Modify contents of Linked List

• Difficulty Level : Medium
• Last Updated : 22 Oct, 2021

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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. Perform 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++

```// 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 */

/* move the head to point to the 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;

/* 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. */
*back_ref = slow->next;
slow->next = NULL;
}

/* Function to reverse the linked list */
void reverseList(struct Node **head_ref)
{
struct Node *current, *prev, *next;
prev = NULL;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
}

// perform 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;

}

// function to modify the contents of the linked list
struct Node* modifyTheList(struct Node *head)
{
// if list is empty or contains only single node

struct Node *front, *back;

// split the list into two halves
// front and back lists

// 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
}

// function to print the linked list
void printList(struct Node *head)
{
return;

while (head->next != NULL)
{
cout << head->data << " -> ";
}
cout << head->data << endl;
}

// Driver program to test above
int main()
{
struct Node *head = NULL;

// creating the linked list

// modify the linked list

// print the modified linked list
cout << "Modified List:" << endl;
return 0;
}
```

Java

```// Java implementation to modify the contents
// of the linked list
class GFG
{

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

/* Function to insert a node at the beginning
of the linked list */
static Node push(Node head_ref, int new_data)
{
/* allocate node */
Node new_node =new Node();
/* put in the data */
new_node.data = new_data;

/* link the old list at the end
of the new node */

/* move the head to point to the new node */

}

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;

/* 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. */
back = slow.next;
slow.next = null;
}

/* Function to reverse the linked list */
static Node reverseList( Node head_ref)
{
Node current, prev, next;
prev = null;
while (current != null)
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
}

// 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;

}

// 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)
front = null; back = null;

// split the list into two halves
// front and back lists

// reverse the 2nd(back) list
back = reverseList(back);

// modify the contents of 1st half
modifyTheContentsOf1stHalf();

// agains reverse the 2nd(back) list
back = reverseList(back);

// concatenating the 2nd list back to the
// end of the 1st list
head = concatFrontAndBackList(front, back);

// pointer to the modified list
}

// function to print the linked list
static void printList( Node head)
{
if (head == null)
return;

while (head.next != null)
{
System.out.print(head.data + " -> ");
}
}

// Driver Code
public static void main(String args[])
{
Node head = null;

// creating the linked list

// modify the linked list

// print the modified linked list
System.out.println( "Modified List:" );
}
}

// This code is contributed by Arnab Kundu
```

Python3

```# Python3 implementation to modify the contents
# of the linked list

# Linked list node
class Node:

def __init__(self, data):
self.data = data
self.next = None

# Function to insert a node at the beginning
# of the linked list

# allocate node
new_node =Node(0)

# put in the data
new_node.data = new_data

# link the old list at the end
#of the new node

# move the head to point to the new node

front = None
back = None

# 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.

global front
global back
slow = None
fast = None

# Advance 'fast' two nodes, and
# advance 'slow' one node
while (fast != None):

fast = fast.next
if (fast != None):
slow = slow.next
fast = fast.next

# 'slow' is before the midpoint in the list,
# so split it in two at that point.
back = slow.next
slow.next = None

# Function to reverse the linked list

current = None
prev = None
next = None
prev = None
while (current != None):

next = current.next
current.next = prev
prev = current
current = next

# perform the required subtraction operation
# on the 1st half of the linked list
def modifyTheContentsOf1stHalf():

global front
global back
front1 = front
back1 = back

# traversing both the lists simultaneously
while (back1 != None):

# 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
def concatFrontAndBackList( front, back):

if(front == None):
return back

while (front.next != None):
front = front.next

front.next = back

# function to modify the contents of the linked list

global front
global back

# if list is empty or contains only single node
if (head == None or head.next == None):
front = None
back = None

# split the list into two halves
# front and back lists

# reverse the 2nd(back) list
back = reverseList(back)

# modify the contents of 1st half
modifyTheContentsOf1stHalf()

# agains reverse the 2nd(back) list
back = reverseList(back)

# concatenating the 2nd list back to the
# end of the 1st list
head = concatFrontAndBackList(front, back)

# pointer to the modified list

# function to print the linked list

if (head == None):
return

while (head.next != None):

print(head.data , " -> ",end="")

# Driver Code

# creating the linked list

# modify the linked list

# print the modified linked list
print( "Modified List:" )

# This code is contributed by Arnab Kundu

