Building a Balanced Multiset: A Step-by-Step Guide for COMP2700 Assignment 2

Introduction to the Multiset ADT

In COMP2700 Assignment 2, you'll implement a Multiset ADT using a balanced binary search tree. A multiset, or bag, allows duplicate elements, each tracked by a count. This is similar to a music streaming service's playlist that counts how many times you've listened to each song—you might have "Blinding Lights" with a count of 150 and "Levitating" with 89. The operations you'll build let you insert, delete, query, and combine multisets efficiently.

This tutorial focuses on the balanced BST implementation and cursor operations, which are essential for achieving O(log n) time complexity. We'll also cover complexity analysis for advanced operations like union and intersection. By the end, you'll be ready to ace the correctness and complexity parts of the assignment.

Part 1: Setting Up the Data Structures

Your MsetStructs.h should define a node for the BST and a multiset struct. The node must store an element, its count, and pointers to left and right children. For balancing, you'll need a height field. The multiset struct holds a pointer to the root and, optionally, the size and total count for O(1) queries.

// MsetStructs.h
typedef struct node {
    int elem;
    int count;
    int height;
    struct node *left;
    struct node *right;
} Node;

typedef struct multiset {
    Node *tree;
    int size;       // number of distinct elements
    int totalCount; // sum of all counts
} Multiset;

The height field is crucial for AVL tree rotations, which keep the tree balanced after insertions and deletions.

Part 2: Balanced BST Operations

For MsetInsert and MsetDelete, you'll implement standard BST insertion/deletion with AVL rotations. After each modification, update heights and check balance factors. If a node becomes unbalanced (balance factor outside [-1,1]), apply the appropriate rotation (left, right, left-right, right-left). This ensures worst-case O(log n) performance, just like a well-organized library where you can find any book quickly because shelves are evenly spaced.

For MsetInsertMany and MsetDeleteMany, you can simply call the single-insert/delete in a loop, but note that this would be O(k log n). The assignment requires O(h) for these operations—meaning you should handle bulk updates in a single traversal. For example, when inserting many copies of the same element, you can search once and update the count. Similarly, when deleting many, you decrement the count and only remove the node if the count reaches zero.

Part 3: Advanced Operations with Complexity Analysis

Advanced operations like MsetUnion and MsetIntersection must be implemented without converting the tree to an array. Instead, you'll traverse both trees simultaneously, similar to merging two sorted lists. For union, take the maximum count of each element; for intersection, take the minimum. The time complexity is O(n + m), where n and m are the numbers of distinct elements in each multiset. In your analysis.txt, explain that each node is visited once, and the recursion stack depth is O(log n) due to the balanced tree.

For MsetMostCommon, you need to find the top k elements by count. A straightforward approach is to traverse the tree (O(n)), collect nodes into an array, and then sort by count (O(n log n)). However, a better way is to use a min-heap of size k, giving O(n log k). In your analysis, discuss the trade-offs and justify your choice.

Part 4: Cursor Operations

Cursors allow iteration over the multiset in sorted order. The tricky part is implementing MsetCursorNext and MsetCursorPrev with O(1) or O(log n) worst-case time. One method is to store a stack of nodes representing the path from the root to the current node. On next, you pop the stack and then push the leftmost nodes of the right child. This is similar to a Spotify playlist cursor that remembers your position in a shuffled list—you can go forward or backward without rescanning the whole playlist.

In your analysis.txt, explain that each cursor operation does a constant number of stack pushes/pops, and the stack depth is O(log n), so the worst-case time is O(log n). Also note that cursor creation initializes the stack by traversing to the smallest element (O(log n)).

Putting It All Together

Testing is critical. Use the provided testMset.c as a starting point, but also write your own tests for edge cases: empty multisets, single elements, duplicates, and large datasets. Check for memory leaks with tools like valgrind. Remember, correctness is 75% of the grade, but complexity analysis (15%) and code style (10%) also matter.

This assignment is your chance to master data structures that power real-world applications—from database indexing to AI recommendation systems. Good luck!