Modify contents of Linked List

2.6

Given a singly linked list containing n nodes. Modify the value of first half nodes such that 1st node’s new value is equal to the last node’s value minus first node’s current value, 2nd node’s new value is equal to the second last node’s value minus 2nd node’s current value, likewise for first half nodes. If n is odd then the value of the middle node remains unchanged.
(No extra memory to be used).

Examples:

Input : 10 -> 4 -> 5 -> 3 -> 6
Output : 4 -> 1 -> 5 -> 3 -> 6

Input : 2 -> 9 -> 8 -> 12 -> 7 -> 10
Output : -8 -> 2 -> -4 -> 12 -> 7 -> 10

Asked in Amazon Interview

Approach: The following steps are:

  1. Split the list from the middle. Perform front and back split. If the number of elements is odd, the extra element should go in the 1st(front) list.
  2. Reverse the 2nd(back) list.
  3. Perfrom the required subtraction while traversing both list simultaneously.
  4. Again reverse the 2nd list.
  5. Concatenate the 2nd list back to the end of the 1st list.
// C++ implementation to modify the contents of 
// the linked list
#include <bits/stdc++.h>
using namespace std;

/* Linked list node */
struct Node
{
    int data;
    struct Node* next;
};

/* function prototype for printing the list */
void printList(struct Node*);

/* Function to insert a node at the beginning of 
   the linked list */
void push(struct Node **head_ref, int new_data)
{
  /* allocate node */
  struct Node* new_node =
            (struct Node*) malloc(sizeof(struct Node));
  
  /* put in the data  */
  new_node->data = new_data;
  
  /* link the old list at the end of the new node */
  new_node->next = *head_ref;    
  
  /* move the head to point to the new node */
  *head_ref = new_node;
} 

/* 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. */
void frontAndBackSplit(struct Node *head, 
               struct Node **front_ref, struct Node **back_ref)
{
    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_ref = head;
    *back_ref = slow->next;
    slow->next = NULL;
}

/* Function to reverse the linked list */
void reverseList(struct Node **head_ref)
{
    struct 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;
}

// perfrom the required subtraction operation on
// the 1st half of the linked list
void modifyTheContentsOf1stHalf(struct Node *front,
                                struct Node *back)
{
    // traversing both the lists simultaneously
    while (back != NULL)
    {
        // subtraction operation and node data
        // modification
        front->data = front->data - back->data;
        
        front = front->next;
        back = back->next;
    }
}

// function to concatenate the 2nd(back) list at the end of
// the 1st(front) list and returns the head of the new list
struct Node* concatFrontAndBackList(struct Node *front,
                                    struct Node *back)
{
    struct Node *head = front;
    
    while (front->next != NULL)
        front = front->next;    
        
    front->next    = back;
    
    return head;
}

// function to modify the contents of the linked list
struct Node* modifyTheList(struct Node *head)
{
    // if list is empty or contains only single node
    if (!head || head->next == NULL)
        return head;
    
    struct Node *front, *back;
    
    // split the list into two halves
    // front and back lists
    frontAndBackSplit(head, &front, &back);    
        
    // reverse the 2nd(back) list
    reverseList(&back);
    
    // modify the contents of 1st half    
    modifyTheContentsOf1stHalf(front, back);
        
    // agains reverse the 2nd(back) list
    reverseList(&back);
    
    // concatenating the 2nd list back to the 
    // end of the 1st list
    head = concatFrontAndBackList(front, back);
    
    // pointer to the modified list
    return head;
}

// function to print the linked list
void printList(struct Node *head)
{
    if (!head)
        return;
    
    while (head->next != NULL)
    {
        cout << head->data << " -> ";
        head = head->next;
    }
    cout << head->data << endl;
}

// Driver program to test above
int main()
{
    struct Node *head = NULL;
    
    // creating the linked list
    push(&head, 10);
    push(&head, 7);
    push(&head, 12);
    push(&head, 8);
    push(&head, 9);
    push(&head, 2);
    
    // modify the linked list
    head = modifyTheList(head);
    
    // print the modified linked list
    cout << "Modified List:" << endl;
    printList(head);
    return 0;
} 

Output:

Modified List:
-8 -> 2 -> -4 -> 12 -> 7 -> 10

Time Complexity: O(n), where n in the number of nodes.

References: https://www.careercup.com/question?id=5657550909341696

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