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

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.

Javascript

<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>

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.

Javascript

<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>

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.

Javascript

<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>

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.

Javascript

<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>

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.

Javascript

<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>

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!