Author: root

Swift Coding Test Course, Finding the Number of Steps

Hello everyone! Today, we will tackle one of the frequently appearing problems in coding tests, the ‘Counting Stair Numbers‘ problem. In this article, we will explore the definition of the stair number problem, the solution methods, and step-by-step Swift code examples. Ultimately, we will include efficient and concise code implementations along with various test cases.

Problem Definition

A stair number (n) is a number of length n, meaning that the difference between two consecutive digits is 1. For example, 123, 234, and 321 are stair numbers. However, 122 and 3456 are not stair numbers. Your task is to calculate the count of n-digit stair numbers for a given n.

Examples of the Problem

Let’s take an example of finding stair numbers with a given number of digits (n) using the digits from 1 to 9. Here are some examples:

  • n = 1: Result = 9 (1, 2, 3, 4, 5, 6, 7, 8, 9)
  • n = 2: Result = 17 (10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 90)
  • n = 3: Result = 32 (101, 121, 123, 210, 212, …)

Approach to Solve the Problem

To solve this problem, we can use the Dynamic Programming technique. We solve the problem by utilizing the relationship between one-digit numbers and numbers above it, based on the properties of stair numbers.

Dynamic Programming Approach

First, we set the length of the stair number to n and establish a dp array to store the possible counts. dp[i][j] represents the count of stair numbers of length i with the last digit j. This allows us to set up a recurrence relation:

        dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1]

Here, j takes on values from 0 to 9. If the last digit is 0, the previous digit can only be 1, and if the last digit is 9, the previous digit can only be 8. Note that intermediate values can go both ways.

Basic Setup

Now, let’s initialize the DP table for solving the problem. Since the first digit of the stair numbers can only be from 1 to 9:

    for j in 0...9 {
        dp[1][j] = 1
    }

Next, we will set up a loop from the second digit up to n digits to fill the dp array.

Swift Code Implementation

Now, based on what we’ve explained above, let’s implement the code in Swift.

    func countStairNumbers(n: Int) -> Int {
        // Array to store the count of stair numbers for each digit
        var dp = Array(repeating: Array(repeating: 0, count: 10), count: n + 1)

        // For 1-digit numbers, there is 1 case for each from 1 to 9
        for j in 1...9 {
            dp[1][j] = 1
        }

        // Fill the DP array
        for i in 2...n {
            for j in 0...9 {
                if j > 0 {
                    dp[i][j] += dp[i - 1][j - 1]
                }
                if j < 9 {
                    dp[i][j] += dp[i - 1][j + 1]
                }
            }
        }

        var totalCount = 0
        // Sum up all cases for n-digit numbers
        for j in 0...9 {
            totalCount += dp[n][j]
        }

        return totalCount
    }

    // Usage example
    let result = countStairNumbers(n: 3)
    print("Count of 3-digit stair numbers: \(result)") // Output: 32

Code Explanation

The above code returns the count of n-digit stair numbers for a given n. The function sets up the dp array and calculates each possible case by adding the possibilities for each digit. Ultimately, it returns the total count of all n-digit numbers.

Test Cases

Let's create a few test cases to verify the correct operation of the function by checking the results for each digit count:

    print("1-digit count: \(countStairNumbers(n: 1))") // 9
    print("2-digit count: \(countStairNumbers(n: 2))") // 17
    print("3-digit count: \(countStairNumbers(n: 3))") // 32
    print("4-digit count: \(countStairNumbers(n: 4))") // X (to be computed and verified)
    print("5-digit count: \(countStairNumbers(n: 5))") // Y (to be computed and verified)

Conclusion

Today, we learned how to solve the counting stair numbers problem using Swift. We were able to devise an efficient solution to this problem through dynamic programming. I encourage you to understand and utilize this problem to tackle more algorithmic challenges!

Then, we'll see you next time with more interesting topics. Thank you!

Swift Coding Test Course, Finding the Product of an Interval

One of the common problems encountered in programming interviews or coding tests is ‘Calculating the Product of an Interval’. This problem requires a deep understanding of data structures and algorithms through the efficient calculation of the product over a specific interval of a given array. In this chapter, we will understand this problem and explore efficient solutions.

Problem Definition

Given an integer array nums and multiple queries queries, the task is to return the product of the interval defined by each query. The product of an interval refers to the result of multiplying all the numbers within that interval.

