vault backup: 2025-03-16 18:59:42

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boris
2025-03-16 18:59:42 +00:00
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commit ae837183f1
188 changed files with 17794 additions and 409 deletions

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# Write a Java method pop() for a stack implemented using an array.
# Write a Java Method pop() for a Stack Implemented Using an Array.
```java
public class Stack {

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@@ -4,6 +4,7 @@ To insert a node into a **binary search tree (BST)** while keeping the tree **ba
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
/ \
@@ -21,6 +22,7 @@ We start by inserting **4** into the BST:
- 4 is greater than 2, so 4 will be placed as the **right child** of 2.
After insertion, the tree looks like this:
```
10
/ \
@@ -44,6 +46,7 @@ Since the tree is balanced at all levels, **no further rebalancing is necessary*
### **Final Balanced Tree**:
The final tree after inserting 4 is:
```
10
/ \
@@ -56,4 +59,4 @@ The final tree after inserting 4 is:
### **Conclusion**:
After inserting 4, the tree remains balanced and no rotations are needed.
After inserting 4, the tree remains balanced and no rotations are needed.

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# Describe a scenario where a stack data structure would be more suitable than a Queue
# 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.
@@ -36,5 +37,5 @@ A **stack data structure** is more suitable than a queue in scenarios where you
- 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
Describe a scenario where a stack data structure would be more suitable than a
Queue

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# What is a binary search tree, and how does it differ from a regular binary tree?
# What is a Binary search Tree, and how Does it Differ from a Regular Binary Tree?
### **Binary Search Tree (BST):**
@@ -38,6 +38,7 @@ A **Binary Tree** is a tree data structure where each node has at most two child
#### Binary Tree:
A binary tree with no ordering constraints:
```
10
/ \
@@ -45,11 +46,13 @@ A binary tree with no ordering constraints:
/
40
```
- Node values are not in any specific order.
#### Binary Search Tree:
A binary search tree:
```
10
/ \
@@ -64,13 +67,13 @@ A binary search tree:
---
### **Advantages of BST Over a General Binary Tree:**
### **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, $h=O(logn)$.
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:**
### **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.
2. Maintaining balance (e.g., in AVL or Red-Black Trees) adds complexity.

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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 and how Does it Differ from a Class?
### **What is an Interface in Object-Oriented Programming?**
### **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.
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:
@@ -12,6 +12,7 @@ Key characteristics of an interface:
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 {
@@ -33,7 +34,7 @@ public class Dog implements Animal {
}
```
### **What is a Class?**
### **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.
@@ -45,6 +46,7 @@ Key characteristics of a class:
4. **Constructors**: Classes can have constructors to initialize objects.
#### Example in Java:
```java
// Define a class
public class Dog {
@@ -73,13 +75,14 @@ public class Dog {
|**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?**
### **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 To Use a Class?**
- When you need to define an object with both state and behavior.
- When you need concrete implementations of methods.

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# Convert the infix expression A * (B + C) / D to a postfix expression
# 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:
@@ -16,19 +16,20 @@ To convert the infix expression `A * (B + C) / D` into postfix notation, we use
#### 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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# Evaluate the postfix expression `6 2 3 + -.`
# Evaluate the Postfix Expression `6 2 3 + -.`
### **Postfix Evaluation Rules**
@@ -13,25 +13,25 @@
**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.
@@ -42,4 +42,3 @@
### **Final Result**:
The result of the postfix expression `6 2 3 + -` is **`1`**.

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# Compare and contrast a queue with a stack.
# Compare and Contrast a Queue with a Stack.
### **Comparison of Queue and Stack**
### **Comparison Of Queue and Stack**
|**Aspect**|**Stack**|**Queue**|
|---|---|---|
@@ -47,4 +47,4 @@
### **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**).
- 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.
# 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:
@@ -8,10 +8,13 @@ In a **min heap**, the root node contains the smallest element, and every parent
### 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
/ \
@@ -19,13 +22,17 @@ This corresponds to the following binary tree:
/ \ / \
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
/ \
@@ -35,6 +42,7 @@ This corresponds to the following binary tree:
/
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:
@@ -42,6 +50,7 @@ Now, we need to **bubble up** the element `3` to restore the heap property. We c
- The parent of `3` is `7` (index 3). Since `3 < 7`, we swap them.
After the swap, the heap becomes:
```
4
/ \
@@ -57,6 +66,7 @@ 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
/ \
@@ -86,6 +96,7 @@ After the swap, the heap becomes:
### **Final Array Representation**:
The final min heap as an array is:
```csharp
[3, 4, 6, 5, 8, 9, 10, 7]
```

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# Create an adjacency matrix for the following weighted undirected graph.
# Create an Adjacency Matrix for the following Weighted Undirected Graph.
```
A-3-B
| |
@@ -17,10 +18,10 @@ To create an **adjacency matrix** for the given **weighted undirected graph**, w
### **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.
- **A -3B**: There is an edge between A and B with weight 3.
- **A -2C**: There is an edge between A and C with weight 2.
- **B -1D**: There is an edge between B and D with weight 1.
- **C -4D**: There is an edge between C and D with weight 4.
---
@@ -49,4 +50,4 @@ Let's map the nodes to indices:
- `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.
This adjacency matrix accurately represents the weighted undirected graph.