def insertNodeAtPosition(llist, data, position): # Write your code here if llist is None: llist = SinglyLinkedList() llist.data = data else: current = llist for i in range(1, position): current = current.next

    new_next = SinglyLinkedList()
    new_next.data = data
    new_next.next = current.next
    current.next = new_next
return llist

Haskell boilerplate contains a mistake which makes this problem unsolvable in Haskell

"This operation requires traversing the linked list up to the node just before the desired position. It’s important to handle edge cases, such as inserting at the head (position 0) or beyond the current length of the list. Proper pointer management is critical to ensure no nodes are lost and the list remains intact." Gold365.site

def insertNodeAtPosition(head, data, position): # Write your code here newnode=SinglyLinkedListNode(data) current=head if head is not None: count=1 while current.next: if count==position: newnode.next=current.next current.next=newnode count+=1 current=current.next return head

My Java solution with linear time complexity and constant space complexity:

public static SinglyLinkedListNode insertNodeAtPosition(SinglyLinkedListNode llist, int data, int position) {
    SinglyLinkedListNode newNode = new SinglyLinkedListNode(data);
    
    // Insertion at head
    if (position == 0) {
        newNode.next = llist;
        return newNode;
    }
    
    SinglyLinkedListNode curr = llist;
    
    // Traverse to the node before the insertion point
    for (int i = 0; i < position - 1; i++) {
        if (curr != null) {
            curr = curr.next;
        }
    }

    // Insert the new node
    newNode.next = curr.next;
    curr.next = newNode;

    return llist;
}