Hay una array dada y se divide desde una posición específica, se mueve la primera parte dividida de la array y luego se agrega al final de la array.
Ejemplos:
Input : arr[] = {12, 10, 5, 6, 52, 36}
k = 2
Output : arr[] = {5, 6, 52, 36, 12, 10}
Explanation : here k is two so first two elements are splitted and they are added at the end of array
Input : arr[] = {3, 1, 2}
k = 1
Output : arr[] = {1, 2, 3}
Explanation :here k is one so first one element are splitted and it is added at the end of array
Solución simple: rotaremos los elementos de la array uno por uno.
Implementación:
C
// CPP program to split array and move first
// part to end.
#include <stdio.h>
void splitArr(int arr[], int n, int k)
{
for (int i = 0; i < k; i++) {
// Rotate array by 1.
int x = arr[0];
for (int j = 0; j < n - 1; ++j)
arr[j] = arr[j + 1];
arr[n - 1] = x;
}
}
// Driver code
int main()
{
int arr[] = { 12, 10, 5, 6, 52, 36 };
int n = sizeof(arr) / sizeof(arr[0]);
int position = 2;
splitArr(arr, n, position);
for (int i = 0; i < n; ++i)
printf("%d ", arr[i]);
return 0;
}
C++
// CPP program to split array and move first
// part to end.
#include <iostream>
using namespace std;
void splitArr(int arr[], int n, int k)
{
for (int i = 0; i < k; i++) {
// Rotate array by 1.
int x = arr[0];
for (int j = 0; j < n - 1; ++j)
arr[j] = arr[j + 1];
arr[n - 1] = x;
}
}
// Driver code
int main()
{
int arr[] = { 12, 10, 5, 6, 52, 36 };
int n = sizeof(arr) / sizeof(arr[0]);
int position = 2;
splitArr(arr, n, position);
for (int i = 0; i < n; ++i)
cout <<" "<< arr[i];
return 0;
}
// This code is contributed by shivanisinghss2110
Java
// Java program to split array and move first
// part to end.
import java.util.*;
import java.lang.*;
class GFG {
public static void splitArr(int arr[], int n, int k)
{
for (int i = 0; i < k; i++) {
// Rotate array by 1.
int x = arr[0];
for (int j = 0; j < n - 1; ++j)
arr[j] = arr[j + 1];
arr[n - 1] = x;
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 12, 10, 5, 6, 52, 36 };
int n = arr.length;
int position = 2;
splitArr(arr, n, position);
for (int i = 0; i < n; ++i)
System.out.print(arr[i] + " ");
}
}
// Code Contributed by Mohit Gupta_OMG <(0_o)>
Python3
# Python program to split array and move first # part to end. def splitArr(arr, n, k): for i in range(0, k): x = arr[0] for j in range(0, n-1): arr[j] = arr[j + 1] arr[n-1] = x # main arr = [12, 10, 5, 6, 52, 36] n = len(arr) position = 2 splitArr(arr, n, position) for i in range(0, n): print(arr[i], end = ' ') # Code Contributed by Mohit Gupta_OMG <(0_o)>
C#
// C# program to split array
// and move first part to end.
using System;
class GFG {
// Function to split array and
// move first part to end
public static void splitArr(int[] arr, int n,
int k)
{
for (int i = 0; i < k; i++)
{
// Rotate array by 1.
int x = arr[0];
for (int j = 0; j < n - 1; ++j)
arr[j] = arr[j + 1];
arr[n - 1] = x;
}
}
// Driver code
public static void Main()
{
int[] arr = {12, 10, 5, 6, 52, 36};
int n = arr.Length;
int position = 2;
splitArr(arr, n, position);
for (int i = 0; i < n; ++i)
Console.Write(arr[i] + " ");
}
}
// This code is contributed by Shrikant13.
PHP
<?php
// PHP program to split array
// and move first part to end.
function splitArr(&$arr, $n, $k)
{
for ($i = 0; $i < $k; $i++)
{
// Rotate array by 1.
$x = $arr[0];
for ($j = 0; $j < $n - 1; ++$j)
$arr[$j] = $arr[$j + 1];
$arr[$n - 1] = $x;
}
}
// Driver code
$arr = array(12, 10, 5, 6, 52, 36);
$n = sizeof($arr);
$position = 2;
splitArr($arr, n, $position);
for ($i = 0; $i < $n; ++$i)
echo $arr[$i]." ";
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// JavaScript program to split
// array and move first
// part to end.
function splitArr(arr, n, k){
for (let i = 0; i < k; i++)
{
// Rotate array by 1.
let x = arr[0];
for (let j = 0; j < n - 1; ++j)
arr[j] = arr[j + 1];
arr[n - 1] = x;
}
}
// Driver code
let arr = [ 12, 10, 5, 6, 52, 36 ];
let n = arr.length;
let position = 2;
splitArr(arr, n, position);
for (let i = 0; i < n; ++i)
document.write(arr[i]+" ");
</script>
Producción
5 6 52 36 12 10
Complejidad temporal: O(n).
