# Move the first element of the linked list to the end

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In this problem, given a linked list, we move the first element of the linked list to the end of the linked list.

```
Example input:
1 -> 2 -> 2 -> 3
Example output:
2 -> 2 -> 3 -> 1
```

### What are linked lists ?

Linked List is a data structure. It is a linear data structure where the elements of the linked list are stored in non-consecutive

memory locations, each node in the linked list contains a data item and pointer to the next node in the linked list.

### Representation of linked lists

In our program we have represented a linked list using a `struct`

construct in C++.

```
struct node {
int data;
struct node* next;
};
```

Here, `data`

represents the data value in the node and the `next`

pointer represents the pointer to the next node in the list.

### Brute Force

In Brute Force approach, we swap the data values of two adjacent nodes one by one till the first node reaches the last position.

Pseudocode:

```
for i = 0 to N-1
swap(node(i), node(i+1))
```

This approach has an O(N) time complexity. We can reduce the number of operations performed in the efficient approach though the Time Complexity will remain same.

### Solution Approach

In this solution approach we maintain two pointers. The algorithm is as follows:

- Two pointers, the
`first`

pointer and the`last`

pointer are maintained. - Both of them initially contain reference to the head.
- The next element after the head is made the new head by moving the head reference to the next of current head.
- The
`next`

of the`last`

points to the previous head and it's`next`

stores

`null`

.

### Implementation of the approach

Here we have implemented the above approach in `C++`

```
#include <bits/stdc++.h>
using namespace std;
//node representation
struct node {
int data;
struct node* next;
};
//pointer declaration
struct node* head;
struct node* tail = new node;
struct node* temp = new node;
struct node* curr;
struct node* pre;
//this is a function to print the data items in the linked list
void print_data(struct node* head){
temp = head;
while(temp != NULL){
cout << temp -> data << " ";
temp = temp -> next;
}
}
//the move to end function which implements the algorithm.
void move_to_end(struct node** head_pointer){
if(*head_pointer == NULL){ //this means our list is empty, we return
return;
}
struct node* first = *head_pointer;
struct node* last = *head_pointer;
//go to the last node of the list and
//store the address of the last node
while(last -> next != NULL){
last = last -> next;
}
//we store the address of the second node here, this is the new head.
*head_pointer = first -> next;
//the previous head has the next set as NULL
first -> next = NULL;
//the previous head now becomes the tail
last -> next = first;
}
int main(){
int n, num, key;
cout << "Enter the number of elements" <<endl;
cin >> n ;
//accepting list from user
cout << "Enter the elements: " <<endl;
do {
cin >> num;
if(head == NULL){
head = new node;
head -> data = num;
head -> next = NULL;
tail = head;
}
else{
//create and initialize a new node
struct node *new_node = new node;
new_node -> data = num;
new_node -> next = NULL;
tail -> next = new_node;
tail = new_node;
}
n = n - 1;
}while(n > 0);
cout << "List before moving" <<endl;
print_data(head);
//pass the reference of head
move_to_end(&head);
cout <<"\nList after moving" <<endl;
print_data(head);
}
```

### A small example to understand the solution

Consider the linked list:

```
head tail
1 -> 3 -> 2 -> 3
```

Initially the `head_pointer`

pointer would be storing the address of the `head`

or the first element of the linked list.

The `first`

and `last`

pointers are also initialized.

```
head tail
1 -> 3 -> 2 -> 3
```

The `head_pointer`

then points to the next element after the `head`

. We make the next of `head`

as our new head now.

```
head tail
1 -> 3 -> 2 -> 3
```

Now `tail`

is moved to the first element and it's next stores `null`

. The structure of the list now becomes as follows

```
head tail
3 -> 2 -> 3 -> 1
```

This is the required output.

Time complexity:

The time complexity of this approach is given by O(N), where, N is the number of nodes in the linked list. As we traverse all the N nodes in one pass, this is done to reach the `tail`

pointer. The space complexity is O(1) as extra space is not allocated during this process.