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DSA Viva Stack Tree Graph Test Practice
- Given the following sequence of letters and asterisks:
(a) Consider the stack data structure, supporting two operations push and pop, as discussed in class. Suppose that for the above sequence, each letter (such as E) corresponds to a push of that letter onto the stack and each asterisk (*) corresponds to a pop operation on the stack. Show the sequence of values returned by the pop operations.
(b) Consider the queue data structure, supporting two operations insert and remove, as discussed in class. Suppose that for the above sequence, each letter (such as E) corresponds to an insert of that letter into the queue and each asterisk (*) corresponds to a remove operation on the queue. Show the sequence of values returned by the remove operations.
Answer : a). Answer for stack Output: SYEUQTSAONIE.
b). Answer for Queue Output: EASY QUESTION
- Suppose you were asked to write a method that will take two sorted stacks A and B (min on top) and create one stack that is sorted (min on top). You are allowed to use only the stack operations such as pop, push, size and top. No other data structure such as arrays are not allowed. You are allowed to use stacks. Note that elements on the stack can be compared using compareTo.
Public Stack mergeSortedStacks(Stack A, Stack B) { …... } Answer :
- Evaluate the postfix expression: 10 5 + 60 6 / * 8 -
your travel by more than 10%.) Describe an efficient algorithm that would
determine an optimal s to t path given your preference for stopping at u along
the way if not too inconvenient. It should either return the shortest path from s
to t, or the shortest path from s to t containing u.
Solution: Run Dijkstra’s algorithm twice, once from s and once from u. The
shortest path from s to t containing u is composed of the shortest path from s
to u and the shortest path from u to t. Compare the length of this path with the
length of the path from s to t, and choose the one to return based on their
lengths.
Question 7 :- Construct a Binary Tree From Inorder and Preorder Traversal
Inorder Traversal: {4,2,17,5,8,3,6}
Preorder Traversal :{1,2,4,3,5,7,8,6}
Answer :
Question 8 :- Write a recursive solution to Check whether a binary tree is a
complete tree or not.
Hint:(A complete binary tree is a binary tree whose all levels except the last level are
completely filled and all the leaves in the last level are all to the left side.)
Solution:
Question 9:-List the nodes of the tree below in preorder, postorder, and breadth-first
order
Answer :-
Preorder: L,K,A,J,B,C,I,H,E,D,F,G
Postorder: A,B,C,J,K,I,D,E,F,G,H,L.
Breadth-first: L,K,I,H,A,J,E,F,G,B,C,D
Question 10 :-In the binary search tree below, carry out the following operations in
sequence: Add 5, add 17, delete 23, delete 9.
Answer : There is more than one way to do the deletes, so the final Answer is not unique,
but here is one:
Answer : Graph 1: Acyclic. Topological sort: A, B, D, C.
Graph 2: Cyclic. Cycle: F,H,G,F
Graph 3: Acyclic. There are several different topological sorts. I,J,K,L,M is one.
Question 13 :-Describe an efficient method to merge two balanced binary
search trees with n elements each into a balanced BST. Give its running time.
Answer :-We can start by doing an in-order walk of both trees concurrently. At
each step, we compare the two tree elements and add the smaller one into a
list, L, before calling that element’s successor. When we finish walking both
trees, L will contain a sorted list of elements from both trees. This takes O(n +
n) = O(n) total time. Now, from the sorted list, we want to create a balanced
binary tree, which is the same problem as described in problem set 2. We can
do this by setting the root as the middle element of the list, and letting the first
half of the list be its left subtree and the second half be its right subtree
(recursively creating the balanced subtrees as well). This also takes O(n + n)
= O(n) time. The total time for this algorithm is therefore O(n).
Question 14 :-Ben Bitdiddle recently learned about heaps and binary search
trees in his algorithms class. He was so excited about getting to know about
them, he thought of an interesting way to combine them to create a new
hybrid binary tree called a treap. The treap T has a tuple value stored at each
node (x, y) where he calls x the key, and y the priority of the node. The keys
follow the BST property (maintain the BST invariant) while the priorities
maintain the min-heap property. An example treap is shown below:
Answer :-Insert the new node according to the standard BST INSERT procedure. This
preserves the BST property, but may result in a violation of the heap property. Now,
we need to fix the heap property by moving the new node up in the treap until it
reaches the correct place. Do this by repeatedly rotating on the node’s parent, moving
the node up one level while preserving the BST invariant. The BST insertion requires
O(h) time. Each rotation takes O(1), and O(h) rotations are required, so the total
insertion time is O(h).
