Category: Python Coding Test

You are now going to take a closer look at one of the algorithm problems, “Determining if Line Segments Intersect.” In this lecture, we will explain the theory and actual implementation process step by step to solve this problem. It will be very beneficial for those who want to improve their algorithm problem-solving skills.

Problem Definition

The problem of determining whether two line segments intersect involves deciding if the given two segments cross each other. We will represent the two segments as A(x1, y1) to B(x2, y2) and C(x3, y3) to D(x4, y4). The problem is as follows:

Given two segments AB and CD, determine whether the two segments intersect.

Input Format

The input consists of four integers, where each integer represents the endpoints of the respective segments:

• A(x1, y1)
• B(x2, y2)
• C(x3, y3)
• D(x4, y4)

Output Format

If the segments intersect, output “YES”; otherwise, output “NO”.

Basic Theory for Problem Solving

One of the methods we can use to determine if segments intersect is a geometric approach. Specifically, we can use the cross product of vectors to ascertain whether the two segments cross each other.

Cross Product

The cross product between two vectors AB(x2-x1, y2-y1) and AC(x3-x1, y3-y1) is defined as follows:

    Cross = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1)
    

This value allows us to determine the direction of the two vectors. If the value of the cross product is 0, it means the vectors are collinear; a positive value indicates one direction, while a negative value indicates the opposite direction.

Intersection Conditions for Line Segments

To determine if two segments AB and CD intersect, we need to check the following conditions:

1. When the two segments are in the “general case” (the endpoints of AB and CD are in different directions)
2. When the two segments “meet at a single point” (intersecting at an internal point)
3. When the two segments “meet at an endpoint” (an endpoint of one segment is on the other segment)

General Case

To confirm that segments AB and CD are in different directions, we compare the signs of the cross products using the points A, B and C, and A, B and D. If the two segments are in different directions, these two signs must be different.

Single/Endpoint Meeting Cases

Unlike the general case, to judge if they meet at a point rather than crossing, we need to check if the endpoints of the segments are on each other’s segments.

Python Code Implementation

Based on the basic theory above, we can write the following Python code.

def orientation(px, py, qx, qy, rx, ry):
    val = (qy - py) * (rx - qx) - (qx - px) * (ry - qy)
    if val == 0: return 0  # Collinear
    return 1 if val > 0 else 2  # Clockwise or Counterclockwise

def on_segment(px, py, qx, qy, rx, ry):
    return min(px, qx) <= rx <= max(px, qx) and min(py, qy) <= ry <= max(py, qy)

def do_intersect(p1, q1, p2, q2):
    o1 = orientation(p1[0], p1[1], q1[0], q1[1], p2[0], p2[1])
    o2 = orientation(p1[0], p1[1], q1[0], q1[1], q2[0], q2[1])
    o3 = orientation(p2[0], p2[1], q2[0], q2[1], p1[0], p1[1])
    o4 = orientation(p2[0], p2[1], q2[0], q2[1], q1[0], q1[1])

    if o1 != o2 and o3 != o4:
        return True
    
    if o1 == 0 and on_segment(p1[0], p1[1], q1[0], q1[1], p2[0], p2[1]):
        return True

    if o2 == 0 and on_segment(p1[0], p1[1], q1[0], q1[1], q2[0], q2[1]):
        return True

    if o3 == 0 and on_segment(p2[0], p2[1], q2[0], q2[1], p1[0], p1[1]):
        return True

    if o4 == 0 and on_segment(p2[0], p2[1], q2[0], q2[1], q1[0], q1[1]):
        return True

    return False

# Test case
A = (0, 0)
B = (10, 10)
C = (0, 10)
D = (10, 0)

if do_intersect(A, B, C, D):
    print("YES")
else:
    print("NO")
    

Code Explanation and Test Cases

In the above code, the orientation function first determines the relative position of three points. The on_segment function checks if a point lies on a given segment. The do_intersect function determines whether two segments intersect. At the end of the code, the actual test case allows us to verify the results.

