Given a linked list and a number n. Find the sum of the last n nodes of the linked list.
Constraints: 0 <= n <= number of nodes in the linked list.
Examples:
Input : 10->6->8->4->12, n = 2 Output : 16 Sum of last two nodes: 12 + 4 = 16 Input : 15->7->9->5->16->14, n = 4 Output : 44
Method 1: (Recursive approach using system call stack):
Recursively traverse the linked list up to the end. Now during the return from the function calls, add up the last n nodes. The sum can be accumulated in some variable passed by reference to the function or to some global variable.
Implementation:
// C++ implementation to find the sum of // last 'n' nodes of the Linked List #include <bits/stdc++.h> using namespace std;
/* A Linked list node */ struct Node {
int data;
struct Node* next;
}; // 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 = new Node;
/* put in the data */
new_node->data = new_data;
/* link the old list to the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} // function to recursively find the sum of last // 'n' nodes of the given linked list void sumOfLastN_Nodes( struct Node* head, int * n,
int * sum)
{ // if head = NULL
if (!head)
return ;
// recursively traverse the remaining nodes
sumOfLastN_Nodes(head->next, n, sum);
// if node count 'n' is greater than 0
if (*n > 0) {
// accumulate sum
*sum = *sum + head->data;
// reduce node count 'n' by 1
--*n;
}
} // utility function to find the sum of last 'n' nodes int sumOfLastN_NodesUtil( struct Node* head, int n)
{ // if n == 0
if (n <= 0)
return 0;
int sum = 0;
// find the sum of last 'n' nodes
sumOfLastN_Nodes(head, &n, &sum);
// required sum
return sum;
} // Driver program to test above int main()
{ struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << "Sum of last " << n << " nodes = "
<< sumOfLastN_NodesUtil(head, n);
return 0;
} |
// Java implementation to find the sum of // last 'n' nodes of the Linked List import java.util.*;
class GFG
{ /* A Linked list node */ static class Node
{ int data;
Node next;
}; static Node head;
static int n, sum;
// function to insert a node at the // beginning of the linked list static void 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head = head_ref;
} // function to recursively find the sum of last // 'n' nodes of the given linked list static void sumOfLastN_Nodes(Node head)
{ // if head = NULL
if (head == null )
return ;
// recursively traverse the remaining nodes
sumOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0 )
{
// accumulate sum
sum = sum + head.data;
// reduce node count 'n' by 1
--n;
}
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil(Node head, int n)
{ // if n == 0
if (n <= 0 )
return 0 ;
sum = 0 ;
// find the sum of last 'n' nodes
sumOfLastN_Nodes(head);
// required sum
return sum;
} // Driver Code public static void main(String[] args)
{ head = null ;
// create linked list 10.6.8.4.12
push(head, 12 );
push(head, 4 );
push(head, 8 );
push(head, 6 );
push(head, 10 );
n = 2 ;
System.out.print( "Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
} } // This code is contributed by 29AjayKumar |
# Python3 implementation to find the sum of # last 'n' nodes of the Linked List # Linked List node class Node:
def __init__( self , data):
self .data = data
self . next = None
head = None
n = 0
sum = 0
# function to insert a node at the # beginning of the linked list def push(head_ref, new_data):
global head
# allocate node
new_node = Node( 0 )
# put in the data
new_node.data = new_data
# link the old list to the new node
new_node. next = head_ref
# move the head to point to the new node
head_ref = new_node
head = head_ref
# function to recursively find the sum of last # 'n' nodes of the given linked list def sumOfLastN_Nodes(head):
global sum
global n
# if head = None
if (head = = None ):
return
# recursively traverse the remaining nodes
sumOfLastN_Nodes(head. next )
# if node count 'n' is greater than 0
if (n > 0 ) :
# accumulate sum
sum = sum + head.data
# reduce node count 'n' by 1
n = n - 1
# utility function to find the sum of last 'n' nodes def sumOfLastN_NodesUtil(head, n):
global sum
# if n == 0
if (n < = 0 ):
return 0
sum = 0
# find the sum of last 'n' nodes
sumOfLastN_Nodes(head)
# required sum
return sum
# Driver Code head = None
# create linked list 10.6.8.4.12 push(head, 12 )
push(head, 4 )
push(head, 8 )
push(head, 6 )
push(head, 10 )
n = 2
print ( "Sum of last " , n ,
