Python Coding Test Course, Traversing Trees

Introduction

Solving algorithm problems is a very important process for software developers. In particular, trees are used as a core structure in many problems, and understanding how to traverse trees is essential for effectively utilizing this structure. In this article, we will explain the basic concepts of trees and various traversal methods using Python, and connect theory to practice by solving actual algorithm problems.

Basics of Trees

A tree is a non-linear data structure consisting of nodes and the connections between them.
Each node can have child nodes, making this structure suitable for representing hierarchies.
The topmost node is called the root node, and there are various traversal methods that define the relationships between nodes.

Types of Trees

– Binary Tree: A tree where each node can have at most two child nodes.
– Binary Search Tree: A sorted binary tree where the left child is smaller than the parent node, and the right child is larger.
– AVL Tree: A balanced binary search tree that maintains a height difference of 1 or less.

Tree Traversal Methods

Tree traversal defines the order in which nodes are visited. The main traversal methods are as follows.

1. Pre-Order Traversal

Pre-order traversal visits the node itself first, then recursively visits the left child node, and finally visits the right child node.
In other words, the visiting order is: Node → Left → Right

2. In-Order Traversal

In-order traversal visits the left child node first, then the node, and finally the right child node.
The order is: Left → Node → Right

3. Post-Order Traversal

Post-order traversal visits the left child node first, then the right child node, and finally the node itself.
The order is: Left → Right → Node

Problem: Print the Results of Binary Tree Traversal

The following is a problem that takes nodes of a binary tree as input and outputs the results of pre-order, in-order, and post-order traversals.
Given a list containing the values of each node, you need to return the results of the tree traversal.

Problem Description

Based on the given list, construct a binary tree and return a list containing the results using each traversal method.
For example, if the list is [1, 2, 3, 4, 5], the following tree can be constructed:

                 1
                / \
               2   3
              / \
             4   5
            

Input / Output Format

  • Input: A list containing each node.
  • Output: Lists of results from pre-order, in-order, and post-order traversals.

Problem-Solving Process

Step 1: Define the Tree Structure

To construct the tree, we will first define a class that represents a node.

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None
            

Step 2: Build the Tree

We will write a function that constructs the tree based on the given list.
In this example, we simply set the first value of the list as the root node and place the rest as the left and right children.

def build_tree(values):
    if not values:
        return None
    root = TreeNode(values[0])
    for value in values[1:]:
        insert_node(root, value)
    return root

def insert_node(root, value):
    if value < root.value:
        if root.left is None:
            root.left = TreeNode(value)
        else:
            insert_node(root.left, value)
    else:
        if root.right is None:
            root.right = TreeNode(value)
        else:
            insert_node(root.right, value)
            

Step 3: Write the Traversal Functions

We will implement each traversal method. The tree will be explored recursively and nodes will be stored in a list.

def preorder_traversal(root):
    result = []
    if root:
        result.append(root.value)
        result.extend(preorder_traversal(root.left))
        result.extend(preorder_traversal(root.right))
    return result

def inorder_traversal(root):
    result = []
    if root:
        result.extend(inorder_traversal(root.left))
        result.append(root.value)
        result.extend(inorder_traversal(root.right))
    return result

def postorder_traversal(root):
    result = []
    if root:
        result.extend(postorder_traversal(root.left))
        result.extend(postorder_traversal(root.right))
        result.append(root.value)
    return result
            

Step 4: Integrate and Call

Finally, we integrate all of the above functions to solve the problem.

def traverse_tree(values):
    root = build_tree(values)
    return {
        'preorder': preorder_traversal(root),
        'inorder': inorder_traversal(root),
        'postorder': postorder_traversal(root),
    }

# Example usage
input_values = [1, 2, 3, 4, 5]
result = traverse_tree(input_values)
print(result)  # Print the results
            

Final Results and Summary

Through the above process, we can traverse the given binary tree list and derive the results of pre-order, in-order, and post-order traversals.
This problem has taught us the concept of tree traversal and how to implement it efficiently.
Additionally, we encourage you to experience and practice with various tree structures and traversal methods.

