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.
C++
// 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 listvoid 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 listvoid 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' nodesint 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 codeint 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;} |
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.
C++
// 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 listvoid 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' nodesint 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 codeint 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;} |
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.
C++
// 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 listvoid 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' nodesint 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 codeint 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;} |
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.
C++
// 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 listvoid 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' nodesint 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 codeint 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;} |
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.
C++
// 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 listvoid 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' nodesint 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 codeint 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;} |
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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