursively calls for both
# the halves of the array
shuffleArrayUtil(arr, start, mid - 1)
shuffleArrayUtil(arr, mid, end)
# Function to shuffle the given array in
# the form of {a1, b1, a2, b2, ....an, bn}
def shuffleArray(arr, N, start, end):
# Function Call
shuffleArrayUtil(arr, start, end)
# Print the modified array
for i in arr:
print(i, end=" ")
# Driver Code
if __name__ == '__main__':
# Given array
arr = [1, 3, 5, 2, 4, 6]
# Size of the array
N = len(arr)
# Shuffles the given array to the
# required permutation
shuffleArray(arr, N, 0, N - 1)
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
public class GFG{
// Function to reverse the array from the
// position 'start' to position 'end'
static void reverse(int[] arr, int start, int end)
{
// Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for (int i = 0; i < mid; i++)
{
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return;
}
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
static void shuffleArrayUtil(int[] arr, int start, int end)
{
// Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4 > 0)
{
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
}
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
static void shuffleArray(int[] arr, int N,
int start, int end)
{
// Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
}
// Driver Code
static public void Main ()
{
// Given array
int[] arr = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = arr.Length;
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
}
}
// This code is contributed by Dharanendra L V
Javascript
<script>
// Javascript program of the above approach
// Function to reverse the array from the
// position 'start' to position 'end'
function reverse(arr, start, end)
{
// Stores mid of start and end
let mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for (let i = 0; i < mid; i++)
{
// Stores arr[start + i]
let temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return;
}
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
function shuffleArrayUtil(arr, start, end)
{
let i;
// Stores the length of the array
let l = end - start + 1;
// If length of the array is 2
if (l == 2)
return;
// Stores mid of the { start, end }
let mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4 > 0)
{
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
let mid1 = start + (mid - start) / 2;
let mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
}
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
function shuffleArray(arr, N,
start, end)
{
// Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for (let i = 0; i < N; i++)
document.write(arr[i] + " ");
}
// Driver Code
// Given array
let arr = [ 1, 3, 5, 2, 4, 6 ];
// Size of the array
let N = arr.length;
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
</script>
Producción:
1 2 3 4 5 6
Complejidad de tiempo : O(N * log(N))
Espacio auxiliar: O(1)