Conclusion

In this lecture, we explored various theoretical backgrounds and Python code implementations to solve the problem of "Determining if Line Segments Intersect." I hope this has provided an opportunity to enhance both your geometric thinking and programming skills while working through a specific algorithm problem. I wish for your continued growth in coding abilities!

Hello! In this post, we will examine one of the coding test problems using Python, titled “Finding the Direction of a Line Segment.” Through this problem, we will develop practical problem-solving skills and practice geometric problems that are often encountered in coding tests.

Problem Description

Given two points A(x1, y1) and B(x2, y2), this problem asks to determine the direction of line segment AB in relation to line segment CD. Based on line segment AB, find the direction of line segment CD and output the following values:

• 1: When line segment CD is to the left of line segment AB
• -1: When line segment CD is to the right of line segment AB
• 0: When line segment CD is parallel to line segment AB

Input Format

The first line contains the coordinates of points A and B separated by a space, and the second line contains the coordinates of points C and D also separated by a space.

A's x y coordinates: x1 y1
B's x y coordinates: x2 y2
C's x y coordinates: x3 y3
D's x y coordinates: x4 y4

Output Format

Output an integer indicating the direction of line segment CD.

Problem-Solving Approach

To solve this problem, we first need to compute the direction vectors of line segments AB and CD. The direction vector can be calculated as follows:

• AB vector = (x2 – x1, y2 – y1)
• CD vector = (x4 – x3, y4 – y3)

Next, we determine the direction by calculating the cross product of the two vectors. The direction of the line segment can be decided based on the result of the cross product.

Cross Product Calculation

The cross product of two vectors (x1, y1) and (x2, y2) is calculated as follows:

cross_product = x1 * y2 - y1 * x2

If this value is > 0, then it is to the left; < 0 means it is to the right; and 0 means it is parallel.

Example

For instance, given A(1, 1), B(4, 4), C(4, 1), D(1, 4), the direction vector of AB is (3, 3). The direction vector of CD is (-3, 3). Calculating the cross product gives:

cross_product = (3 * 3) - (3 * -3) = 9 + 9 = 18 (therefore, line segment CD is located to the left of AB.)

Code Implementation

Now, let’s write a Python code based on the above process:

def direction(a, b, c, d):
    # Vector AB
    ab = (b[0] - a[0], b[1] - a[1])
    # Vector CD
    cd = (d[0] - c[0], d[1] - c[1])
    
    # Cross product calculation
    cross_product = ab[0] * cd[1] - ab[1] * cd[0]
    
    if cross_product > 0:
        return 1    # Left
    elif cross_product < 0:
        return -1   # Right
    else:
        return 0    # Parallel

# Sample input
a = (1, 1)
b = (4, 4)
c = (4, 1)
d = (1, 4)

# Function call
result = direction(a, b, c, d)
print(result)

Result

By executing the code above, since line segment CD is located to the left of AB, the result will be 1.

Conclusion

In this post, we solved an algorithm problem to find the direction of a line segment. This problem requires geometric thinking, and understanding vector cross products is crucial. As you solve more problems, familiarize yourself with such geometric problems, and I hope you achieve good results in coding tests!

Additional Tips

• Be sure to test sufficiently with various inputs.
• Utilizing visual diagrams can be helpful in understanding the problem.
• Debug visually through the results of the cross product.

Problem Description

One day, a special event for gift delivery was held in a village. The purpose of this event is to deliver gifts among different people. There are n participants, and each participant must deliver exactly one gift. However, there are constraints on whom they can deliver gifts to.

Each participant has a list or array that indicates who they can deliver their gifts to. Considering these constraints, we need to determine whether all participants can deliver gifts, meaning whether every participant can receive a gift.

Input Format

The first line contains the number of participants n. The next line contains a list that represents the possible recipients for each of the n participants. Each element of the list expresses the index of the participant that the corresponding participant can deliver their gift to.

Output Format

If all participants can receive gifts, print ‘YES’; otherwise, print ‘NO’.

