LinkedList
Question 1 
Consider the C code fragment given below.
typedef struct node { int data; node* next; } node; void join(node* m, node* n) { node* p = n; while(p→next !=NULL) { p = p→next; } p→next=m; }
Assuming that m and n point to valid NULLterminated linked lists, invocation of join will
append list m to the end of list n for all inputs.  
either cause a null pointer dereference or append list m to the end of list n.  
cause a null pointer dereference for all inputs.  
append list n to the end of list m for all inputs. 
So have to consider both the cases.
⇾ 1. Lists are not null, invocation of JOIN will append list m to the end of list n.
m = 1, 2, 3
n = 4, 5, 6
After join, it becomes 4, 5, 6, 1, 2, 3, NULL.
⇾ 2. If lists are null, if the list n is empty, and itself NULL, then joining & referencing would obviously create a null pointer issue.
Hence, it may either cause a NULL pointer dereference or appends the list m to the end of list n.
Question 2 
N items are stored in a sorted doubly linked list. For a delete operation, a pointer is provided to the record to be deleted. For a decreasekey operation, a pointer is provided to the record on which the operation is to be performed.
An algorithm performs the following operations on the list in this order: Θ(N) delete,O(logN) insert, O(logN) ﬁnd, and Θ(N) decreasekey. What is the time complexity of all these operations put together?
O(log^{2} N)  
O(N)  
O(N^{2})  
Θ(N^{2} logN) 
→ Delete needs O(1) time
→ Insert takes O(N) time
→ Find takes O(N) time
→ Decrease by takes O(N) time
Now number of each operation performed is given, so finally total complexity,
→ Delete = O(1) × O(N) = O(N)
→ Find = O(N) × O(log N) = O(N log N)
→ Insert = O(N) × O(log N) = O(N log N)
→ Decrease key = O(N) × O(N) = O(N^{2})
So, overall time complexity is, O(N^{2}).
Question 3 
The following C function takes a simplylinked list as input argument. It modifies the list by moving the last element to the front of the list and returns the modified list. Some part of the code is left blank.
typedef struct node { int value; struct node *next; }Node; Node *move_to_front(Node *head) { Node *p, *q; if ((head == NULL:  (head>next == NULL)) return head; q = NULL; p = head; while (p> next !=NULL) { q = p; p = p>next; } _______________________________ return head; }
Choose the correct alternative to replace the blank line.
q = NULL; p→next = head; head = p;  
q→next = NULL; head = p; p→next = head;  
head = p; p→next = q; q→next = NULL;  
q→next = NULL; p→next = head; head = p; 
The function modifies the list by moving the last element to the front of the list.
Let the list be 1 → 2 → 3 → 4 & the modified list must be 4 → 1 → 2 → 3.
Algorithm:
Traverse the list till last node. Use two pointers. One to store the address of last node & other for the address of second last node.
After completion of loop. Do these.
(i) Make second last as last
(ii) Set next of last as head
(iii) Make last as head
while (p → !=NULL)
{
q = p;
p = p → next;
}
― p is travelling till the end of list and assigning q to whatever p had visited & p takes next new node, so travels through the entire list.
Now the list looks like
According to the Algorithm new lines must be
q → next = NULL; p → next = head; head = p
Question 4 
The following C function takes a singlelinked list of integers as a parameter and rearranges the elements of the list. The function is called with the list containing the integers 1, 2, 3, 4, 5, 6, 7 in the given order. What will be the contents of the list after the function completes execution?
struct node { int value; struct node *next; }; void rearrange(struct node *list) { struct node *p, * q; int temp; if ((!list)  !list>next)return; p = list; q = list>next; while(q) { temp = p>value;p>value = q>value; q>value = temp;p = q>next; q = p?p>next:0; } }
1,2,3,4,5,6,7  
2,1,4,3,6,5,7  
1,3,2,5,4,7,6  
2,3,4,5,6,7,1 
After 1^{st} Iteration: 2,1,3,4,5,6,7
2^{nd} Iteration: 2,1,4,3,5,6,7
3^{rd} Iteration: 2,1,4,3,6,5,7
After each exchange is done, it starts from unchanged elements due to p=q⟶next;
‘p’ pointing to Null & q pointing to 7.
Hence 2,1,4,3,6,5,7.
Question 5 
A circularly linked list is used to represent a Queue. A single variable p is used to access the Queue. To which node should p point such that both the operations enQueue and deQueue can be performed in constant time?
rear node  
front node  
not possible with a single pointer  
node next to front 
Question 6 
Suppose each set is represented as a linked list with elements in arbitrary order. Which of the operations among union, intersection, membership, cardinality will be the slowest?
union only  
intersection, membership
 
