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CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
OCTOBER15
ASSESSMENT_CODE BCA2020_OCTOBER15
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
11324
QUESTION_TEXT
What are the advantages and disadvantages of linear representation
of binary tree?
SCHEME OF
EVALUATION
Advantages:
1. Node can be accessed directly (2 marks)
2. Data are stored without printer(2 marks)
3. This is very useful where the language(2 marks)
Dis advantages:
1. Memory is wasted here(2 marks)
2. Since the array size is limited (2 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
72861
QUESTION_TEXT
What is post order traversal? Explain the algorithm for the
same
SCHEME OF
EVALUATION
In postorder traversal start with the left most subtree (4 mark)
step 1: Start from the root node (1 mark)
Step 2: Go to the leftmost node(1 mark)
Step 3: When it is reached print it down (1 mark)
Step 4: Visit the left most node’s parent node(1 mark)
Step 5: Visit the first node of the right subtree (1 mark)
Step 6: Proceed in the same way till printing of all the nodes is finished
(1 mark)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
72864
QUESTION_TEXT
Explain the 2 types of binary tree
a. Skewed binary tree
b. Full binary tree
SCHEME OF
EVALUATION
a. Skewed binary tree: A binary tree which has only left subtree is called left
skewed tree and has only right subtree is called right skewed tree. Skewed
trees are not efficient in memory management because generally, an n node
binary tree needs an array whose length is between n+1 and 2n, but while
storing skewed binary tree it wastes most of the memory space.
b.
Full binary tree: A binary tree is a full binary tree if and only if :-

Each non leaf node has exactly two child nodes

All leaf nodes are at the same level
A full binary tree is a binary of depth k having 2k − 1 nodes as referred
in the figure 4.5. If it has < 2k − 1, it is not a full binary tree. For
example, for k = 3, the number of nodes = 2k − 1 = 23 − 1 = 8 − 1 = 7. A
full binary tree with depth k = 3 is shown in Figure 4.5. We use
numbers from 1 to 2k − 1 as labels of the nodes of the tree. If a binary
tree is full, then we can number its nodes sequentially from 1 to 2k−1,
starting from the root node, and at every level numbering the nodes
from left to right.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
72867
QUESTION_TEXT
Explain the algorithm to inserting at the beginning of a list.
Inserting at the beginning of list
1. [OVERFLOW?] if AVAIL = NULL, then: Write: OVERFLOW, and
Exit.
2. [Remove first node from AVAIL list]
Set N := AVAIL and AVAIL := LINK [AVAIL].
3. Set DATA [N] := ITEM. [Copies new data into new node]
4. Set LINK [N] := START.[New node points to the original first
node]
5. Set START := N. [changes START so it points to the new node]
SCHEME OF
EVALUATION
6.
Exit.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
125411
QUESTION_TEXT
SCHEME OF
EVALUATION
Explain bubble sort and merge sort
Bubble sort
Bubble sort is a straightforward and simplistic method of sorting data that is
used very commonly. The algorithm starts at the beginning of the data set. It
compares the first two elements, and if the first is greater than the second,
then it swaps them. It continues doing this for each pair of adjacent elements
to the end of the data set. It then starts again with the first two elements,
repeating until no swaps have occurred on the last pass. This algorithm is
highly inefficient, and is rarely used except as a simplistic example. For
example, if we have 100 elements then the total number of comparisons will
be 10000. Bubble sort may, however, be efficiently used on a list that is
already sorted, except for a very small number of elements. For example, if
only one element is not in order, bubble sort will only take 2n time. If two
elements are not in order, bubble sort will only take at most 3n time. Bubble
sort average case and worst case are both O(n²).
Merge sort
Divide-and conquer is a general algorithm design paradigm:
Divide: divide the input data S in two disjoint subsets S1and S2
Recur: solve the sub problems associated with S1 and S2
Conquer: combine the solutions for S1 and S2 into a solution for S
The base case for the recursion is sub problems of size 0 or 1
Merge-sort on an input sequence S with n elements consists of three steps:
Divide: partition S into two sequences S1 and S2 of about n/2 elements each
Recur: recursively sort S1 and S2
Conquer: merge S1 and S2 into a unique sorted sequence.
(5 Marks for each)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
125413
QUESTION_TEXT
Explain memory allocation and garbage collection.
SCHEME OF EVALUATION
Memory allocation (5 marks)
Garbage collection (5 marks)