Example Input

5
1 2 3 4 0
            

Example Output

YES
            

Solution Method

This problem can be viewed similarly to the “cycle detection” problem in graph theory. Each participant can be represented as a node, and the relationships of whom they can give gifts to can be expressed as edges, forming a directed graph. All nodes must be visited at least once, and if the graph has only one cycle, we can return ‘YES’.

Step-by-Step Solution Process

1. Constructing the Graph

Based on the provided information, we construct the graph in list form. Each participant’s designated recipient is represented by their index in the list.

2. Using DFS or BFS

We can traverse the graph using DFS (Depth First Search) or BFS (Breadth First Search). The goal is to determine whether all participants can receive a gift. This involves checking each node for visit status while traversing the graph.

3. Cycle Detection

We check if a cycle exists that allows every participant to give and receive gifts. This cycle must include all nodes, and if there is exactly one cycle, everyone will be able to receive a gift.

4. Implementation

Based on the above methods, we implement a Python code:

def can_give_gifts(n, connections):
    visited = [False] * n
    count = 0

    def dfs(v):
        nonlocal count
        visited[v] = True
        count += 1
        next_giver = connections[v]
        if not visited[next_giver]:
            dfs(next_giver)

    for i in range(n):
        if not visited[i]:
            count = 0
            dfs(i)
            if count < n:
                return "NO"

    return "YES"

# Test
n = 5
connections = [1, 2, 3, 4, 0]
print(can_give_gifts(n, connections))
        

Code Explanation

The code above is an algorithm to verify gift delivery. The 'can_give_gifts' function takes the number of participants n and the list of possible gift receivers connections as arguments. A visited list is set up to check each participant's visit status. DFS is used to visit each participant, increasing the count. If all participants are visited, it returns 'YES'; otherwise, it returns 'NO'.

Comprehensive Example

Let's examine this algorithm in action through the following example:

6
1 2 3 4 5 0
        

Expected Output

YES
        

Conclusion

This problem is a common type encountered in Python coding tests. It helps in understanding graph concepts and the method of solving problems through cycle detection. By practicing various types of problems, you can enhance your insights into algorithms.

Introduction

Algorithms are one of the most important topics in coding interviews. The ability to understand and implement algorithms is essential for programming and software development. In this course, we will study the Insertion Sort algorithm in depth and increase our understanding through relevant problems.

What is Insertion Sort?

Insertion Sort is a simple sorting algorithm that divides a given data set into a sorted list and an unsorted list, then takes data one by one from the unsorted list and inserts it into the correct position in the sorted list. This algorithm is intuitive and has the advantage of being easy to implement.

How Insertion Sort Works

Insertion Sort works through the following steps:

1. Starting from the second element, each element is compared to the current element (C).
2. If the current element (C) is smaller than the previous element (A) of the sorted list, the current element (C) is inserted into the correct position in the sorted list.
3. This process is repeated until the end of the array is reached, ultimately yielding a sorted array.

Time Complexity of Insertion Sort

The time complexity of Insertion Sort is O(n^2) in the worst case, O(n) in the best case, and O(n^2) on average. It is generally efficient when the data is nearly sorted. However, its performance can degrade when the data is distributed randomly.

Problem: Implement Insertion Sort

Let’s solve the following problem.

Problem: Sort the given list of integers using Insertion Sort.

Input: [5, 2, 9, 1, 5, 6]
Output: [1, 2, 5, 5, 6, 9]

    

Solution Process

To solve the problem above, we will implement the Insertion Sort algorithm. Below is the code that implements Insertion Sort using Python.

def insertion_sort(arr):
    # Start from the second element of the list
    for i in range(1, len(arr)):
        key = arr[i]  # Current element
        j = i - 1  # Previous index of the current element

        # Find the correct position in the sorted list
        while j >= 0 and key < arr[j]:
            arr[j + 1] = arr[j]  # Move the element to the right
            j -= 1  # Decrease the index
        
        arr[j + 1] = key  # Insert the current element in the correct position
    return arr