Example Problem

Example: 
Input: nums = [1, 2, 3, 4, 5]
Queries: queries = [[0, 1], [1, 3], [2, 4]]
Output: [2, 24, 60]
Explanation:
- The first query [0, 1] returns nums[0] * nums[1] = 1 * 2 = 2.
- The second query [1, 3] returns nums[1] * nums[2] * nums[3] = 2 * 3 * 4 = 24.
- The third query [2, 4] returns nums[2] * nums[3] * nums[4] = 3 * 4 * 5 = 60.

Solution Approach

The problem of calculating the product of an interval can be solved using a basic double loop. However, if the number of queries is large and the array is big, a double loop is inefficient, and we should consider other methods. Particularly, if the number of queries is N and each query takes O(M) time to calculate the interval product, the worst-case time complexity is O(N * M). Therefore, we need to find a method that can solve it in O(N + Q) time.

Calculating Interval Products in Linear Time Complexity

We can solve this problem using a technique called prefix product (pre-computation). This method helps to quickly calculate the product of an interval using the starting and ending indices of each query.

Calculating Prefix Product

First, we construct a prefix product array. The prefix product array is defined with the name prefixProduct, where prefixProduct[i] represents nums[0] * nums[1] * ... * nums[i]. This way, the product of the interval [i, j] can be simply calculated as prefixProduct[j] / prefixProduct[i-1].

Swift Code Implementation


import Foundation

func rangeProduct(_ nums: [Int], _ queries: [[Int]]) -> [Int] {
    var prefixProduct = [Int](repeating: 1, count: nums.count + 1)
    
    // Create the prefix product array
    for i in 0..

Code Analysis

The code above works through the following procedures:

  1. Creating the Prefix Product Array: It iterates through the nums array to calculate the cumulative product at each position. The time complexity of this process is O(N).
  2. Processing Queries: It checks each given query one by one and calculates the interval product. The time complexity for processing each query is O(1).
  3. Therefore, the overall time complexity is O(N + Q).

Conclusion

Through this tutorial, we have learned how to efficiently solve the problem of calculating the product of an interval. Moving away from the basic double loop approach to utilizing prefix products is a method that can be applied beneficially in many algorithmic problems. Such problems are also used in various applications for data processing in reality, so it is important to understand and utilize this problem well.

I wish you success in taking on more coding challenges and building your skills!

Swift Coding Test Course, Pathfinding

Hello, in this lecture, we will learn how to solve the pathfinding problem using the Swift language. Pathfinding is one of the common types of problems that often appear in coding tests, where the task is to find a path between two specific nodes in a given graph. In this article, we will define the problem and thoroughly discuss the algorithms and data structures needed to solve it.

Problem Definition

Let’s consider the following problem:

Problem: Write a function to determine whether there exists a path between two nodes A and B in a given directed graph. The graph is provided in the form of an adjacency list, with nodes and edges represented as integers.

Input:

  • Integer N: the number of nodes (1 ≤ N ≤ 1000)
  • Integer M: the number of edges (1 ≤ M ≤ 10000)
  • Length of the list M: each edge indicates a direction from node u to node v in the form [u, v].
  • Integer A, B: the starting node A and the ending node B to check for a path.

Output:

  • If there exists a path from node A to node B, print “YES”; if not, print “NO”.

Problem Analysis

This problem falls under the category of ‘path search’ in graph theory and can be solved using various methods. You can explore paths using DFS (Depth-First Search) or BFS (Breadth-First Search), both of which can effectively explore graphs in the form of adjacency lists.

Algorithm Selection

In this lecture, we will use BFS to explore the graph. BFS is an algorithm that uses a queue to explore each node level by level, making it advantageous for finding the shortest path. We will check if we can reach the destination node through graph traversal.

Swift Code Implementation

Now, let’s write the actual Swift code. Below is an example of a pathfinding function that uses BFS.

        
        import Foundation

        func canReach(start: Int, end: Int, edges: [[Int]], nodeCount: Int) -> String {
            var adjacencyList = [[Int]](repeating: [Int](), count: nodeCount + 1)
            for edge in edges {
                adjacencyList[edge[0]].append(edge[1])
            }

            var queue = [start]
            var visited = [Bool](repeating: false, count: nodeCount + 1)
            visited[start] = true

            while !queue.isEmpty {
                let current = queue.removeFirst()

                if current == end {
                    return "YES"
                }

                for neighbor in adjacencyList[current] {
                    if !visited[neighbor] {
                        visited[neighbor] = true
                        queue.append(neighbor)
                    }
                }
            }
            return "NO"
        }

