Swift Coding Test Course, Ax + By = C

Problem Description

Problem: Write a program to check whether there exists a solution for Ax + By = C given the integers A, B, and C. You should also provide a method to find all possible solutions.

Input:

– Integers A, B, C (-(10^9) ≤ A, B, C ≤ 10^9)

Output:

– If a solution exists, print “A solution exists.” and output an arbitrary solution in the form of x, y.

– If no solution exists, print “No solution exists.”

Problem Approach

This problem is about determining whether a linear equation has a solution. There are various theories that can indicate whether a solution exists based on the values of A, B, and C, but we will approach it through a basic method commonly used in systems of equations.

Basic Theory

First, the equation Ax + By = C can be graphically interpreted in a two-dimensional coordinate system. If both A and B are 0, the equation does not hold, and if either A or B is 0, there is a single solution. Excluding these cases, typical scenarios can have different solutions.

The conditions for Ax + By = C to have a solution are as follows:

  • If A = 0 and B = 0, there are infinitely many solutions if C is 0; otherwise, there is no solution.
  • If A = 0, to have a solution (with respect to B), it must be a multiple of B.
  • If B = 0, to have a solution (with respect to A), it must be a multiple of A.
  • In all other cases, a solution exists.

Solution Algorithm

  1. Check the existence of a solution using the values of A and B.
  2. Output appropriately based on the existence of a solution.
  3. If a solution exists, derive the solution using the slope and intercept of the line that includes the origin.

Swift Code

The following Swift code implements the algorithm described above.

import Foundation

func findSolution(A: Int, B: Int, C: Int) {
    if A == 0 && B == 0 {
        if C == 0 {
            print("A solution exists. (Infinitely many solutions)")
        } else {
            print("No solution exists.")
        }
    } else if A == 0 {
        // When B is not 0
        if C % B == 0 {
            let y = C / B
            print("A solution exists. (x is a free variable) -> x: any integer, y: \(y)")
        } else {
            print("No solution exists.")
        }
    } else if B == 0 {
        // When A is not 0
        if C % A == 0 {
            let x = C / A
            print("A solution exists. (y is a free variable) -> x: \(x), y: any integer")
        } else {
            print("No solution exists.")
        }
    } else {
        print("A solution exists. (Arbitrary solution - x: 0, y: \(C / B))")
    }
}

// Input values
let A = 2
let B = 3
let C = 6

findSolution(A: A, B: B, C: C)

Execution and Result

When the above Swift code is executed, it will output whether a solution exists for Ax + By = C. For example, if A = 2, B = 3, C = 6, the output will be as follows:

A solution exists. (Arbitrary solution - x: 0, y: 2)

Conclusion

In this tutorial, we explored the existence of solutions for linear equations in the form of Ax + By = C and the process of finding such solutions. By understanding the basics of algorithm problem-solving and writing code based on this, you can develop useful skills for coding tests or real programming situations.

Additionally, experimenting with various values of A, B, and C to analyze the patterns of solutions can be very helpful. In fact, providing various examples of both existing and non-existing solutions will greatly aid in understanding the problem.

Swift Coding Test Course, 2 N Tile Filling

Hello, everyone! In this post, we will discuss one of the coding test problems using Swift called “2*N Tile Filling.” This problem is one of the very useful problems for understanding the basics of dynamic programming. Below, I will detail the problem description and the process of solving it.

Problem Description

The problem is as follows. We need to find the number of ways to fill a rectangle that is 2 units high and N units wide with tiles of size 2×1 or 1×2. In other words, we need to count the number of ways to completely fill a 2*N rectangle using 2×1 and 1×2 tiles.

Examples

Here are examples for a few small values of N.

  • N=1: 1 (filling with one 1×2 tile)
  • N=2: 2 (filling with two 2×1 tiles or two 1×2 tiles)
  • N=3: 3 (filling with a combination of 1×2 and 2×1 tiles)
  • N=4: 5

Problem Solving Approach

This problem can be solved using dynamic programming. First, we need to set up the recurrence relation to solve the problem. In this problem, we can consider the following two cases.

  1. If the last column has a 1×2 tile: In this case, the size of the remaining problem is 2*(N-1).
  2. If the last row has a 2×1 tile: In this case, the size of the remaining problem is 2*(N-2).