membership, cardinality  
union, intersection 
Let no. of elements in list 2 be n_{2}.
Union:
To union two lists, for each element in one list we will search in other list, to avoid duplicates. So, time complexity will be O(n_{1}×n_{2}).
Intersection:
To take intersection of two lists, for each element in one list we will search in other list if it is present or not. So time complexity will be O(n_{1} × n_{2}).
Membership:
In this we search if particular element is present in the list or not. So time complexity will be O(n_{1} + n_{2}).
Cardinality:
In this we find the size of set or list. So to find size of list we have to traverse each list once. So time complexity will be O(n_{1}+n_{2}).
Hence, Union and Intersection will be slowest.
Question 7 
Consider the function f defined below.
struct item { int data; struct item * next; }; int f(struct item *p) { return ((p == NULL)  (p>next == NULL)  ((P>data <= p>next>data) && f(p>next))); }
For a given linked list p, the function f returns 1 if and only if
the list is empty or has exactly one element  
the elements in the list are sorted in nondecreasing order of data value
 
the elements in the list are sorted in nonincreasing order of data value
 
not all elements in the list have the same data value 
Question 8 
log n  
n/2  
(log_{2})^{n}  1  
n 
Question 9 
Let p be a pointer as shown in the figure in a single linked list.
What do the following assignment statements achieve?
 q: = p → next
p → next:= q → next
q → next:=(q → next) → next
(p → next) → next:= q
(b) Compute the postfix equivalent of the following Infix expression
3 * log(x+1)  a/2
Theory Explanation. 
Question 10 
The concatenation of two lists is to be performed in O(1) time. Which of the following implementations of a list should be used?
Singly linked list  
Doubly linked list  
Circular doubly linked list  
Array implementation of list 
Question 11 
Linked lists are not suitable data structures of which one of the following problems?
Insertion sort  
Binary search  
Radix sort  
Polynomial manipulation

Question 12 
What is the worst case time complexity of inserting n elements into an empty linked list, if the linked list needs to be maintained in sorted order?
θ(n log n)  
θ(n^{2})  
θ(1)  
θ(n) 
Total number of elements inserted into an empty linked list is O(n). So, it will take O(n) time in the worst case.
After inserting elements into an empty linked list we have to perform sorting operation.
To get minimum time complexity to perform sorting order is merge sort. It will give O(nlogn) time complexity only.
Let head be the first node of the linked list to be sorted and head Reference be the pointer to head.
The head in MergeSort as the below implementation changes next links to sort the linked lists (not data at the nodes), so head node has to be changed if the data at the original head is not the smallest value in the linked list.
Note: There are other sorting methods also will give decent time complexity but quicksort will give O(n^{2}) and heap sort will not be suitable to apply.
Question 13 
Consider a singly linked list having n nodes. The data items d_{1}, d_{2}, ...d_{n} are stored in these n nodes. Let X be a pointer to the jth node (1≤j≤n) in which d_{j} is stored. A new data item d stored in node with address Y is t be inserted. Give an algorithm to insert d into the list to obtain a list having items d_{1}, d_{2}, ..., d_{j1}, d_{j}, ..., d_{n} in the order without using the header.
Theory Explanation. 