Description
Problem 1: AVL Tree
AIM :To have end to end knowledge of the balanced binary search
tree and how it can be used to solve a wide range of problems
efficiently.
TASK: Implement AVL Tree with Following Operations.
Operations to implement:
Operations Complexity
1 Insertion O(log N)
2 Deletion O(log N)
3 Search O(log N)
4 Count occurrences of element O(log N)
5 lower_bound O(log N)
6 upper_bound O(log N)
7 Closest Element to some value O(log N)
8 K-th largest element O(log N)
9 Count the number of elements in the tree whose
values fall into a given range.
O(log N)
IMPORTANT POINTS:
1. Implement it with class or struct. It should be generic.
2. Duplicates are allowed. (We know that AVL tree doesn’t have
duplicates but in this task you have to handle it.)
3. For strings, you can simply compare them but for Class data
type, you have to pass the comparator object so that you can
compare two objects.
Parameter to Judge: Time and space complexity.
Format to submit: RollNo_Q1.cpp
References: https://en.wikipedia.org/wiki/AVL_tree
Problem 2: Hashing
Task: Implement an Unordered Map.
Aim: To learn how Hashing works and importance of Hash Functions.
Also look how Universal Hashing is implemented.
Parameter to Judge: Time and space complexity.
Hashing should be efficient and appropriate reasons must be given on
choice of hash function.
Functions to implement:
1. insert(key, value) – insert key value pair.
2. erase(key) – erase if key is present otherwise do nothing.
3. find(key) – returns true or false.
4. map[key] – returns the value mapped to key.
Note: Unordered Map should be generic.
Format to submit: RollNo_Q2.cpp
References:
https://en.wikipedia.org/wiki/Hash_function
https://en.wikipedia.org/wiki/Universal_hashing
Problem 3: Ordered Map
Task : Implement an Ordered Map which supports the given
Operations
Operations:
1. insert(key, value) – Insert the key value pair. If the key is already
present, update it’s current value. [O(log n)]
2. erase(key) – Erase the given key from the map if it is present.
[O(log n)]
3. find(key) – returns true if key was found else return 0. [O(logn)]
4. map_obj[key] (subscript operator) – Access the element with the
given key(if it is present in the map). Also, you should be able to
modify the value at this key using this operator. If the value is not
present, then insert this key with it’s corresponding assigned value.
[O(log n)]
5. size() – returns the number of keys present in the Map. [O(1)]
6. clear() – removes all the elements from the Map. [O(n)]
Note:
I. You should implement this Ordered Map in a class or struct.
II. Duplicates are not allowed.
III. All the operations should meet their expected time complexity.
IV. You can add constructors and destructors as required.
Parameters to Judge : Correctness, Space and Time Complexity
Submission Format : RollNumber_Q3.cpp
Note : For all the questions, accuracy will be tested on the
basis of test cases passed which will be provided during
evaluation.