" nodes = " , sumOfLastN_NodesUtil(head, n))
# This code is contributed by Arnab Kundu |
// C# implementation to find the sum of // last 'n' nodes of the Linked List using System;
class GFG
{ /* A Linked list node */ public class Node
{ public int data;
public Node next;
}; static Node head;
static int n, sum;
// function to insert a node at the // beginning of the linked list static void 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head = head_ref;
} // function to recursively find the sum of last // 'n' nodes of the given linked list static void sumOfLastN_Nodes(Node head)
{ // if head = NULL
if (head == null )
return ;
// recursively traverse the remaining nodes
sumOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0)
{
// accumulate sum
sum = sum + head.data;
// reduce node count 'n' by 1
--n;
}
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil(Node head, int n)
{ // if n == 0
if (n <= 0)
return 0;
sum = 0;
// find the sum of last 'n' nodes
sumOfLastN_Nodes(head);
// required sum
return sum;
} // Driver Code public static void Main(String[] args)
{ head = null ;
// create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
n = 2;
Console.Write( "Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript implementation to find the sum of // last 'n' nodes of the Linked List /* A Linked list node */
class Node
{
constructor()
{
this .data;
this .next;
}
}
let head;
let n, sum;
// function to insert a node at the
// beginning of the linked list
function push(head_ref,new_data)
{
/* allocate node */
let new_node = new Node();
/* put in the data */
new_node.data = new_data;
/* link the old list to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head = head_ref;
}
// function to recursively find the sum of last
// 'n' nodes of the given linked list
function sumOfLastN_Nodes(head)
{
// if head = NULL
if (head == null )
return ;
// recursively traverse the remaining nodes
sumOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0)
{
// accumulate sum
sum = sum + head.data;
// reduce node count 'n' by 1
--n;
}
}
// utility function to find the sum of last 'n' nodes
function sumOfLastN_NodesUtil(head,n)
{
// if n == 0
if (n <= 0)
return 0;
sum = 0;
// find the sum of last 'n' nodes
sumOfLastN_Nodes(head);
// required sum
return sum;
}
// Driver Code
head = null ;
// create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
n = 2;
document.write( "Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
// This code is contributed by unknown2108 </script> |
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n), if system call stack is being considered.
Method 2 (Iterative approach using user-defined stack):
It is an iterative procedure to the recursive approach explained in Method 1 of this post. Traverses the nodes from left to right. While traversing pushes the nodes to a user-defined stack. Then pops the top n values from the stack and adds them.
Implementation:
// C++ implementation to find the sum of last // 'n' nodes of the Linked List #include <bits/stdc++.h> using namespace std;
/* A Linked list node */ struct Node {
int data;
struct Node* next;
}; // 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 = new Node;
/* put in the data */
new_node->data = new_data;
/* link the old list to the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} // utility function to find the sum of last 'n' nodes int sumOfLastN_NodesUtil( struct Node* head, int n)
{ // if n == 0
if (n <= 0)
return 0;
stack< int > st;
int sum = 0;
// traverses the list from left to right
while (head != NULL) {
// push the node's data onto the stack 'st'
st.push(head->data);
// move to next node
head = head->next;
}
// pop 'n' nodes from 'st' and
// add them
while (n--) {
sum += st.top();
st.pop();
}
// required sum
return sum;
} // Driver program to test above int main()
{ struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << "Sum of last " << n << " nodes = "
<< sumOfLastN_NodesUtil(head, n);
return 0;
} |
// Java implementation to find the sum of last // 'n' nodes of the Linked List import java.util.*;
class GFG
{ /* A 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
return head_ref;
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil(Node head, int n)
{ // if n == 0
if (n <= 0 )
return 0 ;
Stack<Integer> st = new Stack<Integer>();
int sum = 0 ;
// traverses the list from left to right