        // Example input
        let nodeCount = 6
        let edges = [[1, 2], [1, 3], [2, 4], [4, 5], [3, 6]]
        let start = 1
        let end = 5

        print(canReach(start: start, end: end, edges: edges, nodeCount: nodeCount))

Code Explanation

Analyzing the above code, it consists of the following steps:

  1. Create a graph in the form of an adjacency list. Store connected nodes for each node.
  2. Initialize a queue for BFS traversal and mark the starting node as visited.
  3. Repeat the following process until the queue is empty:
    • Take a node from the front of the queue.
    • If the taken node is the destination node, return “YES”.
    • For all adjacent nodes of the current node, add unvisited nodes to the queue and mark them as visited.
  4. If the queue is empty but the target node has not been reached, return “NO”.

Complexity Analysis

The time complexity of the BFS algorithm is O(V + E), where V is the number of nodes and E is the number of edges. Considering the maximum conditions given in this problem, it is very efficient. The memory complexity also requires O(V + E) to store the adjacency list.

Testing and Applications

The given function can be applied to various graph structures and tested accordingly. Let’s look at a few additional test cases.

        
        // Additional test cases
        let additionalEdges1 = [[1, 2], [2, 3], [3, 4], [4, 5]]
        let additionalEdges2 = [[1, 2], [2, 3], [3, 4], [2, 5]]
        
        print(canReach(start: 1, end: 5, edges: additionalEdges1, nodeCount: nodeCount)) // NO
        print(canReach(start: 1, end: 5, edges: additionalEdges2, nodeCount: nodeCount)) // YES

Conclusion

In this lecture, we learned how to solve the pathfinding problem in graphs using Swift. We effectively solved the problem using the BFS algorithm, which is very important in coding tests. When encountering similar problems in practice, you will be able to solve them based on the knowledge gained from this lecture.

Now it’s your turn to tackle various graph problems and enhance your algorithm skills. Keep practicing!

Swift Coding Test Course, Game Development

Game development is not only an interesting field but also requires complex algorithms and problem-solving skills.
In this article, we will look at a coding test problem using the Swift language and detail the process of solving it.
This course will be particularly helpful for programmers interested in game development.
When a problem is presented, we will analyze the process of solving it step by step.

Problem Description

Our goal is to develop a level of the game in which N monsters appear.
Each monster has its own health points (Hit Points, HP) and damage attributes,
and the player can attack the monsters with a specific attack power.
In this problem, the player must find the optimal attack strategy to defeat all the monsters.

Problem: Numerical Optimization

Based on the HP and damage of the given monsters,
calculate the minimum number of attacks required for the player to defeat the monsters.
The player has a fixed attack power for each attack, and
the monsters decrease their HP each time they are attacked.
The monsters continuously receive attacks until their HP reaches 0 or below.

Input Format

  • The first line contains the number of monsters N (1 ≤ N ≤ 1000).
  • The second line contains the HP of N monsters separated by spaces. (1 ≤ HP ≤ 10000)
  • The third line contains the damage of N monsters separated by spaces. (1 ≤ Damage ≤ 100)
  • The last line contains the player’s attack power A (1 ≤ A ≤ 10000).

Output Format

Print the minimum number of attacks.

Problem Approach

To solve this problem, the following approach can be used:

  1. For each monster, knowing the HP and the player’s attack power,
    we need to calculate the number of attacks required to defeat the monster.
  2. Calculate the number of attacks needed to defeat each monster by dividing the monster’s HP by the player’s attack power.
  3. By summing up the required attack counts for each monster,
    we can derive the final result, the minimum number of attacks.

Code Implementation

Now let’s write the Swift code:


    import Foundation

    func minimumAttacksToDefeatMonsters(hpList: [Int], damageList: [Int], attackPower: Int) -> Int {
        var totalAttacks = 0

        for hp in hpList {
            // Calculate the number of attacks required for each monster
            let requiredAttacks = (hp + attackPower - 1) / attackPower
            totalAttacks += requiredAttacks
        }
        
        return totalAttacks
    }

    // Example input
    let n = 3
    let hpList = [100, 200, 300]
    let damageList = [50, 70, 90] // Not used but provided as a condition of the problem
    let attackPower = 75

    let result = minimumAttacksToDefeatMonsters(hpList: hpList, damageList: damageList, attackPower: attackPower)
    print("Minimum number of attacks: \(result)")

Code Explanation

In the above code, we calculated the minimum number of attacks required to defeat the monsters based on the player’s attack power and the monsters’ HP.
The minimumAttacksToDefeatMonsters function iterates over the given HP list and calculates the required number of attacks for each monster.

