Write a function ListNode *removeFirst(ListNode *ptr) that is passed a linked list as parameter, and returns the tail of the list: that is, it removes the first node and returns what is left. The function should deal locate the storage of the removed node. The function returns NULL if the list passed to it is empty.

(First, check if the list is empty, i.e., if the head pointer is NULL. In this case, the function should return NULL.)

(Create a ListNode pointer and assign the head pointer value to it. It will help us later to free the memory occupied by the first node of the list.)

(Update the head pointer of the list to point at the second node in the list. The second node now becomes the head of the new list.)

(Now, using, the pointer created in step 2, deallocate the memory of the first node using the 'delete' operator. This will remove the first node from the list and free its memory.)

(Finally, return the updated list (which is now the tail of the original list) by returning the head pointer.) The final function will look like this: ```cpp struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode *removeFirst(ListNode *ptr) { // Step 1: Handle the empty list case if (ptr == NULL) { return NULL; } // Step 2: Store the head of the list ListNode *temp = ptr; // Step 3: Update the head pointer to point at the second node ptr = ptr->next; // Step 4: Deallocate the memory of the first node delete temp; // Step 5: Return the new list return ptr; } ``` This function will effectively remove the first node of the given singly linked list and return the modified list.