Question 15 :-Demonstrate what happens when we insert the keys 5,28,19,15,20,33,12,17,10into
a hash table with collisions resolved by chaining. Let the table have 9 slots, and let the hash function be h ( k )= k mod9.
Solution: Let us number our slots 0,1,…,8.
Then our resulting hash table will look like following:
Question – 16 ts) Say the following tree was obtained by inserting the
element 42 into an AVL tree. The tree no longer satisfies the AVL invariant,
but the invariant can be reestablished by performing two rotate operations.
Show the resulting tree after this is done
rank to an element and UNION breaks ties by putting the tree whose root has
the larger label as the parent of the other.
for j= 1 to 10
MAKESET(j)
end for
UNION(1,2) ; UNION(1,3); UNION(4,5); UNION(4,6); UNION(1,6); UNION(7,8);
UNION(9,10); UNION(7,10); UNION(3,8); FIND(4); FIND(3);
Give the tree from executing the above steps using union-by-rank with no path-compression Be sure to label the nodes in the final tree including their final ranks.
Give the tree from executing the above steps using union-by-rank with path-compression Be sure to label the nodes in the final tree including their final ranks.
Answer :
Question :19 ) Draw the Huffman tree corresponding to the encoding table
below. char freq encoding
H 3 0110
I? 00
S 15 11
which one or more of the following are possible values for the frequency of the character I. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Answer :
B
F
L
M
6 ≤ freq(I) ≤ 15. Since I is not touched until after merging L with {B, F, H}, freq(I) ≥ freq(L) = 5 and freq(I) ≥ freq({B, F, H}) = 2 + 1 + 3 = 6. • Since I is merged with {B, F, H, L} instead of M or S, freq(I) ≤ freq(M) = 15 and freq(I) ≤ freq(S) = 15
Question 20:-What is the max-heap resulting from performing on the node storing 6?
Answer : will swap 6 with its larger child until 6 reaches a position where it satisfies the max-heap property. The correct heap is:
A common mistake on this problem was to use something resembling , recursively moving the larger child up to replace the deleted parent. This does not result in a valid BST.
Question 22:
- The number of distinct minimum spanning trees for the weighted graph below is
Answer :Highlighted (ingreen) are the edges picked to make a MST. In the right side of MST, we could either pick edge ‘a’ or ‘b’. In the left side, we could either pick ‘c’ or ‘d’ or ‘e’ in MST. There are 2 options for one edge to be picked and 3 options for another edge to be picked. Therefore, total 2*3 possible MSTs.
Question 23:-What is the number of binary search trees with 20 nodes with elements 1, 2, 3,…..
such that the root of tree is 12 and the root of left subtree is 7?
Answer :Number of nodes in left subtree = 11 {1, 2, 3, 4….11} Number of nodes in the right subtree = 8 {13, 14, ….20} Since for the left subtree root is 7 Number of elements in the left part of left subtree = 6 {1, 2, 3..6} Number of elements in the right part of left subtree = 4 {8, 9, 10, 11} We know number of Binary Search trees with n nodes = (C(2n,n)/n+1) Number of BST with 6 nodes = (C(12,6)/7) = 132 Number of BST with 4 nodes = (C(8,4)/5) = 14 Number of BST with 8 nodes = (C(16,8)/9) =1430 Total number of BST = 2642640
Question 24:-Which among the following is max-heap ?Explain the reason why it is and Why it
is not?
Answer :A binary tree is max-heap if it is a complete binary tree (A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible) and it follows the max-heap property (value of each parent is greater than or equal to the values of its children). A) is not a max-heap because it is not a complete binary tree
B) is a max-heap because it is a complete binary tree and follows max-heap property.
C) is not a max-heap because 8 is a child of 5 in this tree, so violates the max-heap property.
D) is not a max-heap because 8 is a child of 5 in this tree, so violates the max-heap property.