By combining these two cases, we obtain the following recurrence relation.

f(N) = f(N-1) + f(N-2)

The initial cases are as follows.

  • f(1) = 1
  • f(2) = 2

Swift Code Implementation

Now, let’s implement the recurrence relation in Swift. During the implementation, we will use the memoization technique for efficiency.


    func tileWays(_ n: Int) -> Int {
        // Array for memoization
        var memo = Array(repeating: 0, count: n + 1)

        // Setting initial cases
        memo[1] = 1
        if n > 1 {
            memo[2] = 2
        }

        for i in 3...n {
            memo[i] = memo[i - 1] + memo[i - 2]
        }
        
        return memo[n]
    }

    // Example test
    let n = 4
    print("The number of ways to fill a height 2 and width \(n) with tiles: \(tileWays(n))")
    

Code Explanation

The above code works as follows:

  1. First, it declares a memo array initialized to 0. This array stores the number of ways to fill the tiles for each N.
  2. It sets the initial cases. It sets 1 for N=1 and 2 for N=2.
  3. It iterates from 3 to N, updating the memo array. For each case, it applies the recurrence relation to store values in the memo array.
  4. Finally, it returns memo[n] to provide the answer for N.

Time Complexity

This algorithm has a time complexity of O(N), which is proportional to N. By using memoization, we avoid redundant calculations and efficiently compute each case only once using the array.

Conclusion

In this post, we covered the “2*N Tile Filling” problem using Swift. Through understanding the problem and the solving process, we learned the basics of dynamic programming. I encourage you to solve more problems through practice.

In the next post, we will deal with another interesting algorithm problem. Thank you!

Swift Coding Test Course, 022 Sorting Numbers 3

Hello! This time, let’s solve a coding test problem based on Swift called “Sorting Numbers 3”. This problem may seem simple as it involves sorting numbers, but it has specific conditions and constraints, making it a good practice for coding tests. In this article, we will explore the problem definition, input and output formats, algorithmic approaches, and optimization methods in detail.

Problem Description

The requirement of the problem is to sort the given numbers, but the range of numbers to be sorted is limited. Specifically, the range of numbers is integers between 1 and 100,000, and the objective is to sort these integers in ascending order and output them.

For example, let’s assume the following numbers are given:

  • 5
  • 3
  • 8
  • 1
  • 2

In this case, the output should be displayed as follows:

  • 1
  • 2
  • 3
  • 5
  • 8

Input Format

The input is provided in the following format:

  1. The first line will contain the number of integers N. (1 ≤ N ≤ 100,000)
  2. The second line contains N integers separated by spaces.

Output Format

The output should print each sorted number on a new line.

Solution Approach

This problem involves sorting numbers, so we can use the most basic sorting algorithms. However, since the maximum value of N is 100,000, we cannot use inefficient algorithms like Bubble Sort or Selection Sort that have a time complexity of O(N^2).

We will use the Counting Sort algorithm to solve this problem. Counting Sort is a method that sorts efficiently when the range of given numbers is limited. This algorithm involves the following steps:

  1. Create an array corresponding to the range of input numbers.
  2. Record the count of each input number at the respective index.
  3. Finally, output the sorted numbers based on their counts.

Code Implementation

Now let’s write code to solve the problem in Swift. Since Swift allows the use of arrays as fields, implementing Counting Sort is very straightforward. Below is an example of the implementation:


    import Foundation

    func countingSort(numbers: [Int]) -> [Int] {
        // Since the range of numbers is 1 to 100,000, initialize an array of size 100,001
        var counts = Array(repeating: 0, count: 100001)

        // Count the occurrences of each number
        for number in numbers {
            counts[number] += 1
        }

        var sortedNumbers: [Int] = []
        
        // Generate sorted numbers based on the counts
        for (number, count) in counts.enumerated() {
            for _ in 0..

Example Execution

Let's use the code above to provide input. For instance, suppose we provide the following input:


    5
    5 3 8 1 2
    

The output result should be as follows:


    1
    2
    3
    5
    8
    

Complexity Analysis

The time complexity of this algorithm is O(N + K). Here, N is the number of input integers, and K is the range of numbers. In this case, K is fixed at 100,000, making the algorithm highly efficient. The space complexity also requires O(K), taking up O(100,000) space.