```

C#

```// C# implementation to modify the
// contents of the linked list
using System;

class GFG
{

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

/* Function to insert a node at
the beginning of the linked list */
static Node push(Node head_ref,
int new_data)
{
/* allocate node */
Node new_node = new Node();

/* put in the data */
new_node.data = new_data;

/* link the old list at the end
of the new node */

/* move the head to point to the new node */

}

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;

/* 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. */
back = slow.next;
slow.next = null;
}

/* Function to reverse the linked list */
static Node reverseList(Node head_ref)
{
Node current, prev, next;
prev = null;
while (current != null)
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
}

// 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;

}

// 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)
front = null; back = null;

// split the list into two halves
// front and back lists

// reverse the 2nd(back) list
back = reverseList(back);

// modify the contents of 1st half
modifyTheContentsOf1stHalf();

// agains reverse the 2nd(back) list
back = reverseList(back);

// concatenating the 2nd list back to the
// end of the 1st list
head = concatFrontAndBackList(front, back);

// pointer to the modified list
}

// function to print the linked list
static void printList( Node head)
{
if (head == null)
return;

while (head.next != null)
{
Console.Write(head.data + " -> ");
}
}

// Driver Code
public static void Main()
{
Node head = null;

// creating the linked list

// modify the linked list

// print the modified linked list
Console.WriteLine( "Modified List:" );
}
}

// This code is contributed by PrinciRaj1992
```

Javascript

```<script>

// JavaScript implementation to modify the
// contents of the linked list
/* Linked list node */
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}

/* Function to insert a node at
the beginning of the linked list */
function push(head_ref, new_data) {
/* allocate node */
var new_node = new Node();

/* put in the data */
new_node.data = new_data;

/* link the old list at the end
of the new node */

/* move the head to point to the new node */

}

var 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. */
var slow, fast;

/* 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. */
back = slow.next;
slow.next = null;
}

/* Function to reverse the linked list */
var current, prev, next;
prev = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
}

// perform the required subtraction operation
// on the 1st half of the linked list
function modifyTheContentsOf1stHalf() {
var 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
function concatFrontAndBackList(front, back) {
var head = front;

if (front == null) return back;

while (front.next != null) front = front.next;

front.next = back;

}

// function to modify the contents of
// the linked list
// 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

// reverse the 2nd(back) list
back = reverseList(back);

// modify the contents of 1st half
modifyTheContentsOf1stHalf();

// agains reverse the 2nd(back) list
back = reverseList(back);

// concatenating the 2nd list back to the
// end of the 1st list
head = concatFrontAndBackList(front, back);

// pointer to the modified list
}

// function to print the linked list
if (head == null) return;

while (head.next != null) {
document.write(head.data + " -> ");
}
}

// Driver Code
var head = null;

// creating the linked list

// modify the linked list

// print the modified linked list
document.write("Modified List: <br>");

</script>```

Output:

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

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

Another approach (Using Stack) :
1. Find the starting point of second half Linked List.
2. Push all elements of second half list into stack s.
3. Traverse list starting from head using temp until stack is not empty and do Modify temp->data by subtracting the top element of stack for every node.

Below is the implementation using stack.

C++

```// 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

// move the head to point to the new node
}

// function to print the linked list
void printList(struct Node *head)
{
return;

while (head->next != NULL)
{
cout << head->data << " -> ";
}
cout << head->data << endl;
}

// Function to middle node of list.
{
Node *temp = head, *slow = head, *fast = head ;

while(fast && fast->next)
{

// Advance 'fast' two nodes, and
// advance 'slow' one node
slow = slow->next ;
fast = fast->next->next ;
}

// If number of nodes are odd then update slow
// by slow->next;
if(fast)
slow = slow->next ;

return slow ;
}

// function to modify the contents of the linked list.
void modifyTheList(struct Node *head, struct Node *slow)
{
// Create Stack.
stack <int> s;
Node *temp = head ;

while(slow)
{
s.push( slow->data ) ;
slow = slow->next ;
}

// Traverse the list by using temp until stack is empty.
while( !s.empty() )
{
temp->data = temp->data - s.top() ;
temp = temp->next ;
s.pop() ;
}

}

// Driver program to test above
int main()
{
struct Node *head = NULL, *mid ;

// creating the linked list

// Call Function to Find the starting point of second half of list.
mid = find_mid(head) ;

// Call function to modify the contents of the linked list.

// print the modified linked list
cout << "Modified List:" << endl;
return 0;
}

// This is contributed by Mr. Gera
```

Java

```// Java implementation to modify the
// contents of the linked list
import java.util.*;

class GFG
{

// Linked list node
static class Node
{

int data;
Node next;
};

// Function to insert a node at the
// beginning of the linked list
static Node push(Node head_ref, int new_data)
{

// allocate node
Node new_node = new Node();

// put in the data
new_node.data = new_data;

// link the old list at the end of the new node

// move the head to point to the new node
}

// function to print the linked list
static void printList(Node head)
{
if (head == null)
{
return;
}

while (head.next != null)
{
}
}

// Function to middle node of list.
static Node find_mid(Node head)
{
Node temp = head, slow = head, fast = head;

while (fast != null && fast.next != null)
{

// Advance 'fast' two nodes, and
// advance 'slow' one node
slow = slow.next;
fast = fast.next.next;
}

// If number of nodes are odd then update slow
// by slow.next;
if (fast != null)
{
slow = slow.next;
}

return slow;
}

// function to modify the contents of the linked list.
static void modifyTheList(Node head, Node slow)
{
// Create Stack.
Stack<Integer> s = new Stack<Integer>();
Node temp = head;

while (slow != null)
{
slow = slow.next;
}

// Traverse the list by using temp until stack is empty.
while (!s.empty())
{
temp.data = temp.data - s.peek();
temp = temp.next;
s.pop();
}

}

// Driver program to test above
public static void main(String[] args)
{
Node head = null, mid;

// creating the linked list

// Call Function to Find the starting
// point of second half of list.

