### 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 i^{th} node will be equal to the value of (N-i-1)^{th} node minus the value of the i^{th} 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 4â†’1â†’10â†’42â†’5â†’NULL.

- After changing the values of the first half of nodes in the fashion mentioned above, our output linked list will be 1â†’41â†’10â†’42â†’5â†’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
**tempâ†’data**by doing**tempâ†’data = (s.top() – tempâ†’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 â†’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.

- If it is NULL then the number of nodes in the list is
- 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 stackâ€™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.

#### Helpful Observations

- 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 nodesâ€™ 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 frontâ€™s data to the difference of backâ€™s and frontâ€™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

#includeusing namespace std; class Node { public: int data; Node* next; Node(int x){ data = x; next = NULL; } }; void midPartition(Node *head, Node **front_ref, Node **back_ref) { Node *slow, *fast; slow = head; fast = head->next; /* 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. */ *front_ref = head; *back_ref = slow->next; slow->next = NULL; } void reverseList(Node **head_ref) { Node *current, *prev, *next; current = *head_ref; prev = NULL; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head_ref = prev; } void modifyContent(Node *head1,Node *head2) { Node *front = head1, *back = head2; // traverse both the lists simultaneously while (back != NULL) { front->data = back->data - front->data; front = front->next; back = back->next; } } void printList(Node *head) { if (!head) return; while (head->next != NULL) { cout << head->data << " "; head = head->next; } cout << head->data << endl; } Node* concatFrontAndBackList(Node *front, Node *back) { Node *head = front; while (front->next != NULL) front = front->next; front->next = back; return head; } Node* solve(Node *head) { // if list is empty or contains only single node then return it if (!head || head->next == NULL) return head; Node *front, *back; // split the list into two parts midPartition(head, &front, &back); // 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 head = concatFrontAndBackList(front, back); // pointer to the modified list return head; } int main(void) { Node* head = NULL; head = new Node(3); head->next = new Node(12); head->next->next = new Node(1); head->next->next->next = new Node(5); head->next->next->next->next = new Node(18); head->next->next->next->next->next = new Node(22); Node *tmp = solve(head); printList(tmp); return 0; }

#include#include #include struct Node { int data; struct Node* prev; struct Node* next; }; void push(struct Node** head_ref, int new_c) { struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); new_node->data = new_c; new_node->prev = NULL; new_node->next = (*head_ref); if ((*head_ref) != NULL) (*head_ref)->prev = new_node; *head_ref = new_node; } bool isCircular(struct Node *head){ struct Node *temp=head; while(temp!=NULL) { //if temp points to head then it has completed a circle,thus a circular linked list. if(temp->next==head) return true; temp=temp->next; } return false; } int main(void) { struct Node* head = NULL; push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); if(isCircular(head)) printf("Yes\n"); else printf("No\n"); }

class Modify { static class Node { int data; Node next; }; static Node push(Node head_ref, int new_data) { Node new_node =new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } static Node front,back; /* Split the nodes of the given list into front and back halves, and return the two lists using the reference parameters. Uses the fast/slow pointer strategy. */ static void frontAndBackSplit( Node head) { Node slow, fast; slow = head; fast = head.next; /* Advance 'fast' two nodes, and advance 'slow' one node */ 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. */ front = head; back = slow.next; slow.next = null; } /* Function to reverse the linked list */ static Node reverseList( Node head_ref) { Node current, prev, next; current = head_ref; prev = null; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } head_ref = prev; return head_ref; } // perform the required subtraction operation on the 1st half of the linked list static void modifyTheContentsOf1stHalf() { Node front1 = front, back1 = back; // traversing both the lists simultaneously while (back1 != null) { // subtraction operation and node data modification front1.data = front1.data - back1.data; front1 = front1.next; back1 = back1.next; } } // function to concatenate the 2nd(back) list // at the end of the 1st(front) list and // returns the head of the new list static Node concatFrontAndBackList(Node front,Node back) { Node head = front; if(front == null)return back; while (front.next != null) front = front.next; front.next = back; return head; } // function to modify the contents of the linked list static Node modifyTheList( Node head) { // if list is empty or contains only single node if (head == null || head.next == null) return head; front = null; back = null; // split the list into two halves front and back lists frontAndBackSplit(head); back = reverseList(back); // modify the contents of 1st half modifyTheContentsOf1stHalf(); // agains reverse the 2nd(back) list back = reverseList(back); head = concatFrontAndBackList(front, back); return head; } static void printList( Node head) { if (head == null) return; while (head.next != null) { System.out.print(head.data + " -> "); head = head.next; } System.out.println(head.data ); } // Driver Code public static void main(String args[]) { Node head = null; // creating the linked list head = push(head, 10); head = push(head, 7); head = push(head, 12); head = push(head, 8); head = push(head, 9); head = push(head, 2); // modify the linked list head = modifyTheList(head); // print the modified linked list System.out.println( "Modified List:" ); printList(head); } } // This code is contributed by Arnab Kundu

class Node: def __init__(self, data): self.data = data self.next = None def push(head_ref, new_data): new_node =Node(0) new_node.data = new_data new_node.next = head_ref head_ref = new_node return head_ref front = None back = None def frontAndBackSplit( head): global front global back slow = None fast = None slow = head fast = head.next while (fast != None): fast = fast.next if (fast != None): slow = slow.next fast = fast.next front = head back = slow.next slow.next = None return head def reverseList( head_ref): current = None prev = None next = None current = head_ref prev = None while (current != None): next = current.next current.next = prev prev = current current = next head_ref = prev return head_ref def modifyTheContentsOf1stHalf(): global front global back front1 = front back1 = back while (back1 != None): front1.data = (back1.data - front1.data) front1 = front1.next back1 = back1.next def concatFrontAndBackList( front, back): head = front if(front == None): return back while (front.next != None): front = front.next front.next = back return head def modifyTheList( head): global front global back if (head == None or head.next == None): return head front = None back = None frontAndBackSplit(head) back = reverseList(back) modifyTheContentsOf1stHalf() back = reverseList(back) head = concatFrontAndBackList(front, back) return head def printList( head): if (head == None): return while (head.next != None): print(head.data ,end=" ") head = head.next print(head.data ) head = None head = push(head, 22) head = push(head, 18) head = push(head, 5) head = push(head, 1) head = push(head, 12) head = push(head, 3) head = modifyTheList(head) printList(head)

#### Output

19 6 4 5 18 22

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

[forminator_quiz id=”4649″]

So, in this blog, we have tried to explain how you can modify the contents of a linked list in the most optimal way. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.