There is a given an array and split it from a specified position, and move the first part of array add 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 = 2SplitAndAdd(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
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