Split the array and add the first part to the end
Last Updated :
17 Oct, 2022
There is a given array and split it from a specified position, and move the first splitted part of the array and then add to the end of array
Examples:
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
Simple Solution: We will rotate the elements of the array one by one.
Implementation:
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;
}
// 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 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)>
# 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# 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 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
?>
<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>
Time complexity: O(n*k).
Space complexity: O(1), since no extra space has been taken.
Another approach: Another approach is to make a temporary array with double the size and copy our array element into a new array twice and then copy the element from the new array to our array by taking the rotation as starting index up to the length of our array.
Below is the implementation of the above approach.
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 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 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# 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
<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>
Time complexity: O(n)
Space complexity: O(2*n)
An efficient O(n) solution is discussed in the following post: Split the array and add the first part to the end | Set 2
This problem is noted but array rotation problem and we can apply the optimized O(n) array rotation methods here.
Program for array rotation
Block swap algorithm for array rotation
Reversal algorithm for array rotation
Quickly find multiple left rotations of an array | Set 1
Print left rotation of array in O(n) time and O(1) space
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