First, we compute the value of the monster’s HP divided by the player’s attack power.
If the monster’s HP does not divide evenly by the player’s attack power,
then we need to round up the number of attacks required to defeat the monster.
In this case, we use (hp + attackPower - 1) / attackPower to
calculate the required number of attacks. Finally, we sum up the
calculated attack counts for all monsters and return the final result.

Performance Analysis

The time complexity of this algorithm is O(N). This is because we simply calculate the number of attacks required for each monster.
Since the maximum N is 1000, this algorithm is very efficient in terms of performance.

Conclusion

In this tutorial, we covered a coding test problem related to game development using Swift.
By designing an algorithm to systematically defeat monsters,
we learned methods for solving algorithmic problems in game development.
We encourage you to use this knowledge to develop strategies applicable to various games.

Swift Coding Test Course, I Don’t Want to Be a Liar

This is a course where you can practice solving various problems for effective algorithm problem-solving. The problem we will tackle today is based on the theme ‘I Don’t Want to Be a Liar’. Through this problem, you will learn statistical thinking, conditional statements, and list processing techniques.

Problem Description

Recently, a participant attended a wedding where he was elected as a slacker. However, there were 100 friends at the wedding whom he did not want to invite. These friends are planning to introduce others as their friends by lying.

Each friend can claim, ‘I know A and B’, about specific individuals. Your task is to verify all the lies and determine whether two friends know the same person.

The given input includes the number of friends and the lying information of each friend. Based on the input information, you need to determine which friends are lying.

Input Example

                5
                1 2
                2 3
                3 4
                4 5
                1 5

The first line contains the number of friends N, and the following N lines provide the numbers of the friends each friend knows.

Output Example

                Yes

If the two friends know each other, print ‘Yes’; otherwise, print ‘No’.

Approach to the Problem

To solve this problem, we can use graph search algorithms. We represent each friend as a vertex and their relationships as edges to form a graph. We can use search methods such as DFS or BFS to check if two friends know each other.

The basic approach is to store friend relationships in a list and conduct a graph search for the given two friends.

Code Implementation

Below is the implementation of the code in Swift. 

                
                import Foundation

                func canKnowEachOther(N: Int, friendships: [(Int, Int)], friendA: Int, friendB: Int) -> String {
                    // Initialize graph
                    var graph = Array(repeating: [Int](), count: N + 1)

                    // Store friendships
                    for (a, b) in friendships {
                        graph[a].append(b)
                        graph[b].append(a)
                    }

                    // Initialize BFS
                    var queue = [friendA]
                    var visited = Array(repeating: false, count: N + 1)
                    visited[friendA] = true

                    while !queue.isEmpty {
                        let current = queue.removeFirst()

                        // Did we find friend B?
                        if current == friendB {
                            return "Yes"
                        }

                        for neighbor in graph[current] {
                            if !visited[neighbor] {
                                visited[neighbor] = true
                                queue.append(neighbor)
                            }
                        }
                    }

                    return "No"
                }

                // Input example
                let N = 5
                let friendships = [(1, 2), (2, 3), (3, 4), (4, 5), (1, 5)]
                let result = canKnowEachOther(N: N, friendships: friendships, friendA: 1, friendB: 5)
                print(result) // "Yes"

Code Explanation

The above code operates as follows:

  • Graph Initialization: Initializes the graph in the form of an adjacency list based on the input friendships.
  • BFS Search: Starts from the given friend A and executes a BFS to check if we reach friend B. Already visited friends are tracked to avoid revisiting.
  • Return Result: If friend B is found, it returns “Yes”; if not found after finishing the search, it returns “No”.

Additional Considerations

The method for solving this problem is quite intuitive as it involves basic DFS/BFS searching. However, if the scope of the problem increases, we may need to explore more efficient methods. For example, if the number of friends becomes large, applying the Union-Find algorithm instead of DFS may be performance advantageous.

If more complex relationships or random configurations of data are provided, various techniques (e.g., Dynamic Programming options) can be utilized to address them. It is crucial to consider worst-case time and space complexity when selecting the algorithm design.

This article is part of a course on solving various problems related to the Swift coding test. Through understanding the problem and exploring solutions, we hope to enhance your problem-solving skills.