while (head != null )
{
// push the node's data onto the stack 'st'
st.push(head.data);
// move to next node
head = head.next;
}
// pop 'n' nodes from 'st' and
// add them
while (n-- > 0 )
{
sum += st.peek();
st.pop();
}
// required sum
return sum;
} // Driver program to test above public static void main(String[] args)
{ Node head = null ;
// create linked list 10.6.8.4.12
head = push(head, 12 );
head = push(head, 4 );
head = push(head, 8 );
head = push(head, 6 );
head = push(head, 10 );
int n = 2 ;
System.out.print( "Sum of last " + n+ " nodes = "
+ sumOfLastN_NodesUtil(head, n));
} } // This code is contributed by 29AjayKumar |
# Python3 implementation to find the sum of # last 'n' nodes of the Linked List # Linked List node class Node:
def __init__( self , data):
self .data = data
self . next = None
head = None
n = 0
sum = 0
# function to insert a node at the # beginning of the linked list def push(head_ref, new_data):
global head
# allocate node
new_node = Node( 0 )
# put in the data
new_node.data = new_data
# link the old list to the new node
new_node. next = head_ref
# move the head to point to the new node
head_ref = new_node
head = head_ref
# utility function to find the sum of last 'n' nodes def sumOfLastN_NodesUtil(head, n):
global sum
# if n == 0
if (n < = 0 ):
return 0
st = []
sum = 0
# traverses the list from left to right
while (head ! = None ):
# push the node's data onto the stack 'st'
st.append(head.data)
# move to next node
head = head. next
# pop 'n' nodes from 'st' and
# add them
while (n):
n - = 1
sum + = st[ 0 ]
st.pop( 0 )
# required sum
return sum
# Driver Code head = None
# create linked list 10.6.8.4.12 push(head, 12 )
push(head, 4 )
push(head, 8 )
push(head, 6 )
push(head, 10 )
n = 2
print ( "Sum of last" , n ,
"nodes =" , sumOfLastN_NodesUtil(head, n))
# This code is contributed by shubhamsingh10 |
// C# implementation to find the sum of last // 'n' nodes of the Linked List using System;
using System.Collections.Generic;
class GFG
{ /* A Linked list node */ 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
return head_ref;
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil(Node head, int n)
{ // if n == 0
if (n <= 0)
return 0;
Stack< int > st = new Stack< int >();
int sum = 0;
// traverses the list from left to right
while (head != null )
{
// push the node's data onto the stack 'st'
st.Push(head.data);
// move to next node
head = head.next;
}
// pop 'n' nodes from 'st' and
//.Add them
while (n-- >0)
{
sum += st.Peek();
st.Pop();
}
// required sum
return sum;
} // Driver code public static void Main(String[] args)
{ Node head = null ;
// create linked list 10.6.8.4.12
head = push(head, 12);
head = push(head, 4);
head = push(head, 8);
head = push(head, 6);
head = push(head, 10);
int n = 2;
Console.Write( "Sum of last " + n+ " nodes = "
+ sumOfLastN_NodesUtil(head, n));
} } // This code is contributed by 29AjayKumar |
<script> // Javascript implementation to find the sum of last // 'n' nodes of the Linked List /* A Linked list node */
class Node
{
constructor()
{
let data,next;
}
}
// function to insert a node at the
// beginning of the linked list function push(head_ref,new_data)
{
/* allocate node */
let new_node = new Node();
/* put in the data */
new_node.data = new_data;
/* link the old list to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
return head_ref;
}
// utility function to find the sum of last 'n' nodes
function sumOfLastN_NodesUtil(head,n)
{
// if n == 0
if (n <= 0)
return 0;
let st = [];
let sum = 0;
// traverses the list from left to right
while (head != null )
{
// push the node's data onto the stack 'st'
st.push(head.data);
// move to next node
head = head.next;
}
// pop 'n' nodes from 'st' and
// add them
while (n-- >0)
{
sum += st[st.length-1];
st.pop();
}
// required sum
return sum;
}
// Driver program to test above
let head = null ;
// create linked list 10.6.8.4.12
head = push(head, 12);
head = push(head, 4);
head = push(head, 8);
head = push(head, 6);
head = push(head, 10);
let n = 2;
document.write( "Sum of last " + n+ " nodes = "
+ sumOfLastN_NodesUtil(head, n));
// This code is contributed by patel2127 </script> |
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n), stack size
Method 3 (Reversing the linked list):
Following are the steps:
- Reverse the given linked list.
- Traverse the first n nodes of the reversed linked list.