# Example list
example_list = [5, 2, 9, 1, 5, 6]
sorted_list = insertion_sort(example_list)
print(sorted_list)

    

Explanation of the Code

The code above has the following structure:

1. def insertion_sort(arr): - Defines the Insertion Sort function.
2. for i in range(1, len(arr)): - Begins the iteration from the second element.
3. key = arr[i] - Stores the current element.
4. while j >= 0 and key < arr[j]: - Looks for an element in the sorted list that is larger than the current element.
5. arr[j + 1] = arr[j] - Moves the element to the right to make space.
6. arr[j + 1] = key - Inserts the current element in the correct position.
7. Finally, it returns the sorted array.

Advantages and Disadvantages of Insertion Sort

Advantages

• It is simple and intuitive to implement.
• It is particularly efficient when the data is nearly sorted.
• As an in-place sorting algorithm, it uses little additional space.

Disadvantages

• Its time complexity of O(n^2) does not offer good performance in absolute terms.
• It is inefficient when the data is distributed randomly.

Conclusion

In this course, we learned about the Insertion Sort algorithm. Insertion Sort is a simple yet useful sorting algorithm that can often appear in coding tests. Understanding the principles of the algorithm and practicing its rigorous implementation is important to solve given problems. In the next course, we will cover another sorting algorithm.

References

• Insertion Sort - Wikipedia
• Insertion Sort - GeeksforGeeks

Problem Description

A person wants to find a word in a dictionary. This dictionary contains words sorted in alphabetical order. Write a program to find the first occurrence of a given word in the dictionary.

The input consists of a sorted list of words and the word to be searched. If the word does not exist in the dictionary, return -1.

Input Format

• List of words: words (in list form, each word as a string)
• The word to search for: target (string)

Output Format

Return the index of the word to be searched (0-based index). If the word does not exist, return -1.

Example

    Input: 
    words = ["apple", "banana", "cherry", "date", "fig", "grape"]
    target = "cherry"

    Output: 2
    

Approach to the Problem

This problem can be solved using the Binary Search algorithm. Binary search is an efficient algorithm for finding a specific value in sorted data, with a time complexity of O(log n).

Overview of the Binary Search Algorithm

Binary search proceeds as follows:

1. Find the middle element of the list.
2. Check if the middle element matches the value being searched for.
3. If it matches, return the index of the middle element.
4. If it does not match, continue searching in the left part if the value to be searched is smaller than the middle element, or in the right part if it is larger.
5. Repeat this process until the condition is satisfied.

Implementation

Here we will implement the binary search algorithm using Python. Below is the dictionary-finding function using binary search.

def binary_search(words, target):
    left, right = 0, len(words) - 1

    while left <= right:
        mid = (left + right) // 2
        if words[mid] == target:
            return mid
        elif words[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

Code Explanation

The above binary_search function operates as follows:

1. Use left and right variables to set the search range. The initial values are 0 and len(words) - 1, respectively.
2. Using the while left <= right loop, repeat until the search range is valid.
3. Calculate the middle index mid (through integer division).
4. If words[mid] is equal to target, return mid to return the index.
5. If not, update left to mid + 1 if words[mid] is less than target, or update right to mid - 1 if it is larger.
6. If the search ends without finding the target word, return -1.

Test Cases

Let’s add a few cases to test the functionality of the code.

if __name__ == "__main__":
    words = ["apple", "banana", "cherry", "date", "fig", "grape"]
    target = "cherry"
    result = binary_search(words, target)
    print(f"'{target}' is at index {result}.")  # Output: 'cherry' is at index 2.

    target = "orange"
    result = binary_search(words, target)
    print(f"'{target}' is at index {result}.")  # Output: 'orange' is at index -1.
    

Conclusion

In this lesson, we solved the problem of finding a specific word in a given dictionary using Python’s binary search algorithm. This algorithm is useful for quickly finding values in a sorted list and can be applied to various problems. Understanding and utilizing binary search is highly beneficial for coding tests and real programming tasks.

References

• Wikipedia – Binary Search
• Geeks for Geeks – Binary Search