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Python Program To Find The Sum Of Last N Nodes Of The Given Linked List

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  • Last Updated : 18 May, 2022

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.

Python3




# 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

Output:  

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.

Python3




# 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 

Output:  

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:  

  1. Reverse the given linked list.
  2. Traverse the first n nodes of the reversed linked list.
  3. While traversing add them.
  4. Reverse the linked list back to its original order.
  5. Return the added sum.

Python3




# 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 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
    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 is contributed by Princi Singh

Output: 

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: 

  1. Calculate the length of the given Linked List. Let it be len.
  2. First, traverse the (len – n) nodes from the beginning.
  3. Then traverse the remaining n nodes and while traversing add them.
  4. Return the added sum.

Python3




# 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 

Output:  

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.

Python3




# Python3 implementation to find the sum 
# of last 'n' nodes of the Linked List
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 code
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

 
 

Output: 

 

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)

Please refer complete article on Find the sum of last n nodes of the given Linked List for more details!
 


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