Python Coding Test Course, Try

1. Introduction

Hello! Today, we will explore the very useful Trie data structure for job seekers. To solve various algorithm problems, one must understand and utilize different data structures, among which the Trie data structure is very powerful in effectively solving string-related problems. In this post, we will cover the concept, characteristics, examples of use, and the problems that can be solved using Tries.

2. What is the Trie Data Structure?

A Trie is a tree structure optimized for storing multiple strings. It is primarily used for prefix searches. Each node typically represents one character of a string, and by starting from the root node and progressing down through child nodes, one can interpret each character of the string step by step. A Trie has the following characteristics:

  • Each node in the Trie has characters and its child nodes.
  • Strings are constructed through the path of the nodes.
  • Duplicate strings are merged into a single path.
  • Each node can mark the end of a string to indicate its termination.

3. Trie Structure

The Trie structure can be designed as follows. Each node corresponds to a single character, and as one descends from the root node, strings are formed. For example, adding the strings ‘cat’ and ‘car’ would create the following structure:

        Root
        └── c
            ├── a
            │   ├── t (end node)
            │   └── r (end node)
        

This structure allows for easy exploration of strings with the prefix ‘ca’.

4. Implementing a Trie

To implement a Trie, we will start by defining classes and methods. The basic implementation involves defining nodes and creating a Trie class with functionalities for insertion, search, and deletion.

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end_of_word = False

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word: str) -> None:
        node = self.root
        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.is_end_of_word = True

    def search(self, word: str) -> bool:
        node = self.root
        for char in word:
            if char not in node.children:
                return False
            node = node.children[char]
        return node.is_end_of_word

    def starts_with(self, prefix: str) -> bool:
        node = self.root
        for char in prefix:
            if char not in node.children:
                return False
            node = node.children[char]
        return True
        

The above code defines the TrieNode class that represents a Trie node and the Trie class that includes basic Trie functionalities.

5. Problem: Finding Prefixes in a List of Strings

Now, let’s use the Trie to solve the ‘prefix search’ problem. The problem is as follows.

Problem Description

Given a list of strings, write a program to verify if a specific string is a prefix of any string in this list.

Input

- List of strings: ["apple", "banana", "app", "apricot"]
- Prefix: "app" 

Output

'app' is a prefix of 'apple' and 'app' in the list.

Now let’s use a Trie to solve this problem.

6. Problem Solving Process

We will use the Trie to solve this problem. First, we will insert the list of strings into the Trie, then check if the given prefix is included in the Trie.

def find_prefix_words(words: List[str], prefix: str) -> List[str]:
    trie = Trie()
    for word in words:
        trie.insert(word)

    result = []
    for word in words:
        if word.startswith(prefix):
            result.append(word)
    return result

# Example usage
words = ["apple", "banana", "app", "apricot"]
prefix = "app"
print(find_prefix_words(words, prefix))
        

The code above implements a function that takes a list of strings and finds the prefixes of that list. It is designed to return all strings that start with the user-provided prefix.

7. Conclusion

Tries offer significant performance improvements in string data processing and searching. They are a very powerful tool for solving problems like prefix searches. As discussed in this post, using Tries allows for the effective handling of character-based data. I hope today’s post has helped you understand the Trie data structure.

Look forward to more posts covering various algorithm problems!

python coding test course, two pointers

Hello, everyone! Today we will take a closer look at the “Two Pointer” algorithm and spend some time solving problems using this algorithm. The two-pointer technique is a very useful method for solving problems in data structures with a fixed size, such as arrays or lists. Each pointer is used to identify a position in the data structure, which helps reduce time complexity.

1. What is the Two Pointer Algorithm?

The two-pointer algorithm is a representative technique for efficiently solving problems when using two points simultaneously in a given array. It is mainly used to solve problems like sub-sums, maximum/minimum values, or to find pairs of data that satisfy specific conditions in sorted arrays.

The advantage of the two-pointer approach is that it allows you to obtain the necessary results by traversing the array only once with a single loop. This enables solving problems with a time complexity of O(n).

2. Problem Description

The problem we will solve this time is “Finding Ideal Pairs in an Integer Array.” Given an integer array arr and an integer K, find two numbers that sum up to K without a exclamation mark (!). Write a function that returns the indices of these two numbers. If such a pair does not exist, return -1.