- While traversing add them.
- Reverse the linked list back to its original order.
- Return the added sum.
Implementation:
// C++ implementation to find the sum of last // 'n' nodes of the Linked List #include <bits/stdc++.h> using namespace std;
/* A Linked list node */ struct Node {
int data;
struct Node* next;
}; // 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 = new Node;
/* put in the data */
new_node->data = new_data;
/* link the old list to the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} 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;
} // utility function to find the sum of last 'n' nodes int sumOfLastN_NodesUtil( struct Node* head, int n)
{ // if n == 0
if (n <= 0)
return 0;
// reverse the linked list
reverseList(&head);
int sum = 0;
struct Node* current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and add them
while (current != NULL && n--) {
// accumulate node's data to 'sum'
sum += current->data;
// move to next node
current = current->next;
}
// reverse back the linked list
reverseList(&head);
// required sum
return sum;
} // Driver program to test above int main()
{ struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << "Sum of last " << n << " nodes = "
<< sumOfLastN_NodesUtil(head, n);
return 0;
} |
// Java implementation to find the sum of last // 'n' nodes of the Linked List import java.util.*;
class GFG
{ /* A Linked list node */ static class Node
{ int data;
Node next;
}; static Node head;
// function to insert a node at the // beginning of the linked list static void 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head=head_ref;
} static 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;
head = head_ref;
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil( int n)
{ // if n == 0
if (n <= 0 )
return 0 ;
// reverse the linked list
reverseList(head);
int sum = 0 ;
Node current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and add them
while (current != null && n-- > 0 )
{
// accumulate node's data to 'sum'
sum += current.data;
// move to next node
current = current.next;
}
// reverse back the linked list
reverseList(head);
// required sum
return sum;
} // Driver code public static void main(String[] args)
{ // create linked list 10.6.8.4.12
push(head, 12 );
push(head, 4 );
push(head, 8 );
push(head, 6 );
push(head, 10 );
int n = 2 ;
System.out.println( "Sum of last " + n + " nodes = "
+ sumOfLastN_NodesUtil(n));
} } /* This code is contributed by PrinciRaj1992 */ |
# Python implementation to find the sum of last # 'n' Nodes of the Linked List ''' A Linked list Node ''' # A Linked list Node class Node:
def __init__( self , x):
self .data = x
self . next = None
head = None
# Function to insert a Node at the # beginning of the linked list def push(head_ref, new_data):
# Allocate Node
new_Node = Node(new_data)
# Put in the data
new_Node.data = new_data
# Link the old list to the new Node
new_Node. next = head_ref
# Move the head to point to the new Node
head_ref = new_Node
head = head_ref
return head
def reverseList():
global head;
current, prev, next = None , None , None ;
current = head;
prev = None ;
while (current ! = None ):
next = current. next ;
current. next = prev;
prev = current;
current = next ;
head = prev;
# utility function to find the sum of last 'n' Nodes def sumOfLastN_NodesUtil(n):
# if n == 0
if (n < = 0 ):
return 0 ;
# reverse the linked list
reverseList();
sum = 0 ;
current = head;
# traverse the 1st 'n' Nodes of the reversed
# linked list and add them
while (current ! = None and n > 0 ):
# accumulate Node's data to 'sum'
sum + = current.data;
# move to next Node
current = current. next ;
n - = 1 ;
# reverse back the linked list
reverseList();
# required sum
return sum ;
# Driver code if __name__ = = '__main__' :
# create linked list 10.6.8.4.12
# create linked list 10.6.8.4.12
head = push(head, 12 )
head = push(head, 4 )
head = push(head, 8 )
head = push(head, 6 )
head = push(head, 10 )
n = 2 ;
print ( "Sum of last " , n , " Nodes = " , sumOfLastN_NodesUtil(n));
# This code contributed by Princi Singh |
// C# implementation to find the sum of last // 'n' nodes of the Linked List using System;
class GFG
{ /* A Linked list node */ public class Node
{ public int data;
public Node next;
}; static Node head;
// function to insert a node at the // beginning of the linked list static void 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head=head_ref;
} static 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;
head = head_ref;
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil( int n)
{ // if n == 0
if (n <= 0)
return 0;
// reverse the linked list
reverseList(head);
int sum = 0;
Node current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and add them
while (current != null && n-- >0)
{
// accumulate node's data to 'sum'
sum += current.data;
// move to next node
current = current.next;
}
// reverse back the linked list
reverseList(head);
// required sum
return sum;
} // Driver code public static void Main(String[] args)
{ // create linked list 10->6->8->4->12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
int n = 2;
Console.WriteLine( "Sum of last " + n + " nodes = "
+ sumOfLastN_NodesUtil(n));
} } // This code is contributed by Rajput-Ji |
<script> // javascript implementation to find the sum of last // 'n' nodes of the Linked List /* A Linked list node */ class Node {
constructor() {
this .data = 0;
this .next = null ;
}
}
var head;
// 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head=head_ref;
} function reverseList(head_ref)
{ var current, prev, next;
current = head_ref;
prev = null ;
while (current != null )
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head_ref = prev;
head = head_ref;
} // utility function to find the sum of last 'n' nodes function sumOfLastN_NodesUtil(n)
{ // if n == 0
if (n <= 0)
return 0;
// reverse the linked list
reverseList(head);
var sum = 0;
var current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and add them
while (current != null && n-- >0)
{
// accumulate node's data to 'sum'
sum += current.data;
// move to next node
current = current.next;
}
// reverse back the linked list
reverseList(head);
// required sum
return sum;
} // Driver code // create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
var n = 2;
document.write( "Sum of last " + n + " nodes = "
+ sumOfLastN_NodesUtil(n));
// This code contributed by umadevi9616 </script> |
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Method 4 (Using the length of linked list):
Following are the steps:
- Calculate the length of the given Linked List. Let it be len.
- First, traverse the (len – n) nodes from the beginning.
- Then traverse the remaining n nodes and while traversing add them.
- Return the added sum.
Implementation:
// C++ implementation to find the sum of last // 'n' nodes of the Linked List #include <bits/stdc++.h> using namespace std;
/* A Linked list node */ struct Node {
int data;
struct Node* next;
}; // 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 = new Node;
/* put in the data */
new_node->data = new_data;
/* link the old list to the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} // utility function to find the sum of last 'n' nodes int sumOfLastN_NodesUtil( struct Node* head, int n)
{ // if n == 0
if (n <= 0)
return 0;
int sum = 0, len = 0;
struct Node* temp = head;
// calculate the length of the linked list
while (temp != NULL) {
len++;
temp = temp->next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != NULL && c--)
// move to next node
temp = temp->next;
// now traverse the last 'n' nodes and add them
while (temp != NULL) {
// accumulate node's data to sum
sum += temp->data;
// move to next node
temp = temp->next;
}
// required sum
return sum;
} // Driver program to test above int main()
{ struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << "Sum of last " << n << " nodes = "
<< sumOfLastN_NodesUtil(head, n);
return 0;
} |
// Java implementation to find the sum of last // 'n' nodes of the Linked List class GFG
{ /* A Linked list node */ static class Node
{ int data;
Node next;
}; static Node head;
// function to insert a node at the // beginning of the linked list static void 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head = head_ref;
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil(Node head, int n)
{ // if n == 0
if (n <= 0 )
return 0 ;
int sum = 0 , len = 0 ;
Node temp = head;
// calculate the length of the linked list
while (temp != null )
{
len++;
temp = temp.next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null &&c-- > 0 )
{
// move to next node
temp = temp.next;
}
// now traverse the last 'n' nodes and add them
while (temp != null )
{
// accumulate node's data to sum
sum += temp.data;
// move to next node
temp = temp.next;
}
// required sum
return sum;
} // Driver code public static void main(String[] args)
{ // create linked list 10.6.8.4.12
push(head, 12 );
push(head, 4 );
push(head, 8 );
push(head, 6 );
push(head, 10 );
int n = 2 ;
System.out.println( "Sum of last " + n + " nodes = "
+ sumOfLastN_NodesUtil(head, n));
} } // This code is contributed by 29AjayKumar |
# Python3 implementation to find the sum # of last 'n' Nodes of the Linked List # A Linked list Node class Node:
def __init__( self , x):
self .data = x
self . next = None
head = None
# Function to insert a Node at the # beginning of the linked list def push(head_ref, new_data):
# Allocate Node
new_Node = Node(new_data)
# Put in the data
new_Node.data = new_data
# Link the old list to the new Node
new_Node. next = head_ref
# Move the head to point to the new Node
head_ref = new_Node
head = head_ref
return head
# Utility function to find the sum of # last 'n' Nodes def sumOfLastN_NodesUtil(head, n):
# If n == 0
if (n < = 0 ):
return 0
sum = 0
len = 0
temp = head
# Calculate the length of the linked list
while (temp ! = None ):
len + = 1
temp = temp. next
# Count of first (len - n) Nodes
c = len - n
temp = head
# Just traverse the 1st 'c' Nodes
while (temp ! = None and c > 0 ):
# Move to next Node
temp = temp. next
c - = 1
# Now traverse the last 'n' Nodes
# and add them
while (temp ! = None ):
# Accumulate Node's data to sum
sum + = temp.data
# Move to next Node
temp = temp. next
# Required sum
return sum
# Driver code if __name__ = = '__main__' :
# Create linked list 10->6->8->4->12
head = push(head, 12 )
head = push(head, 4 )
head = push(head, 8 )
head = push(head, 6 )
head = push(head, 10 )
n = 2
print ( "Sum of last " , n, " Nodes = " ,
sumOfLastN_NodesUtil(head, n))
# This code is contributed by Princi Singh |
// C# implementation to find the sum of last // 'n' nodes of the Linked List using System;
class GFG
{ /* A Linked list node */ public class Node
{ public int data;
public Node next;
}; static Node head;
// function to insert a node at the // beginning of the linked list static void 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 to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
head = head_ref;
} // utility function to find the sum of last 'n' nodes static int sumOfLastN_NodesUtil(Node head, int n)
{ // if n == 0
if (n <= 0)
return 0;
int sum = 0, len = 0;
Node temp = head;
// calculate the length of the linked list
while (temp != null )
{
len++;
temp = temp.next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null &&c-- >0)
{
// move to next node
temp = temp.next;
}
// now traverse the last 'n' nodes and add them
while (temp != null )
{
// accumulate node's data to sum
sum += temp.data;
// move to next node
temp = temp.next;
}
// required sum
return sum;
} // Driver code public static void Main(String[] args)
{ // create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
int n = 2;
Console.WriteLine( "Sum of last " + n + " nodes = "
+ sumOfLastN_NodesUtil(head, n));
} } // This code is contributed by Princi Singh |
<script> // Javascript implementation to // find the sum of last // 'n' nodes of the Linked List /* A Linked list node */
class Node { constructor() {
this .data = 0;
this .next = null ;
}
} var head;
// function to insert a node at the
// beginning of the linked list
function push( head_ref , new_data)
{
/* allocate node */
new_node = new Node();
/* put in the data */
new_node.data = new_data;
/* link the old list to
the new node */
new_node.next = head_ref;
/* move the head to point
to the new node */
head_ref = new_node;
head = head_ref;
}
// utility function to find
// the sum of last 'n' nodes
function sumOfLastN_NodesUtil( head , n)
{
// if n == 0
if (n <= 0)
return 0;
var sum = 0, len = 0;
temp = head;
// calculate the length of the linked list
while (temp != null ) {
len++;
temp = temp.next;
}
// count of first (len - n) nodes
var c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null && c-- > 0) {
// move to next node
temp = temp.next;
}
// now traverse the last 'n'
// nodes and add them
while (temp != null ) {
// accumulate node's data to sum
sum += temp.data;
// move to next node
temp = temp.next;
}
// required sum
return sum;
}
// Driver code
// create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
var n = 2;
document.write( "Sum of last " + n
+ " nodes = " + sumOfLastN_NodesUtil(head, n));
// This code contributed by umadevi9616 </script> |
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Method 5 (Use of two pointers requires single traversal):
Maintain two pointers – reference pointer and main pointer. Initialize both reference and main pointers to head. First, move reference pointer to n nodes from head and while traversing accumulate node’s data to some variable, say sum.
Now move both pointers simultaneously until the reference pointer reaches the end of the list and while traversing accumulate all node’s data to sum pointed by the reference pointer and accumulate all node’s data to some variable, say, temp, pointed by the main pointer. Now, (sum – temp) is the required sum of the last n nodes.
Implementation:
// C++ implementation to find the sum of last // 'n' nodes of the Linked List #include <bits/stdc++.h> using namespace std;
/* A Linked list node */ struct Node {
int data;
struct Node* next;
}; // 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 = new Node;
/* put in the data */
new_node->data = new_data;
/* link the old list to the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} // utility function to find the sum of last 'n' nodes int sumOfLastN_NodesUtil( struct Node* head, int n)
{ // if n == 0
if (n <= 0)
return 0;
int sum = 0, temp = 0;
struct Node* ref_ptr, *main_ptr;
ref_ptr = main_ptr = head;
// traverse 1st 'n' nodes through 'ref_ptr' and
// accumulate all node's data to 'sum'
while (ref_ptr != NULL && n--) {
sum += ref_ptr->data;
// move to next node
ref_ptr = ref_ptr->next;
}
// traverse to the end of the linked list
while (ref_ptr != NULL) {
// accumulate all node's data to 'temp' pointed
// by the 'main_ptr'
temp += main_ptr->data;
// accumulate all node's data to 'sum' pointed by
// the 'ref_ptr'
sum += ref_ptr->data;
// move both the pointers to their respective
// next nodes
main_ptr = main_ptr->next;
ref_ptr = ref_ptr->next;
}
// required sum
return (sum - temp);
} // Driver program to test above int main()
{ struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << "Sum of last " << n << " nodes = "
<< sumOfLastN_NodesUtil(head, n);
return 0;
} |
// Java implementation to find the sum of last // 'n' nodes of the Linked List class GfG
{ // Defining structure
static class Node
{
int data;
Node next;
}
static Node head;
static void printList(Node start)
{
Node temp = start;
while (temp != null )
{
System.out.print(temp.data + " " );
temp = temp.next;
}
System.out.println();
}
// Push function
static void push(Node start, int info)
{
// Allocating node
Node node = new Node();
// Info into node
node.data = info;
// Next of new node to head
node.next = start;
// head points to new node
head = node;
}
private static int sumOfLastN_NodesUtil(Node head, int n)
{
// if n == 0
if (n <= 0 )
return 0 ;
int sum = 0 , temp = 0 ;
Node ref_ptr, main_ptr;
ref_ptr = main_ptr = head;
// traverse 1st 'n' nodes through 'ref_ptr' and
// accumulate all node's data to 'sum'
while (ref_ptr != null && (n--) > 0 )
{
sum += ref_ptr.data;
// move to next node
ref_ptr = ref_ptr.next;
}
// traverse to the end of the linked list
while (ref_ptr != null )
{
// accumulate all node's data to 'temp' pointed
// by the 'main_ptr'
temp += main_ptr.data;
// accumulate all node's data to 'sum' pointed by
// the 'ref_ptr'
sum += ref_ptr.data;
// move both the pointers to their respective
// next nodes
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
// required sum
return (sum - temp);
}
// Driver code
public static void main(String[] args)
{
head = null ;
// Adding elements to Linked List
push(head, 12 );
push(head, 4 );
push(head, 8 );
push(head, 6 );
push(head, 10 );
printList(head);
int n = 2 ;
System.out.println( "Sum of last " + n +
" nodes = " + sumOfLastN_NodesUtil(head, n));
}
} // This code is contributed by shubham96301 |
# Python3 implementation to find the sum of last # 'n' nodes of the Linked List # include class Node:
def __init__( self , x):
self .data = x
self . next = None
# function to insert a node at the # beginning of the linked list def push(head_ref,new_data):
# /* allocate node */
new_node = Node(new_data)
#/* link the old list to the new node */
new_node. next = head_ref
#/* move the head to point to the new node */
head_ref = new_node
return head_ref
# utility function to find the sum of last 'n' nodes def sumOfLastN_NodesUtil(head, n):
# if n == 0
if (n < = 0 ):
return 0
sum = 0
temp = 0
ref_ptr = None
main_ptr = None
ref_ptr = main_ptr = head
# traverse 1st 'n' nodes through 'ref_ptr' and
# accumulate all node's data to 'sum'
while (ref_ptr ! = None and n):
sum + = ref_ptr.data
# move to next node
ref_ptr = ref_ptr. next
n - = 1
# traverse to the end of the linked list
while (ref_ptr ! = None ):
# accumulate all node's data to 'temp' pointed
# by the 'main_ptr'
temp + = main_ptr.data
# accumulate all node's data to 'sum' pointed by
# the 'ref_ptr'
sum + = ref_ptr.data
# move both the pointers to their respective
# next nodes
main_ptr = main_ptr. next
ref_ptr = ref_ptr. next
# required sum
return ( sum - temp)
# Driver program to test above if __name__ = = '__main__' :
head = None
# create linked list 10.6.8.4.12
head = push(head, 12 )
head = push(head, 4 )
head = push(head, 8 )
head = push(head, 6 )
head = push(head, 10 )
n = 2
print ( "Sum of last " ,n, " nodes = " ,sumOfLastN_NodesUtil(head, n))
# This code is contributed by mohit kumar 29 |
// C# implementation to find the sum of last // 'n' nodes of the Linked List using System;
class GfG
{ // Defining structure
public class Node
{
public int data;
public Node next;
}
static Node head;
static void printList(Node start)
{
Node temp = start;
while (temp != null )
{
Console.Write(temp.data + " " );
temp = temp.next;
}
Console.WriteLine();
}
// Push function
static void push(Node start, int info)
{
// Allocating node
Node node = new Node();
// Info into node
node.data = info;
// Next of new node to head
node.next = start;
// head points to new node
head = node;
}
private static int sumOfLastN_NodesUtil(Node head, int n)
{
// if n == 0
if (n <= 0)
return 0;
int sum = 0, temp = 0;
Node ref_ptr, main_ptr;
ref_ptr = main_ptr = head;
// traverse 1st 'n' nodes through 'ref_ptr' and
// accumulate all node's data to 'sum'
while (ref_ptr != null && (n--) > 0)
{
sum += ref_ptr.data;
// move to next node
ref_ptr = ref_ptr.next;
}
// traverse to the end of the linked list
while (ref_ptr != null )
{
// accumulate all node's data to 'temp' pointed
// by the 'main_ptr'
temp += main_ptr.data;
// accumulate all node's data to 'sum' pointed by
// the 'ref_ptr'
sum += ref_ptr.data;
// move both the pointers to their respective
// next nodes
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
// required sum
return (sum - temp);
}
// Driver code
public static void Main(String[] args)
{
head = null ;
// Adding elements to Linked List
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
printList(head);
int n = 2;
Console.WriteLine( "Sum of last " + n +
" nodes = " + sumOfLastN_NodesUtil(head, n));
}
} // This code contributed by Rajput-Ji |
<script> // Javascript implementation to find the sum of last // 'n' nodes of the Linked List // Defining structure
class Node
{
constructor()
{
let node,next;
}
}
let head;
function printList(start)
{
let temp = start;
while (temp != null )
{
document.write(temp.data + " " );
temp = temp.next;
}
document.write( "<br>" );
}
// Push function
function push(start,info)
{
// Allocating node
let node = new Node();
// Info into node
node.data = info;
// Next of new node to head
node.next = start;
// head points to new node
head = node;
}
function sumOfLastN_NodesUtil(head,n)
{
// if n == 0
if (n <= 0)
return 0;
let sum = 0, temp = 0;
let ref_ptr, main_ptr;
ref_ptr = main_ptr = head;
// traverse 1st 'n' nodes through 'ref_ptr' and
// accumulate all node's data to 'sum'
while (ref_ptr != null && (n--) > 0)
{
sum += ref_ptr.data;
// move to next node
ref_ptr = ref_ptr.next;
}
// traverse to the end of the linked list
while (ref_ptr != null )
{
// accumulate all node's data to 'temp' pointed
// by the 'main_ptr'
temp += main_ptr.data;
// accumulate all node's data to 'sum' pointed by
// the 'ref_ptr'
sum += ref_ptr.data;
// move both the pointers to their respective
// next nodes
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
// required sum
return (sum - temp);
}
// Driver code
head = null ;
// Adding elements to Linked List
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
let n = 2;
document.write( "Sum of last " + n +
" nodes = " + sumOfLastN_NodesUtil(head, n));
// This code is contributed by avanitrachhadiya2155 </script> |
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)