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76
Data Structures/GPT Answers to Past Paper/Question 1.md Normal file
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# Write a Java method pop() for a stack implemented using an array.
```java
public class Stack {
private int[] stackArray; // Array to hold stack elements
private int top; // Index of the top element
private int capacity; // Maximum capacity of the stack
// Constructor to initialize the stack
public Stack(int size) {
stackArray = new int[size];
capacity = size;
top = -1; // Indicates the stack is empty
}
// Method to add an element to the stack
public void push(int item) {
if (isFull()) {
throw new RuntimeException("Stack overflow");
}
stackArray[++top] = item; // Increment top and store the item
}
// Method to remove and return the top element of the stack
public int pop() {
if (isEmpty()) {
throw new RuntimeException("Stack underflow");
} return stackArray[top--]; // Return the top element and decrement top
}
// Method to check if the stack is empty
public boolean isEmpty() {
return top == -1;
}
// Method to check if the stack is full
public boolean isFull() {
return top == capacity - 1;
}
// Method to view the top element without removing it
public int peek() {
if (isEmpty()) {
throw new RuntimeException("Stack is empty");
} return stackArray[top];
}
public static void main(String[] args) {
Stack stack = new Stack(5); // Create a stack with capacity 5
// Example usage
stack.push(10);
stack.push(20);
stack.push(30);
System.out.println(stack.pop()); // Output: 30
System.out.println(stack.pop()); // Output: 20
System.out.println(stack.pop()); // Output: 10
}
}
```
### Explanation:
1. **Array Storage**: The stack elements are stored in an array (`stackArray`).
2. **Pointer**: The `top` variable keeps track of the current top of the stack.
3. **Pop Operation**: The `pop()` method checks if the stack is empty using the `isEmpty()` method. If not, it returns the element at the `top` index and decrements `top`.
4. **Edge Cases**:
- **Underflow**: If the stack is empty, it throws a `RuntimeException`.
- **Overflow**: The `push()` method checks for overflow before adding elements.
This ensures a robust implementation of the `pop()` operation.
- The expression `stackArray[top--]` first retrieves the value at `stackArray[top]` and then **decrements `top`** by 1.
- This uses the **post-decrement operator (`--`)**, which means:
1. The value of `top` is used to access the current top element.
2. After accessing the value, `top` is decremented.
### Why `top` Controls the Stack?
The value of `top` acts as a pointer to the last valid element in the stack. By decrementing it, you effectively "remove" the topmost element from the stack, even though the value remains in the array until overwritten by a `push()` operation.
### Conclusion:
Yes, calling `pop()` **does update `top`**, which is crucial for maintaining the stack's state. This is a standard approach for implementing a stack using an array.

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Data Structures/GPT Answers to Past Paper/Question 10.md Normal file
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To insert a node into a **binary search tree (BST)** while keeping the tree **balanced**, we need to perform the following steps:
1. **Insert the node** in the appropriate place according to the **BST properties** (left child < parent < right child).
2. After insertion, **rebalance the tree** (if necessary) to ensure that it remains balanced. A tree is considered balanced if the height difference between the left and right subtrees of any node is at most 1 (this is the **balanced binary tree** or **AVL tree** property).
### **Given Binary Search Tree**:
```
10
/ \
5 15
/ / \
2 12 20
```
### **Step 1: Insert 4 into the BST**
We start by inserting **4** into the BST:
- 4 is less than 10, so we move to the left of 10.
- 4 is less than 5, so we move to the left of 5.
- 4 is greater than 2, so 4 will be placed as the **right child** of 2.
After insertion, the tree looks like this:
```
10
/ \
5 15
/ / \
2 12 20
\
4
```
### **Step 2: Balance the Tree (If Needed)**
At this point, the tree is not balanced. Let's check the balance factors of the nodes:
- **Node 2** has a left subtree height of 0 and a right subtree height of 1. Its balance factor is `0 - 1 = -1`, which is acceptable.
- **Node 5** has a left subtree height of 1 and a right subtree height of 1. Its balance factor is `1 - 1 = 0`, which is balanced.
- **Node 10** has a left subtree height of 2 (because of nodes 5, 2, and 4) and a right subtree height of 2 (because of nodes 15, 12, and 20). Its balance factor is `2 - 2 = 0`, which is balanced.
Since the tree is balanced at all levels, **no further rebalancing is necessary**.
### **Final Balanced Tree**:
The final tree after inserting 4 is:
```
10
/ \
5 15
/ / \
2 12 20
\
4
```
### **Conclusion**:
After inserting 4, the tree remains balanced and no rotations are needed.