There are many other nodes in this tree which violate max-heap property in this tree.
Question 25:-Consider a hash table with 9 slots. The hash function is h(k) = k mod 9. The
collisions are resolved by chaining. The following 9 keys are inserted in the order: 5, 28, 19, 15, 20, 33, 12, 17, 10. What is the maximum, minimum, and average chain lengths in the hash table.
Answer :-Following are values of hash function for all keys 5 ----- > 28 -----> 19 ----- >1{chained with 28} 15 ------ > 20 ----- > 33 ----- >6{chained with 15} 12 ----- > 17 -------- >
CE No
CF Yes
FE No
DF No
BF No
Resultant MST:
Note: As there are edges with equal weights, there may be multiple correct Answer s
possible
Question 27:-
a) Insert 84 in the given below AVL tree. Re-draw the tree. b) Delete 66 from the resultant tree in (a). Re-draw the tree.
Solution:-
or (for deletion using predecessor)
Question 28 Perform topological sort for below DAG. Provide at least two orderings by
showing the discovery and finishing times of visit for each node.
Solution Ordering 1: G A B D C E F
A B C D E F G
D(t) 1 2 3 9 4 5 13
F(t) 12 11 8 10 7 6 14
Ordering 2: A B C G D E F
A B C D E F G
D(t) 9 10 11 2 3 4 1
F(t) 14 13 12 7 6 5 8
(Optional) Other possible ordering: GABCDEF)
Question 29:-
Both Correct 3 marks
Mentioned any wrong one with D and F 2.5 marks
c) What single edge could be removed from the graph such that Dijkstra’s algorithm would happen to compute correct Answer s for all vertices in the remaining graph.
Solution:
c) The edge E to G
Question 30:-Delhi Traffic Police has made all roads one way. The CM claims that there is still
a legal way to drive from one point of the city to another. The opposition is not convinced and argues the CM’s claim to be false. Formulate this problem as a graph problem and give a linear time algorithm to check the validity of the claim. Explain the time complexity of your algorithm.
Answer :-This problem could be represented as a directed graph with the vertices as the points in the city and edge from point A to B representing the oneway road in direction from A to B. The task is to check if for all pairs of vertices there exists a path or not. This is clearly a problem on SCC. If the entire road network(the graph) is one SCC then the CM’s claim is true, else false.
Algorithm1. Compute finishing time for every vertex by calling DFS on graph. Store these vertices in decreasing order of their finishing time in a List L.
- Reverse the direction of every edge to get a graph G’
- Run DFS on G’, selecting vertices as stored in order in List L. Maintain a counter(initialized to 0) and for a depth first tree obtained from every white vertex, increment the counter by 1.
- If after step 3, counter =1 => CM’s claim is true, else false
Time Complexity: O(V+E) as DFS has time complexity of O(V+E) step 1 and step 3 take O(V+E) , step 2 is O(E), step 4 is O(1). Note that this is linear time in terms of size of graph(including the edges and vertices)
Question 31:-Assume this tree is a binary search tree even though you cannot see what the
keys and values are at the nodes (the letters we write below are just “names” for the nodes for the
purpose of Answer ing the Question s).
(a) What node is the successor of node A? (b) What node is the successor of node B? (c) What node is the successor of node C? (d) What is the height of the tree? (e) What is the maximum number of nodes that could be added to the tree without increasing its height? (f) Is the tree an AVL tree? (g) If we remove only node H, is the result an AVL tree? (h) If we remove only node J, is the result an AVL tree? (i) If we remove only node K, is the result an AVL tree?
Answer :-(a) J (b) E (c) K (d) 4 (e) 18 (f) yes (g) yes (h) no (i) no
Question 32:- Suppose there is a binary min-heap with exactly 4 nodes, containing items with
priorities 3, 9, 11, and 15. (a) Show every possible binary min-heap that could match this description. For each, draw the appropriate tree and the array representation. (You can show just the priorities, not the corresponding items.) (b) For one of your Answer s to part (a), show what happens with 4 deleteMin operations. Clearly indicate which heap you are starting with and show the heap after each deleteMin. You can just draw the tree (not the array) after each step.
Solution:
