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Split the array and add the first part to the end
  • Difficulty Level : Easy
  • Last Updated : 06 Dec, 2019
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There is a given an array and split it from a specified position, and move the first part of array add to the end.
Split the array and add the first part to the end

Examples:

Input : arr[] = {12, 10, 5, 6, 52, 36}
            k = 2
Output : arr[] = {5, 6, 52, 36, 12, 10}
Explanation : Split from index 2 and first 
part {12, 10} add to the end .

Input : arr[] = {3, 1, 2}
           k = 1
Output : arr[] = {1, 2, 3}
Explanation : Split from index 1 and first
part add to the end.

Simple Solution
We one by one rotate array.

C++




// CPP program to split array and move first
// part to end.
#include <bits/stdc++.h>
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, 6, position);
  
    for (int i = 0; i < n; ++i)
        printf("%d ", arr[i]);
  
    return 0;
}

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, 6, 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 spilt 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, 6, 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, 6, $position);
  
for ($i = 0; $i < $n; ++$i)
    echo $arr[$i]." ";
  
// This code is contributed
// by ChitraNayal
?>


Output:



5 6 52 36 12 10

Time complexity of above solution is O(nk).
Another approach: An another approach is to make a temporary array with double the size and copy our array element in to new array twice .and then copy element from new array to our array by taking the rotation as starting index upto the length of our array.

Below is the implementation of above approach.

C++




// CPP program to split array and move first
// part to end.
#include <bits/stdc++.h>
using namespace std;
  
// Function to spilt 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 spilt 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 spilt 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 spilt 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


Output:
5 6 52 36 12 10

An efficient O(n) solution is discussed the following post: Split the array and add the first part to the end | Set 2

This problem is noting but array rotation problem and we can apply 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

This article is contributed by Aditya Ranjan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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