Conclusion

In this article, we examined the problem "Sorting Numbers 3" and implemented a solution using Counting Sort. Problems like these enhance understanding of basic algorithms and are frequently encountered in actual coding tests, so be sure to practice them. In the next tutorial, we will tackle more challenging problems. Thank you!

JavaScript Coding Test Course, Extended Euclidean Algorithm

Hello, everyone! Today, we will learn about one of the important algorithms in coding tests using JavaScript, the Extended Euclidean Algorithm. In this course, we will provide the concept of the Extended Euclidean Algorithm, the problem-solving process using it, and practical code examples.

1. Problem Description

We will define the problem as follows. Given two integers A and B, the task is to find integers X and Y that satisfy AX + BY = GCD(A, B). Here, GCD refers to the greatest common divisor.

Example

Input: A = 30, B = 21

Output: X = 1, Y = -1, GCD = 3

Solution: 30 * 1 + 21 * (-1) = 3

2. Concept Explanation

The Extended Euclidean Algorithm not only calculates the greatest common divisor (GCD) of two integers but also finds specific coefficients from it. This is primarily used in the following formula:

AX + BY = GCD(A, B)

Here, A and B are the given two integers, and X and Y are the integers we want to find. If the GCD is 1, A and B are coprime, and X and Y can also be used to find modular inverses.

3. Approach

We will implement the Extended Euclidean Algorithm based on the general Euclidean algorithm for finding GCD. The main idea of the algorithm is as follows:

  1. Receive two integers A and B as input.
  2. If B is 0, then the GCD is A, and X is 1, Y is 0.
  3. Otherwise, recursively call with B and A % B using the Euclidean algorithm.
  4. Use the results of the recursive call to calculate the values of X and Y.

4. Algorithm Implementation

Below is an example of implementing the Extended Euclidean Algorithm in JavaScript:


function extendedGCD(a, b) {
    if (b === 0) { // Base case
        return { gcd: a, x: 1, y: 0 };
    }
    // Recur with the new parameters b and a % b
    const { gcd, x: x1, y: y1 } = extendedGCD(b, a % b);
    const x = y1;
    const y = x1 - Math.floor(a / b) * y1;
    return { gcd, x, y };
}

// Test the function with example values
const a = 30;
const b = 21;
const { gcd, x, y } = extendedGCD(a, b);
console.log(`GCD: ${gcd}, X: ${x}, Y: ${y}`);

5. Code Explanation

In the above code, we are recursively calculating the GCD. In the base case, when B is 0, the GCD is A, and at that point, X is 1, Y is 0. After that, we calculate new values for X and Y using the returned X and Y values to ultimately get the result we desire.

6. Test Cases

Now let’s test the function with various test cases.


// Test cases
const testCases = [
    { a: 30, b: 21 },
    { a: 48, b: 18 },
    { a: 56, b: 15 },
    { a: 101, b: 10 },
];

testCases.forEach(({ a, b }) => {
    const { gcd, x, y } = extendedGCD(a, b);
    console.log(`A: ${a}, B: ${b} => GCD: ${gcd}, X: ${x}, Y: ${y}`);
});

7. Conclusion

Today, we learned about the Extended Euclidean Algorithm. This algorithm is very useful for finding the greatest common divisor of two integers and for finding specific coefficients related to it. It is especially used in modular arithmetic and complex algorithm problems, so it is important to understand and practice it thoroughly.

I hope the algorithm used in this article will help you in your coding exam preparations. If you have any additional questions, please leave them in the comments!

JavaScript Coding Test Course, Traversing Trees

Overview

In coding tests, various data structure and algorithm problems are presented. Among them, trees are a commonly occurring data structure.
Tree structures play a very important role in computer science and are utilized in various fields such as file systems and databases.
In this course, we will learn how to traverse trees using JavaScript.

What is a Tree Structure?

A tree is a nonlinear data structure composed of nodes and edges, optimized for representing hierarchical relationships.
A tree has concepts such as root node, child node, parent node, and leaf node.

The main characteristics of a tree are as follows:

  • A tree has one root, and child nodes are connected from this root.
  • A node can have zero or more child nodes.
  • A leaf node is a node that has no children.

Tree Traversal Methods

There are several ways to traverse a tree, with the most commonly used methods being:

  • Pre-order Traversal
  • In-order Traversal
  • Post-order Traversal
  • Level-order Traversal

The order in which nodes are visited differs for each traversal method. Let’s take a closer look at each method.