Complejidad del espacio : O(1)
Otro enfoque: otro enfoque es hacer una array temporal con el doble de tamaño y copiar nuestro elemento de array en una nueva array dos veces. Y luego copiar el elemento de la nueva array a nuestra array tomando la rotación como índice inicial hasta la longitud de nuestra array.
A continuación se muestra la implementación del enfoque anterior.
C++
// CPP program to split array and move first
// part to end.
#include <bits/stdc++.h>
using namespace std;
// Function to split array and
// move first part to end
void splitArr(int arr[], int length, int rotation)
{
int tmp[length * 2] = {0};
for(int i = 0; i < length; i++)
{
tmp[i] = arr[i];
tmp[i + length] = arr[i];
}
for(int i = rotation; i < rotation + length; i++)
{
arr[i - rotation] = tmp[i];
}
}
// Driver code
int main()
{
int arr[] = { 12, 10, 5, 6, 52, 36 };
int n = sizeof(arr) / sizeof(arr[0]);
int position = 2;
splitArr(arr, n, position);
for (int i = 0; i < n; ++i)
printf("%d ", arr[i]);
return 0;
}
// This code is contributed by YashKhandelwal8
Java
// Java program to split array and move first
// part to end.
import java.util.*;
import java.lang.*;
class GFG {
// Function to split array and
// move first part to end
public static void SplitAndAdd(int[] A,int length,int rotation){
//make a temporary array with double the size
int[] tmp = new int[length*2];
// copy array element in to new array twice
System.arraycopy(A, 0, tmp, 0, length);
System.arraycopy(A, 0, tmp, length, length);
for(int i=rotation;i<rotation+length;i++)
A[i-rotation]=tmp[i];
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 12, 10, 5, 6, 52, 36 };
int n = arr.length;
int position = 2;
SplitAndAdd(arr, n, position);
for (int i = 0; i < n; ++i)
System.out.print(arr[i] + " ");
}
}
Python3
# Python3 program to split array and # move first part to end. # Function to split array and # move first part to end def SplitAndAdd(A, length, rotation): # make a temporary array with double # the size and each index is initialized to 0 tmp = [ 0 for i in range(length * 2)] # copy array element in to new array twice for i in range(length): tmp[i] = A[i] tmp[i + length] = A[i] for i in range(rotation, rotation + length, 1): A[i - rotation] = tmp[i]; # Driver code arr = [12, 10, 5, 6, 52, 36] n = len(arr) position = 2 SplitAndAdd(arr, n, position); for i in range(n): print(arr[i], end = " ") print() # This code is contributed by SOUMYA SEN
C#
// C# program to split array
// and move first part to end.
using System;
class GFG
{
// Function to split array and
// move first part to end
public static void SplitAndAdd(int[] A,
int length,
int rotation)
{
// make a temporary array with double the size
int[] tmp = new int[length * 2];
// copy array element in to new array twice
Array.Copy(A, 0, tmp, 0, length);
Array.Copy(A, 0, tmp, length, length);
for (int i = rotation; i < rotation + length; i++)
{
A[i - rotation] = tmp[i];
}
}
// Driver code
public static void Main(string[] args)
{
int[] arr = new int[] {12, 10, 5, 6, 52, 36};
int n = arr.Length;
int position = 2;
SplitAndAdd(arr, n, position);
for (int i = 0; i < n; ++i)
{
Console.Write(arr[i] + " ");
}
}
}
// This code is contributed by kumar65
Javascript
<script>
// Javascript program to split
// array and move first
// part to end.
// Function to split array and
// move first part to end
function SplitAndAdd(A,Length,rotation)
{
// make a temporary array
// with double the size
let tmp = new Array(Length*2);
for(let i=0;i<tmp.length;i++)
{
tmp[i]=0;
}
// copy array element in to new array twice
for(let i=0;i<Length;i++)
{
tmp[i] = A[i]
tmp[i + Length] = A[i]
}
for(let i=rotation;i<rotation+Length;i++)
{
A[i - rotation] = tmp[i];
}
}
// Driver code
let arr=[12, 10, 5, 6, 52, 36];
let n = arr.length;
let position = 2;
SplitAndAdd(arr, n, position);
for (let i = 0; i < n; ++i)
document.write(arr[i] + " ");
// This code is contributed by avanitrachhadiya2155
</script>
Producción
5 6 52 36 12 10
Complejidad temporal: O(n)
Complejidad espacial: O(2n)
Una solución O(n) eficiente se analiza en la siguiente publicación: Divida la array y agregue la primera parte al final | Conjunto 2
Se observa este problema, pero el problema de rotación de array y podemos aplicar los métodos optimizados de rotación de array O (n) aquí.
Programa para rotación de
arreglos Algoritmo de intercambio de bloques para rotación de arreglos
Algoritmo de inversión para rotación de arreglos
Encuentre rápidamente múltiples rotaciones a la izquierda de un arreglo | Conjunto 1
Imprime la rotación a la izquierda de la array en el tiempo O (n) y el espacio O (1)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA