Python Program To Find The Sum Of Last N Nodes Of The Given Linked List
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
class Node:
def __init__(self, data):
self.data = data
self.next = None
head = None
n = 0
sum = 0
def push(head_ref, new_data):
global head
new_node = Node(0)
new_node.data = new_data
new_node.next = head_ref
head_ref = new_node
head = head_ref
def sumOfLastN_Nodes(head):
global sum
global n
if (head == None):
return
sumOfLastN_Nodes(head.next)
if (n > 0) :
sum = sum + head.data
n = n - 1
def sumOfLastN_NodesUtil(head, n):
global sum
if (n <= 0):
return 0
sum = 0
sumOfLastN_Nodes(head)
return sum
head = None
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))
|
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
class Node:
def __init__(self, data):
self.data = data
self.next = None
head = None
n = 0
sum = 0
def push(head_ref, new_data):
global head
new_node = Node(0)
new_node.data = new_data
new_node.next = head_ref
head_ref = new_node
head = head_ref
def sumOfLastN_NodesUtil(head, n):
global sum
if (n <= 0):
return 0
st = []
sum = 0
while (head != None):
st.append(head.data)
head = head.next
while (n):
n -= 1
sum += st[0]
st.pop(0)
return sum
head = None
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))
|
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:
- 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.
Python3
class Node:
def __init__(self, x):
self.data = x
self.next = None
head = None
def push(head_ref, new_data):
new_Node = Node(new_data)
new_Node.data = new_data
new_Node.next = head_ref
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;
def sumOfLastN_NodesUtil(n):
if (n <= 0):
return 0;
reverseList();
sum = 0;
current = head;
while (current != None and n > 0):
sum += current.data;
current = current.next;
n -= 1;
reverseList();
return sum;
if __name__ == '__main__':
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));
|
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:
- 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.
Python3
class Node:
def __init__(self, x):
self.data = x
self.next = None
head = None
def push(head_ref, new_data):
new_Node = Node(new_data)
new_Node.data = new_data
new_Node.next = head_ref
head_ref = new_Node
head = head_ref
return head
def sumOfLastN_NodesUtil(head, n):
if (n <= 0):
return 0
sum = 0
len = 0
temp = head
while (temp != None):
len += 1
temp = temp.next
c = len - n
temp = head
while (temp != None and c > 0):
temp = temp.next
c -= 1
while (temp != None):
sum += temp.data
temp = temp.next
return sum
if __name__ == '__main__':
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))
|
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
class Node:
def __init__(self, x):
self.data = x
self.next = None
def push(head_ref,new_data):
new_node = Node(new_data)
new_node.next = head_ref
head_ref = new_node
return head_ref
def sumOfLastN_NodesUtil(head, n):
if (n <= 0):
return 0
sum = 0
temp = 0
ref_ptr = None
main_ptr = None
ref_ptr = main_ptr = head
while (ref_ptr != None and n):
sum += ref_ptr.data
ref_ptr = ref_ptr.next
n -= 1
while (ref_ptr != None):
temp += main_ptr.data
sum += ref_ptr.data
main_ptr = main_ptr.next
ref_ptr = ref_ptr.next
return (sum - temp)
if __name__ == '__main__':
head = None
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))
|
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!
Similar Reads
Python Program For Inserting Node In The Middle Of The Linked List
Given a linked list containing n nodes. The problem is to insert a new node with data x at the middle of the list. If n is even, then insert the new node after the (n/2)th node, else insert the new node after the (n+1)/2th node. Examples: Input : list: 1->2->4->5 x = 3 Output : 1->2->
4 min read
Python Program For Finding The Length Of Loop In Linked List
Write a function detectAndCountLoop() that checks whether a given Linked List contains loop and if loop is present then returns count of nodes in loop. For example, the loop is present in below-linked list and length of the loop is 4. If the loop is not present, then the function should return 0. Re
4 min read
Python Program To Delete N Nodes After M Nodes Of A Linked List
Given a linked list and two integers M and N. Traverse the linked list such that you retain M nodes then delete next N nodes, continue the same till end of the linked list.Difficulty Level: Rookie Examples: Input: M = 2, N = 2 Linked List: 1->2->3->4->5->6->7->8 Output: Linked L
3 min read
Python Program To Check Whether The Length Of Given Linked List Is Even Or Odd
Given a linked list, the task is to make a function which checks whether the length of the linked list is even or odd. Examples: Input : 1->2->3->4->NULL Output : Even Input : 1->2->3->4->5->NULL Output : OddRecommended: Please solve it on "PRACTICE" first, before moving o
4 min read
Python Program For Removing Every K-th Node Of The Linked List
Given a singly linked list, Your task is to remove every K-th node of the linked list. Assume that K is always less than or equal to length of Linked List.Examples : Input: 1->2->3->4->5->6->7->8 k = 3 Output: 1->2->4->5->7->8 As 3 is the k-th node after its deletion list would be 1->2->4->5->6->7->
3 min read
Python3 Program to Rotate Doubly linked list by N nodes
Given a doubly linked list, rotate the linked list counter-clockwise by N nodes. Here N is a given positive integer and is smaller than the count of nodes in linked list.  N = 2Rotated List:  Examples:  Input : a b c d e N = 2Output : c d e a b Input : a b c d e f g h N = 4Output : e f g h a b c
4 min read
Python Program To Delete Alternate Nodes Of A Linked List
Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3. Recomm
3 min read
Python Program For Finding Intersection Of Two Sorted Linked Lists
Given two lists sorted in increasing order, create and return a new list representing the intersection of the two lists. The new list should be made with its own memory — the original lists should not be changed. Example: Input: First linked list: 1->2->3->4->6 Second linked list be 2->4->6->8, Ou
4 min read
Python Program For Finding The Middle Element Of A Given Linked List
Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3. If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list
4 min read
Python Program To Delete Middle Of Linked List
Given a singly linked list, delete the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the linked list should be modified to 1->2->4->5 If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if g
4 min read