// Call function to modify
// the contents of the linked list.

// print the modified linked list
System.out.print("Modified List:" + "\n");
}
}

// This code is contributed by Rajput-Ji

```

Python3

```# Python3 implementation to modify the
# contents of the linked list

# Linked list node
class Node:

def __init__(self):

self.data = 0
self.next = None

# Function to insert a node at the
# beginning of the linked list

# Allocate node
new_node = Node()

# Put in the data
new_node.data = new_data

# Link the old list at the end
# of the new node

# Move the head to point to the new node

# Function to print the linked list

return;

while (head.next != None):
print(head.data, end = ' -> ')

# Function to middle node of list.

while (fast and fast.next):

# Advance 'fast' two nodes, and
# advance 'slow' one node
slow = slow.next
fast = fast.next.next

# If number of nodes are odd then
# update slow by slow.next;
if (fast):
slow = slow.next

return slow

# Function to modify the contents of
# the linked list.

# Create Stack.
s = []

while (slow):
s.append(slow.data)
slow = slow.next

# Traverse the list by using
# temp until stack is empty.
while (len(s) != 0):
temp.data = temp.data - s[-1]
temp = temp.next
s.pop()

# Driver code
if __name__=='__main__':

# creating the linked list

# Call Function to Find the
# starting point of second half of list.

# Call function to modify the
# contents of the linked list.

# Print the modified linked list
print("Modified List:")

# This code is contributed by rutvik_56```

C#

```// C# implementation to modify the
// contents of the linked list
using System;
using System.Collections.Generic;

class GFG
{

// Linked list node
public class Node
{

public int data;
public Node next;
};

// Function to insert a node at the
// beginning of the linked list
static Node push(Node head_ref, int new_data)
{

// allocate node
Node new_node = new Node();

// put in the data
new_node.data = new_data;

// link the old list at the end of the new node

// move the head to point to the new node
}

// function to print the linked list
static void printList(Node head)
{
if (head == null)
{
return;
}

while (head.next != null)
{
}
}

// Function to middle node of list.
static Node find_mid(Node head)
{
Node temp = head, slow = head, fast = head;

while (fast != null && fast.next != null)
{

// Advance 'fast' two nodes, and
// advance 'slow' one node
slow = slow.next;
fast = fast.next.next;
}

// If number of nodes are odd then update slow
// by slow.next;
if (fast != null)
{
slow = slow.next;
}

return slow;
}

// function to modify the contents of the linked list.
static void modifyTheList(Node head, Node slow)
{
// Create Stack.
Stack<int> s = new Stack<int>();
Node temp = head;

while (slow != null)
{
s.Push(slow.data);
slow = slow.next;
}

// Traverse the list by using temp until stack is empty.
while (s.Count != 0)
{
temp.data = temp.data - s.Peek();
temp = temp.next;
s.Pop();
}

}

// Driver code
public static void Main(String[] args)
{
Node head = null, mid;

// creating the linked list

// Call Function to Find the starting
// point of second half of list.

// Call function to modify
// the contents of the linked list.

// print the modified linked list
Console.Write("Modified List:" + "\n");
}
}

// This code is contributed by PrinciRaj1992

```

Javascript

```<script>

// JavaScript implementation to modify the
// contents of the linked list

// Linked list node
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}

// Function to insert a node at the
// beginning of the linked list
function push(head_ref , new_data) {

// allocate node
var new_node = new Node();

// put in the data
new_node.data = new_data;

// link the old list at the end of the new node

// move the head to point to the new node
}

// function to print the linked list
if (head == null) {
return;
}

while (head.next != null) {
}
}

// Function to middle node of list.
var temp = head, slow = head, fast = head;

while (fast != null && fast.next != null) {

// Advance 'fast' two nodes, and
// advance 'slow' one node
slow = slow.next;
fast = fast.next.next;
}

// If number of nodes are odd then update slow
// by slow.next;
if (fast != null) {
slow = slow.next;
}

return slow;
}

// function to modify the contents of the linked list.
function modifyTheList(head,  slow) {
// Create Stack.
var s = [];
var temp = head;

while (slow != null) {
s.push(slow.data);
slow = slow.next;
}

// Traverse the list by using temp until stack is empty.
while (s.length!=0) {
temp.data = temp.data - s.pop();
temp = temp.next;

}

}

// Driver program to test above

var head = null, mid;

// creating the linked list

// Call Function to Find the starting
// point of second half of list.

// Call function to modify
// the contents of the linked list.

// print the modified linked list
document.write("Modified List:" + "<br/>");

// This code contributed by Rajput-Ji

</script>
```

Output:

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

Time Complexity : O(n)
Space Complexity : O(n/2)
References: https://www.careercup.com/question?id=5657550909341696

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