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Data Structures/GPT Answers to Past Paper/Question 2.md Normal file
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# Describe a scenario where a stack data structure would be more suitable than a Queue
A **stack data structure** is more suitable than a queue in scenarios where you need to process items in a **Last-In, First-Out (LIFO)** order
### Scenario: **Undo Operation in Text Editors**
- **Description**: In text editors, when a user types, deletes, or performs other actions, these operations are stored so they can be undone in reverse order of execution.
- **Why a Stack?**:
- The most recent action (the "last-in") is the first one that should be undone (the "first-out").
- A stack efficiently tracks these operations, allowing actions to be added to the stack as they occur and popped off when the user undoes them.
- **How It Works**:
1. User types "Hello" → each character is pushed onto the stack.
2. User presses "Undo" → the stack pops the last character added.
### Scenario: **Call Stack in Recursion**
- **Description**: When a program uses recursion, the function calls are nested, and each call must finish before the previous one can resume.
- **Why a Stack?**:
- The current function (most recent call) must finish execution (be "popped" off the stack) before the previous function call can resume.
- The stack keeps track of the state and local variables for each function call.
- **How It Works**:
1. Function `A` calls `B`, which calls `C`.
2. The stack stores the execution state of `A`, then `B`, then `C` (last-in).
3. When `C` finishes, its state is removed (first-out), resuming `B`.
### Scenario: **Browser Backtracking**
- **Description**: When navigating a web browser, the "Back" button relies on a stack to revisit the most recently visited page.
- **Why a Stack?**:
- The most recent page (last-in) should be revisited first (first-out) when the user clicks "Back".
- **How It Works**:
1. User navigates to pages: A → B → C. Each page is pushed onto the stack.
2. Clicking "Back" pops the stack, returning to page `B`, then `A`.
### Key Contrast with a Queue:
- A **queue** is more suitable when processing items in a **First-In, First-Out (FIFO)** order, such as serving customer requests or scheduling tasks.
- A stack is better when you need **reversal** or to **process the most recent element first**, such as undo operations, recursion, or browser backtracking.
Describe a scenario where a stack data structure would be more suitable than a
Queue

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Data Structures/GPT Answers to Past Paper/Question 3.md Normal file
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# What is a binary search tree, and how does it differ from a regular binary tree?
### **Binary Search Tree (BST):**
A **Binary Search Tree (BST)** is a type of binary tree where nodes are organized based on their key values to maintain a specific order. It adheres to the following properties:
1. **Left Subtree Rule**: For any node, all values in its left subtree are less than the value of the node.
2. **Right Subtree Rule**: For any node, all values in its right subtree are greater than the value of the node.
3. **Uniqueness**: Typically, BSTs do not allow duplicate values (though this can vary depending on implementation).
These properties ensure that a **BST is sorted**, which allows for efficient searching, insertion, and deletion operations.
---
### **Binary Tree (General Definition):**
A **Binary Tree** is a tree data structure where each node has at most two children. It does not impose any ordering constraints on the values of the nodes.
- **Structure**: Each node can have a **left child**, a **right child**, or both (or no children).
- **No Value Order**: The values in a binary tree are not arranged in any specific order.
---
### **Key Differences Between Binary Tree and Binary Search Tree:**
|**Aspect**|**Binary Tree**|**Binary Search Tree**|
|---|---|---|
|**Node Value Order**|No specific order for node values.|Nodes are organized by the BST property (left < root < right).|
|**Searching Efficiency**|Searching is linear in worst case (O(n)).|Searching is efficient (O(log n)) if balanced.|
|**Insertion Rule**|Nodes can be added in any order.|Nodes are inserted based on their value to maintain the BST property.|
|**Duplicates**|Allows duplicates freely.|May or may not allow duplicates, depending on implementation.|
|**Applications**|Used for general tree structures.|Used for sorted collections or when efficient searching is required.|
---
### **Example:**
#### Binary Tree:
A binary tree with no ordering constraints:
```
10
/ \
20 30
/
40
```
- Node values are not in any specific order.
#### Binary Search Tree:
A binary search tree:
```
10
/ \
5 20
/ \ / \
2 7 15 25
```
- For every node:
- Left subtree values < Node value.
- Right subtree values > Node value.
---
### **Advantages of BST Over a General Binary Tree:**
1. **Efficient Searching**: Searching in a BST takes $O(h)$, where $h$ is the height of the tree. In a balanced BST, $h=O(log⁡n)$.
2. **Ordered Traversal**: In-order traversal of a BST produces the nodes in sorted order.
3. **Efficient Range Queries**: BSTs can efficiently find all elements within a specific range.
### **Disadvantages of BST:**
1. If unbalanced (e.g., when nodes are inserted in sorted order), the BST can degrade to a linked list, resulting in $O(n)$ time for operations.
2. Maintaining balance (e.g., in AVL or Red-Black Trees) adds complexity.

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Data Structures/GPT Answers to Past Paper/Question 4.md Normal file
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# What is an interface in object-oriented programming and how does it differ from a class?
### **What is an Interface in Object-Oriented Programming?**
An **interface** is a contract in object-oriented programming (OOP) that defines a set of methods (and sometimes properties) that a class must implement. It specifies _what_ a class should do, but not _how_ it should do it.
Key characteristics of an interface:
1. **Method Declaration**: An interface only contains method signatures (names, return types, and parameters) without implementation.
2. **Multiple Inheritance**: Many OOP languages, like Java, allow a class to implement multiple interfaces, enabling multiple inheritance of behavior.
3. **Abstraction**: Interfaces are entirely abstract, meaning they cannot have any concrete (implemented) methods in many languages (e.g., Java before version 8).
4. **No State**: Interfaces do not contain instance variables or state but may include constants.
#### Example in Java:
```java
// Define an interface
public interface Animal {
void eat(); // Abstract method
void sleep(); // Abstract method
}
// Class implementing the interface
public class Dog implements Animal {
@Override
public void eat() {
System.out.println("Dog is eating.");
}
@Override
public void sleep() {
System.out.println("Dog is sleeping.");
}
}
```
### **What is a Class?**
A **class** is a blueprint for creating objects in OOP. It defines the structure (fields or attributes) and behavior (methods) of objects. Unlike interfaces, classes can have implementations, state, and behavior.
Key characteristics of a class:
1. **State and Behavior**: Classes define fields to hold data (state) and methods to operate on that data (behavior).
2. **Concrete Methods**: Methods in a class can have concrete implementations.
3. **Inheritance**: A class can inherit from another class (single inheritance in Java) to reuse or extend functionality.
4. **Constructors**: Classes can have constructors to initialize objects.
#### Example in Java:
```java
// Define a class
public class Dog {
private String name; // Field (state)
// Constructor
public Dog(String name) {
this.name = name;
}
// Method (behavior)
public void bark() {
System.out.println(name + " is barking.");
}
}
```
### **Key Differences Between an Interface and a Class**
|**Aspect**|**Interface**|**Class**|
|---|---|---|
|**Purpose**|Defines a contract for behavior.|Defines both state (fields) and behavior (methods).|
|**Implementation**|Does not provide method implementations (in most cases).|Can provide both concrete (implemented) and abstract methods.|
|**Fields/State**|Cannot contain instance variables (only constants).|Can contain fields (instance and static variables).|
|**Inheritance**|A class can implement multiple interfaces.|A class can inherit from only one superclass in languages like Java.|
|**Constructors**|Cannot have constructors.|Can have constructors to initialize objects.|
|**Abstract**|Interfaces are entirely abstract (unless extended in modern versions).|Can be abstract or concrete.|
|**Default Methods**|Starting from Java 8, interfaces can have default methods (methods with a body).|Classes always support method implementation.|
### **When to Use an Interface?**
- **Polymorphism**: When you want different classes to provide their specific implementation of the same behavior.
- **Decoupling**: To decouple code by relying on abstractions rather than concrete classes.
- **Multiple Inheritance**: When you need a class to inherit behavior from multiple sources.
### **When to Use a Class?**
- When you need to define an object with both state and behavior.
- When you need concrete implementations of methods.
- When abstraction is not required.
### **Summary:**
An **interface** defines the "what" (a contract) without defining the "how," while a **class** provides the actual implementation and can store both state and behavior.

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Data Structures/GPT Answers to Past Paper/Question 5.md Normal file
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# Convert the infix expression A * (B + C) / D to a postfix expression
To convert the infix expression `A * (B + C) / D` into postfix notation, we use the following rules:
1. **Operator Precedence**: Parentheses have the highest precedence, followed by multiplication (`*`) and division (`/`), and then addition (`+`).
2. **Associativity**: Operators like `*` and `/` are left-associative, meaning operations are performed left to right if they have the same precedence.
3. **Conversion Rules**:
- Operands (like `A`, `B`, `C`, `D`) are output immediately.
- Operators are pushed onto a stack.
- Parentheses dictate the priority: Solve what's inside the parentheses first.
---
### **Step-by-Step Conversion**
#### Infix Expression: `A * (B + C) / D`
1. Start with the subexpression inside the parentheses `(B + C)`:
- Convert `B + C` to postfix: `BC+`.
2. Substitute the postfix for `(B + C)` back into the original expression:
- The expression becomes `A * BC+ / D`.
3. Process the multiplication (`*`) and division (`/`):
- `A * BC+` becomes `ABC+*`.
- `ABC+* / D` becomes `ABC+*D/`.
---
### **Final Postfix Expression**:
```
ABC+*D/
```

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Data Structures/GPT Answers to Past Paper/Question 6.md Normal file
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# Evaluate the postfix expression `6 2 3 + -.`
### **Postfix Evaluation Rules**
1. Operands are pushed onto a stack.
2. Operators pop the required number of operands from the stack, perform the operation, and push the result back onto the stack.
3. Continue until the expression is fully evaluated, and the final value is the only element on the stack.
---
### **Step-by-Step Evaluation**
**Expression**: `6 2 3 + -`
1. **Read `6`**:
- Push `6` onto the stack.
- **Stack**: `[6]`
2. **Read `2`**:
- Push `2` onto the stack.
- **Stack**: `[6, 2]`
3. **Read `3`**:
- Push `3` onto the stack.
- **Stack**: `[6, 2, 3]`
4. **Read `+`**:
- Pop the top two operands (`3` and `2`).
- Perform `2 + 3 = 5`.
- Push the result (`5`) onto the stack.
- **Stack**: `[6, 5]`
5. **Read `-`**:
- Pop the top two operands (`5` and `6`).
- Perform `6 - 5 = 1`.
- Push the result (`1`) onto the stack.
- **Stack**: `[1]`
---
### **Final Result**:
The result of the postfix expression `6 2 3 + -` is **`1`**.

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Data Structures/GPT Answers to Past Paper/Question 7.md Normal file
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# Compare and contrast a queue with a stack.
### **Comparison of Queue and Stack**
|**Aspect**|**Stack**|**Queue**|
|---|---|---|
|**Definition**|A stack is a collection of elements that follows the **Last-In, First-Out (LIFO)** principle: the last element added is the first one removed.|A queue is a collection of elements that follows the **First-In, First-Out (FIFO)** principle: the first element added is the first one removed.|
|**Primary Operations**|- **Push**: Add an element to the top of the stack. <br>- **Pop**: Remove the top element of the stack. <br>- **Peek/Top**: View the top element without removing it.|- **Enqueue**: Add an element to the back of the queue. <br>- **Dequeue**: Remove the element from the front of the queue. <br>- **Front**: View the element at the front without removing it.|
|**Order of Removal**|LIFO: Last element inserted is the first to be removed.|FIFO: First element inserted is the first to be removed.|
|**Structure**|Operates on a single end of the collection (top).|Operates on both ends: elements are added at the back and removed from the front.|
|**Real-World Analogy**|A stack of plates: the last plate added is the first one removed.|A line (queue) at a ticket counter: the first person in line is the first one served.|
|**Applications**|- Function call stack <br>- Undo operations in software <br>- Backtracking algorithms (e.g., maze solving) <br>- Expression evaluation and parsing|- Task scheduling (e.g., printer queue) <br>- Breadth-first search (BFS) in graphs <br>- Data buffers (e.g., IO buffers) <br>- Handling requests in web servers|
|**Performance**|Both `push` and `pop` are O(1)O(1)O(1) operations in most implementations.|Both `enqueue` and `dequeue` are O(1)O(1)O(1) operations in most implementations.|
|**Types/Variants**|- Simple Stack <br>- Double Stack <br>- Min/Max Stack (keeps track of min/max values)|- Simple Queue <br>- Circular Queue <br>- Priority Queue (elements removed based on priority, not order) <br>- Deque (Double-Ended Queue)|
---
### **Key Differences**
|**Aspect**|**Stack**|**Queue**|
|---|---|---|
|**Access Pattern**|LIFO (Last-In, First-Out)|FIFO (First-In, First-Out)|
|**Addition (Insert)**|Always at the top of the stack.|Always at the back of the queue.|
|**Removal (Delete)**|Always from the top of the stack.|Always from the front of the queue.|
|**Flexibility in Ends**|Operates only on one end (top).|Operates on both ends (front and back).|
---
### **Real-World Examples**
#### **Stack Examples**:
1. **Undo Feature in Text Editors**:
- When you perform actions, they are stored in a stack. Undoing removes the last action first.
2. **Browser Back Button**:
- Tracks visited pages in a stack; the last page visited is the first to be returned when you press "Back."
#### **Queue Examples**:
1. **Ticket Counter**:
- People stand in line (queue), and the first person in line is served first.
2. **Task Scheduling**:
- Jobs sent to a printer are handled in the order they are received (FIFO).
---
### **Summary**:
- A **stack** is suitable for tasks where the most recent action needs to be undone or processed first (**LIFO**).
- A **queue** is ideal for tasks where the first action needs to be processed first (**FIFO**).

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# Consider a min heap with these elements: [4, 5, 6, 7, 8, 9, 10]. Insert 3 into this min heap.
In a **min heap**, the root node contains the smallest element, and every parent node is less than its children. When we insert a new element into a min heap, we follow these steps:
1. **Insert the element at the end** of the heap (maintaining the complete binary tree property).
2. **Bubble up (or sift up)** the inserted element to restore the heap property by comparing it with its parent and swapping if necessary.
### Given Min Heap:
The given min heap is represented as an array:
```csharp
[4, 5, 6, 7, 8, 9, 10]
```
This corresponds to the following binary tree:
```
4
/ \
5 6
/ \ / \
7 8 9 10
```
### **Step 1: Insert 3 into the Heap**
First, insert 3 at the end of the heap:
```csharp
[4, 5, 6, 7, 8, 9, 10, 3]
```
This corresponds to the following binary tree:
```
4
/ \
5 6
/ \ / \
7 8 9 10
/
3
```
### **Step 2: Bubble Up**
Now, we need to **bubble up** the element `3` to restore the heap property. We compare `3` with its parent and swap if necessary:
- The parent of `3` is `7` (index 3). Since `3 < 7`, we swap them.
After the swap, the heap becomes:
```
4
/ \
5 6
/ \ / \
3 8 9 10
/
7
```
Now, we continue the bubbling up process:
- The parent of `3` is `5` (index 1). Since `3 < 5`, we swap them.
After the swap, the heap becomes:
```
4
/ \
3 6
/ \ / \
5 8 9 10
/
7
```
Finally, we compare `3` with its parent `4` (index 0):
- Since `3 < 4`, we swap them.
After the swap, the heap becomes:
```
3
/ \
4 6
/ \ / \
5 8 9 10
/
7
```
### **Final Array Representation**:
The final min heap as an array is:
```csharp
[3, 4, 6, 5, 8, 9, 10, 7]
```

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Data Structures/GPT Answers to Past Paper/Question 9.md Normal file
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# Create an adjacency matrix for the following weighted undirected graph.
```
A-3-B
| |
2 1
| |
C-4-D
```
To create an **adjacency matrix** for the given **weighted undirected graph**, we need to follow these steps:
1. **Identify the nodes**: We have 4 nodes: `A`, `B`, `C`, and `D`.
2. **Create a matrix**: The matrix will be 4x4, where rows and columns represent the nodes, and the elements in the matrix represent the weights of the edges between nodes. If there's no edge between two nodes, the matrix will have a value of `0` (or infinity, depending on the context, but we'll use `0` here).
3. **Place the edge weights in the matrix**: Since the graph is undirected, the matrix will be symmetric, meaning if there's an edge from node `A` to node `B`, the matrix entry at `[A][B]` and `[B][A]` will have the same value.
---
### **Graph Details**:
- **A -3- B**: There is an edge between A and B with weight 3.
- **A -2- C**: There is an edge between A and C with weight 2.
- **B -1- D**: There is an edge between B and D with weight 1.
- **C -4- D**: There is an edge between C and D with weight 4.
---
### **Adjacency Matrix**
Let's map the nodes to indices:
- `A` = 0
- `B` = 1
- `C` = 2
- `D` = 3
#### Matrix:
| | A | B | C | D |
| --- | --- | --- | --- | --- |
| A | 0 | 3 | 2 | 0 |
| B | 3 | 0 | 0 | 1 |
| C | 2 | 0 | 0 | 4 |
| D | 0 | 1 | 4 | 0 |
### **Explanation**:
- `A-B` has a weight of 3, so `matrix[0][1] = 3` and `matrix[1][0] = 3` (since the graph is undirected).
- `A-C` has a weight of 2, so `matrix[0][2] = 2` and `matrix[2][0] = 2`.
- `B-D` has a weight of 1, so `matrix[1][3] = 1` and `matrix[3][1] = 1`.
- `C-D` has a weight of 4, so `matrix[2][3] = 4` and `matrix[3][2] = 4`.
This adjacency matrix accurately represents the weighted undirected graph.