Advanced Data Structures Assignment Sample: Expert Solutions for Master-Level Programming Students

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Master-Level Assignment Question 1

Question:
Design and implement a balanced Binary Search Tree (AVL Tree) in C++ that supports insertion and automatically maintains balance using rotations. Analyze the time complexity of insertion.

Solution:
An AVL Tree is a self-balancing BST where the height difference between left and right subtrees is at most one. After every insertion, rotations are applied to maintain balance.

#include iostream
using namespace std;

struct Node {
int key;
Node* left;
Node* right;
int height;
};

int height(Node* n) {
return n ? n-height : 0;
}

int getBalance(Node* n) {
return n ? height(n-left) - height(n-right) : 0;
}

Node* rightRotate(Node* y) {
Node* x = y-left;
Node* T2 = x-right;

x-right = y;
y-left = T2;

y-height = max(height(y-left), height(y-right)) + 1;
x-height = max(height(x-left), height(x-right)) + 1;

return x;
}

Node* leftRotate(Node* x) {
Node* y = x-right;
Node* T2 = y-left;

y-left = x;
x-right = T2;

x-height = max(height(x-left), height(x-right)) + 1;
y-height = max(height(y-left), height(y-right)) + 1;

return y;
}

Node* insert(Node* node, int key) {
if (!node) {
Node* n = new Node();
n-key = key;
n-left = n-right = nullptr;
n-height = 1;
return n;
}

if (key node-key)
node-left = insert(node-left, key);
else if (key node-key)
node-right = insert(node-right, key);
else
return node;

node-height = 1 + max(height(node-left), height(node-right));
int balance = getBalance(node);

if (balance 1 key node-left-key)
return rightRotate(node);

if (balance -1 key node-right-key)
return leftRotate(node);

return node;
}

Explanation:
Insertion in an AVL Tree takes O(log n) time because the tree height is always logarithmic. Rotations ensure balance without compromising BST properties. Students who ask us to Do My Programming assignment often appreciate such step-by-step clarity.

Master-Level Assignment Question 2

Question:
Implement Dijkstra’s Algorithm using a priority queue to find the shortest path from a source vertex in a weighted graph.

Solution:

#include bits/stdc++.h
using namespace std;

void dijkstra(int src, vectorvectorpairint,int graph) {
int V = graph.size();
vectorint dist(V, INT_MAX);
priority_queuepairint,int, vectorpairint,int, greater pq;

dist[src] = 0;
pq.push({0, src});

while (!pq.empty()) {
int u = pq.top().second;
pq.pop();

for (auto edge : graph[u]) {
int v = edge.first;
int weight = edge.second;

if (dist[u] + weight dist[v]) {
dist[v] = dist[u] + weight;
pq.push({dist[v], v});
}
}
}

for (int i = 0; i V; i++)
cout "Distance from source to " i " is " dist[i] endl;
}

Explanation:
Using a priority queue optimizes Dijkstra’s Algorithm to O(E log V) time complexity. This approach is widely expected in master-level coursework and exams.

Final Thoughts

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