Example Problem

  • Input: arr = [10, 15, 3, 7], K = 17
  • Output: (0, 2) // Indices are 0-based
  • Input: arr = [10, 15, 3, 7], K = 20
  • Output: -1

3. Approach

We can use a mathematical approach to solve this problem. Since the sum of the two numbers needs to be K, the first pointer points to the start of the array, and the second pointer points to the end of the array. The sum of the values pointed to by the two pointers is then calculated and compared with K.

1. First, place the first pointer at the far left of the array and the second pointer at the far right.

2. Repeat until the two pointers cross each other:

  • If the sum is equal to K, return the indices of the two pointers.
  • If the sum is less than K, move the first pointer to the right to increase the sum.
  • If the sum is greater than K, move the second pointer to the left to decrease the sum.

3. If the two pointers cross each other without finding a pair, return -1.

4. Code Implementation

  
def two_sum(arr, K):
    left = 0
    right = len(arr) - 1
    
    while left < right:
        current_sum = arr[left] + arr[right]
        
        if current_sum == K:
            return (left, right)  # Found a pair
        elif current_sum < K:
            left += 1  # Move left pointer to increase the sum
        else:
            right -= 1  # Move right pointer to decrease the sum

    return -1  # Pair not found
  
  

5. Code Explanation

The above code is a function that uses the two-pointer technique to find a pair of indices such that their sum equals K. The left and right variables are used to start the pointers from both ends of the array.

First, in the while loop, we check if the two pointers are improving in direction. After calculating the sum of the current two values, if this value equals K, we return the corresponding indices. If not, we adjust the pointers based on the size of the sum and repeat.

6. Time Complexity Analysis

The time complexity of this algorithm is O(n). Since the two pointers start at both ends of the array, it ends with a single iteration, not requiring a full traversal of the array to satisfy the conditions. The space complexity is O(1), making it a very efficient method that uses only constant space.

7. Additional Examples

You can verify the correctness of the code through test cases.

  
print(two_sum([10, 15, 3, 7], 17))  # Output: (0, 2)
print(two_sum([10, 15, 3, 7], 20))  # Output: -1
print(two_sum([1, 2, 3, 4, 6], 10))  # Output: (3, 4)
  
  

8. Conclusion

Today we solved problems using the two-pointer technique. When solving Python algorithm problems, utilizing such techniques can save time and solve problems efficiently. Practice on various problems to accumulate more know-how. Next time, we will discuss dynamic programming. Thank you for your hard work today!

Python Coding Test Course, Preparing for Resignation

Transitioning to a new job after leaving your previous one can be a challenge for many people. If you wish to work in the IT and software fields, problem-solving skills in algorithms are essential. This course will introduce a method to improve your skills by solving algorithm problems frequently presented in developer recruitment tests.

Problem Description

Problem: Intersection of Two Arrays

Given two arrays, write a function to find the intersection of the two arrays. The intersection refers to the set of elements that exist in both arrays.

Input

  • Integer array A (length: m, 1 ≤ m ≤ 10^5)
  • Integer array B (length: n, 1 ≤ n ≤ 10^5)

Output

Return an array of the intersection elements sorted in ascending order. If there are no intersections, return an empty array.

Example

Input: A = [1, 2, 2, 1], B = [2, 2]
Output: [2]

Input: A = [4, 9, 5], B = [9, 4, 9, 8, 4]
Output: [4, 9]

Solution Process

Problem Analysis

The task is to find the common elements that exist in both given arrays A and B. After identifying the identical elements among the elements of both arrays, the intersection should be returned, ensuring that each duplicate element is included only once.

Approach

There are several ways to solve this problem, but an efficient method is to use a hash table (in Python, you can use dictionaries or sets). With this approach, you can convert both arrays into sets and easily find the intersection.

