# Modify contents of Linked List

### Problem Statement

In this problem, we would be given a linked list, and we need to modify the values of the first half nodes such that the value of the ith node will be equal to the value of (N-i-1)th node minus the value of the ith node.

Note: If the number of nodes is odd, then the value of the middle node will remain unchanged.

### Problem Statement Understanding

To understand this problem statement, let us take examples by referring the best online programming courses.

If the given linked list is : then according to the problem statement:

• At first, focus on the first and last node and change the value of the first node to (22 - 3) = 19.
• Then for the second node, its value will be updated to the value equal to the last-second node value minus the value of the second node, i.e., (18-12) = 6.
• Similarly, the value of the third node will be (5-1) = 4.
• So, the final linked list after modification will be: Let us take another example:
If the linked list is 4→1→10→42→5→NULL.

• After changing the values of the first half of nodes in the fashion mentioned above, our output linked list will be 1→41→10→42→5→NULL
• Note that the number of nodes in the linked list is odd, so we will not change the middle node, i.e., node 10.

Now, I think the problem statement is clear, so let's see how we can approach it. Any ideas? If not, it's okay, and we will see how we can approach it in the next section.

### Approach 1

• Find the first element of the second half of the linked list, i.e., find the middle node of the linked list.
• Push all elements starting from the node found above till the last of the linked list into a stack.
• Now using a temporary pointer, temp iterate the linked list starting from the head until the stack is not empty and modify temp→data by doing temp→data = (s.top() - temp→data) and then pop the element from the stack.
• Finally, when the traversal is over, we will have our final modified list as per the problem statement.

### Algorithm 1

• Initialize two pointers with the head of the list and name them slow and fast.
• Find the mid-node of the list by moving the slow pointer one node at a time and the fast pointer two nodes at a time, till fast or fast →next are not NULL.
• Now once you reach the end, check whether the fast pointer is NULL or not
• If it is NULL then the number of nodes in the list is even, then no need to update the slow pointer as it will point to the first element of the second half of the list.
• If it is not NULL that means the number of nodes in the list is odd so, to point to the first element of the second half of the list, we need to move the slow pointer by one node.
• Now, we have the starting point of the second half of the list so, we will start iterating the list from here and push all data of nodes in the stack till the end of the list.
• Now, we will begin an iteration from the starting of the list and change the value of each node to the difference of the stack’s top element and the current node at which we are in the iteration, and then we will pop the element from the stack.
• We will repeat the above step till the stack is not empty.

We will get our desired list by following the above steps.

Time complexity: O(n), Where n is the number of nodes in the list.
Space Complexity: O(n), Where n is the number of nodes in the list

As you can see that in the above approach, we used an extra space of O(n) because we were pushing the elements of the second half of the list in the stack.

Can we improve this to constant extra space?

• The answer is yes, we can improve our algorithm, refer below to get an idea of the constant extra space approach.

• Notice that we cannot move backward in a linked list, so we need to devise a way by which we can traverse the list in forward and backward directions.
• This can be only done if we first break the list from the middle into two parts and then reverse the second half.
• After doing this, we can treat each half as two different lists, and moving forward in the second half of the list would eventually mean to iterate in the list in the backward direction as it is reversed.
• Then, we can keep one pointer at the starting of each half, modify the nodes’ values that belong to the first half, and advance each pointer to the next node.

### Approach 2

Our approach will be simple:

• Find the middle node of the linked list and then split the linked list from the middle with the help of a fast and a slow pointer, where the slow will move one node at a time and the fast one will move 2 nodes at a time, so that when fast will be at the end of the list slow will be at the middle.
• Now, reverse the second half of the list obtained in the above step.
• Initialize two pointers named front and back with the starting of each list, respectively.
• Iterate over both the lists simultaneously and update the values of the first half nodes as directed in the problem statement.
• Then again, reverse the second half of the list.
• Then join the two parts (first half and second half) of the list as they were originally given to us.

### Algorithm 2

• At first, we will find the middle node of our linked list and then break the list from there.
• Then, we will have two different linked lists.
• Then, we will reverse the second linked list.
• Initialize two pointers, front and back, where the front will point to the first node of the first list and the back will point to the first node of the second list.
• While the end of the second list is not reached i.e., back is not equal to NULL pointer, we need to iterate the list and do the following:
a) Change front’s data to the difference of back’s and front’s data (front->data = back->data - front->data).
b) Advance front pointer by one node (front = front->next).
c) Advance back pointer by one node (front = front->next).
• After the while loop ends, reverse the second list again.
• Now, attach the second list to the end of the first list.
• Return the head of the newly created list.
• The first half nodes of the returned list will be modified according to the problem statement.

### Dry Run   ### Code Implementation

```#include
using namespace std;
class Node
{
public:
int data;
Node* next;
Node(int x){
data = x;
next = NULL;
}
};

Node **front_ref, Node **back_ref)
{
Node *slow, *fast;

/* Advance 'fast' pointer two nodes at a time and
advance 'slow' pointer one node at a time */
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}

/* 'slow' is before the midpoint in the list,
so split it in two at that point. */
*back_ref = slow->next;
slow->next = NULL;
}

{
Node *current, *prev, *next;
prev = NULL;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
}
{
// traverse both the lists simultaneously
while (back != NULL)
{
front->data = back->data - front->data;

front = front->next;
back = back->next;
}
}

{
return;

{
cout << head->data << " ";
}
}

Node* concatFrontAndBackList(Node *front,
Node *back)
{

while (front->next != NULL)
front = front->next;

front->next = back;

}

{
// if list is empty or contains only single node then return it

Node *front, *back;

// split the list into two parts

// reverse the 2nd half of the list
reverseList(&back);

// modify the contents of 1st half of the list
modifyContent(front, back);

// again reverse the 2nd half of list
reverseList(&back);

// concatenate both the halves to get the original list
// that was provided initially

// pointer to the modified list
}
int main(void)
{