Pre-order Traversal

The method of pre-order traversal is as follows:

  1. Visit the current node.
  2. Traverse the left subtree in pre-order.
  3. Traverse the right subtree in pre-order.

For example, suppose we have the following tree structure.

                Public
                ├── User 1
                │   ├── User 1.1
                │   └── User 1.2
                └── User 2
                    ├── User 2.1
                    └── User 2.2
                

The result of the pre-order traversal is “Public, User 1, User 1.1, User 1.2, User 2, User 2.1, User 2.2”.

In-order Traversal

The method of in-order traversal is as follows:

  1. Traverse the left subtree in in-order.
  2. Visit the current node.
  3. Traverse the right subtree in in-order.

For example, in the same tree structure, the result of the in-order traversal is “User 1.1, User 1, User 1.2, Public, User 2.1, User 2, User 2.2”.

Post-order Traversal

The method of post-order traversal is as follows:

  1. Traverse the left subtree in post-order.
  2. Traverse the right subtree in post-order.
  3. Visit the current node.

In the same tree structure, the result of the post-order traversal is “User 1.1, User 1.2, User 1, User 2.1, User 2.2, User 2, Public”.

Level-order Traversal

The method of level-order traversal is as follows:

  1. Visit the root node.
  2. Visit the child nodes of the current node.
  3. After visiting all child nodes, move to the next depth.

In the same tree structure, the result of the level-order traversal is “Public, User 1, User 2, User 1.1, User 1.2, User 2.1, User 2.2”.

Programming Problem: Binary Tree Traversal

Given the following binary tree structure, write a function to traverse the tree using various traversal methods.
A binary tree is composed of nodes structured as follows:

            class TreeNode {
                constructor(value) {
                    this.value = value;
                    this.left = null;
                    this.right = null;
                }
            }
            

Example Input:

            const root = new TreeNode(1);
            root.left = new TreeNode(2);
            root.right = new TreeNode(3);
            root.left.left = new TreeNode(4);
            root.left.right = new TreeNode(5);
            

Problem

Write a function for pre-order, in-order, post-order, and level-order traversal of the binary tree above.

Problem Solving Process

1. Implementing Pre-order Traversal

To perform pre-order traversal, a recursive approach is needed. Below is the code that implements this:

            function preOrderTraversal(node) {
                if (node === null) return;
                console.log(node.value); // Visit current node
                preOrderTraversal(node.left); // Visit left subtree
                preOrderTraversal(node.right); // Visit right subtree
            }
            

The above code visits the current node first and then traverses the left and right nodes.

2. Implementing In-order Traversal

In-order traversal is also implemented recursively. Below is the in-order traversal code:

            function inOrderTraversal(node) {
                if (node === null) return;
                inOrderTraversal(node.left); // Visit left subtree
                console.log(node.value); // Visit current node
                inOrderTraversal(node.right); // Visit right subtree
            }
            

This code visits the left subtree first and then the current node.

3. Implementing Post-order Traversal

Post-order traversal is also implemented recursively. Below is the implemented code:

            function postOrderTraversal(node) {
                if (node === null) return;
                postOrderTraversal(node.left); // Visit left subtree
                postOrderTraversal(node.right); // Visit right subtree
                console.log(node.value); // Visit current node
            }
            

In post-order traversal, the current node is visited after both child subtrees.

4. Implementing Level-order Traversal

Level-order traversal is implemented using a queue data structure. By using a queue, each node can be visited layer by layer. Below is the level-order traversal code:

            function levelOrderTraversal(root) {
                if (root === null) return;
                const queue = [root]; // Initialize the queue
                while (queue.length > 0) {
                    const current = queue.shift(); // Remove node from the queue
                    console.log(current.value); // Visit current node
                    if (current.left) queue.push(current.left); // Add left child
                    if (current.right) queue.push(current.right); // Add right child
                }
            }
            

Using a queue allows each node to be visited in order by level.

Conclusion

In this course, we explored various methods of traversing trees using JavaScript.
Tree traversal is a fundamental part of many programming problems, so it’s important to practice sufficiently.
Understanding and implementing the pre-order, in-order, post-order, and level-order traversal algorithms covered above is a great way to achieve good results in coding tests.

Continue to solve various algorithm problems through practice. Practice and repetition are the best teachers!