Implementation

In step 1, convert both arrays into sets; in step 2, find the intersection between the sets.
In step 3, sort the intersection result and return it.


def intersection(A, B):
    # Step 1: Convert arrays to sets
    set_A = set(A)
    set_B = set(B)
    
    # Step 2: Find intersection
    intersection_set = set_A.intersection(set_B)
    
    # Step 3: Sort and convert to list
    return sorted(list(intersection_set))

# Example execution
A = [1, 2, 2, 1]
B = [2, 2]
print(intersection(A, B))  # Output: [2]

A = [4, 9, 5]
B = [9, 4, 9, 8, 4]
print(intersection(A, B))  # Output: [4, 9]

Complexity Analysis

Time complexity: O(m + n) – The time taken to convert the two arrays into sets, and the time required to calculate the intersection.
Space complexity: O(m + n) – In the worst case, if both arrays are composed of different elements, space is needed to store the sets.

Conclusion

This algorithm problem frequently appears in coding tests, and can be solved efficiently by utilizing Python’s sets. It is important to systematically approach algorithm problems by dividing them into stages of problem analysis, approach, implementation, and complexity analysis.

Preparing for coding tests may be a challenging journey, but consistent practice and tackling various problems can help build confidence. I hope you can develop the ability to solve algorithm problems alongside your job transition preparation.

python coding test course, fast track with time machine

Hello! In this lesson, we will tackle an interesting algorithm problem called ‘Quick Travel with a Time Machine.’ This problem can be solved using various data structures and algorithms, and it is frequently featured in coding tests. Below, we will discuss the problem description, approach, and code implementation in detail.

Problem Description

There is a time machine that allows for time travel. This time machine has the ability to move to a specific time. However, the following conditions are given for the operation of the time machine:

  • The current time starts at 0.
  • The time machine has the ability to move to the time t - 1 years later. In other words, if a time t is given, it can move to t - 1.
  • We need to find the minimum number of movements to quickly reach the given time n.
  • The movements of the time machine are defined as follows:
    • A direct move that takes +1 second from the current time.
    • It is possible to move to t - 1 years using the time machine.

Input

  • Integer n (0 ≤ n ≤ 1,000,000): The target time to reach

Output

  • Output the minimum number of movements to reach the time n.

Approach

To solve this problem, we will use the BFS (Breadth-First Search) algorithm. BFS is effective for finding the shortest path. Due to its property of evenly accessing all vertices of a network graph, we can try all possible moves from the current position once.

Approach Using BFS

  • The current time starts from 0.
  • Use a queue to store the current time and number of moves.
  • Try two movements by removing time from the queue:
    • +1: Current time + 1
    • t - 1: Current time – 1 (check to ensure it is not less than 0)
  • When the target time is reached, output the number of moves at that point.

Algorithm Implementation for Problem Solving


from collections import deque

def min_move_to_time(n):
    visited = [False] * (2 * n + 1) # Set to store visit status
    queue = deque([(0, 0)]) # Starting point (current time, number of moves)

    while queue:
        current_time, moves = queue.popleft()
        
        # Check if the target has been reached
        if current_time == n:
            return moves
        
        # Try two movements: +1 and t - 1
        next_pos = current_time + 1
        if next_pos <= 2 * n and not visited[next_pos]:
            visited[next_pos] = True
            queue.append((next_pos, moves + 1))

        next_pos = current_time - 1
        if 0 <= next_pos and not visited[next_pos]:
            visited[next_pos] = True
            queue.append((next_pos, moves + 1))

Code Explanation

In the code above, the min_move_to_time function returns the minimum number of moves to reach the given input n. Here is the structure of the code:

  • The visited list is used to record the visited times to avoid infinite exploration.
  • Remove the current time and move count from the queue, and check if the target has been reached.
  • Add the next time to the queue using the two movement options of the time machine, and update the visited list accordingly.

Testing the Results

Now, let’s run some test cases using this algorithm.


# Test case 1
print(min_move_to_time(5))  # Output: 5

# Test case 2
print(min_move_to_time(10)) # Output: 10

# Test case 3
print(min_move_to_time(100)) # Output: 100

Conclusion

The ‘Quick Travel with a Time Machine’ problem is a good example of learning the process of making optimal decisions using the BFS algorithm. Through this, you can enhance your algorithm problem-solving skills. I wish you the best results in your coding tests!