Category: Swift Coding Test

Swift Coding Test Course, ‘Finding Good Numbers’

In the Subrimari coding test, problems frequently involve finding a “good number.” In this session, we will solve a problem that involves finding specific numbers that satisfy the conditions referred to as “good numbers.” This process will illustrate how to approach and solve problems when using the Swift language.

Problem Description

The problem is as follows:

For a given natural number N, a friend has defined a good number. Specifically, among the numbers from 1 to N, if the sum of any two numbers x and y equals N, then x and y are considered good numbers. You need to find and output the pairs of good numbers for the given N. Note that x must be less than or equal to y, and x + y must equal N.

Input Format

  • The first line contains the natural number N (1 ≤ N ≤ 1000).

Output Format

  • Output all pairs of good numbers (x, y) one pair per line.

Approach to the Problem

To find good numbers, we can use the following approach:

  1. Use two loops to generate all possible pairs from 1 to N.
  2. Check if the sum of each pair is N.
  3. Output the pairs whose sum equals N.

Algorithm Implementation

In Swift, we can solve the problem using the following structure:

import Foundation

func findGoodNumbers(N: Int) {
    var goodPairs: [(Int, Int)] = []
    
    for x in 1...N {
        for y in x...N {
            if x + y == N {
                goodPairs.append((x, y))
            }
        }
    }
    
    // Output result
    for pair in goodPairs {
        print("Good number pair: \(pair.0), \(pair.1)")
    }
}

let N = 10 // Example input
findGoodNumbers(N: N)

Code Explanation

In the Swift code above, we can see the following important details:

  1. The function findGoodNumbers takes a natural number N as a parameter and finds the pairs of good numbers for it.
  2. A nested for loop is used to generate all pairs of x and y.
  3. If the sum equals N, that pair is added to the goodPairs array.
  4. Finally, the pairs of good numbers stored in the array are printed.

Execution Result

If the input N is 10, the program produces the following output:

Good number pair: 1, 9
Good number pair: 2, 8
Good number pair: 3, 7
Good number pair: 4, 6
Good number pair: 5, 5

Complexity Analysis

The time complexity of this algorithm is O(N^2). Since we use a nested loop for N, the execution time may increase for larger values of N. However, since the maximum value of N is limited to 1000, this level of time complexity is practically feasible.

Conclusion

In this article, we explored the approach to solve the “good number” problem and implemented the solution in Swift code. We learned how to resolve the problem through a simple yet effective nested loop. I hope this helps improve algorithmic thinking through similar problems in the future.

Additional Considerations

Various methods can be explored to solve this problem. For example:

  • Using a hashmap to find good numbers with O(N) time complexity
  • When N is even, restricting the range of x to up to N/2
  • Solutions through recursive methods

Through these various approaches, a deeper understanding of algorithms can be achieved. In the next session, we will address additional problems related to this topic. Thank you!

Swift Coding Test Course, Finding the Kth Shortest Path

Problem Description

This is a problem of finding the K-th shortest path between two nodes A and B in a given graph.
The graph is a weighted directed graph. The K-th shortest path refers to the path from A to B
whose sum of weights is the K-th smallest among all paths.
K is a positive integer and K is always greater than or equal to 1.

Input Description

The first line of the input contains two integers N (1 ≤ N ≤ 100) and M (1 ≤ M ≤ 1000).
Here, N represents the number of nodes, and M represents the number of edges.
Following this, M lines contain three integers U, V, W, which represent an edge with weight W from node U to node V.
The last line contains two integers K, A, B. K indicates which K-th shortest path to find,
A is the starting node, and B is the destination node.

Output Description

Output the sum of weights of the K-th shortest path. If the K-th shortest path does not exist, output -1.

Problem Solving Process

There are various methods to solve this problem, but a variant of the ‘Dijkstra’s Algorithm’ can be used.
Dijkstra’s algorithm is typically used to find the shortest path but can utilize a priority queue
and path list to find a certain number of shortest paths.
The following describes that process.

1. Graph Representation

Represent the graph in the form of an adjacency list. Store the connected nodes and their weights for each node.

2. Use of Priority Queue

Use a priority queue to manage the weights of paths from the current node to find the shortest path.
To find the K-th path, maintain a path list for each node.

3. Path Calculation

Start from the starting node and explore paths for each node.
The lower the sum of weights of each path, the higher the priority is set,
and this process continues until the K-th shortest path is found.

4. Result Output

Once the sum of weights for the K-th shortest path is successfully calculated, that value is printed.
If no path exists, -1 is printed.

Swift Code Implementation

Below is the Swift code that uses the method described above.


import Foundation

struct Edge {
    let to: Int
    let weight: Int
}

func kthShortestPath(n: Int, edges: [[Edge]], k: Int, start: Int, end: Int) -> Int {
    var paths = [[Int]](repeating: [], count: n + 1)
    var minHeap = [(weight: Int, node: Int)]()
    
    // Insert (0, start node)
    minHeap.append((0, start))
    
    while !minHeap.isEmpty {
        // Remove the node with the smallest weight
        minHeap.sort { $0.weight < $1.weight }
        let current = minHeap.removeFirst()
        
        // Save the path
        paths[current.node].append(current.weight)
        
        // If the k-th path is found, terminate
        if paths[end].count == k {
            return current.weight
        }
        
        // Explore adjacent nodes
        for edge in edges[current.node] {
            minHeap.append((current.weight + edge.weight, edge.to))
        }
    }
    
    // K-th path does not exist
    return -1
}

// Input
let n = 5 // Number of nodes
let m = 7 // Number of edges
let edges = [
    1: [Edge(to: 2, weight: 10), Edge(to: 3, weight: 5)],
    2: [Edge(to: 3, weight: 2), Edge(to: 4, weight: 1)],
    3: [Edge(to: 2, weight: 3), Edge(to: 4, weight: 9)],
    4: [Edge(to: 5, weight: 2)],
    5: []
]
let k = 3 // K-th path
let start = 1 // Starting node
let end = 5 // Destination node

// Function call
let result = kthShortestPath(n: n, edges: edges, k: k, start: start, end: end)
print(result) // Output result

Conclusion

The K-th shortest path problem is one of the important techniques in algorithm problem solving.
Utilizing Dijkstra's algorithm and priority queues can effectively solve the problem.
Additionally, problems like this are often encountered in actual interviews, so sufficient practice is necessary.
Through this course, I hope it serves as a foundation for understanding the K-th shortest path problem.

Swift Coding Test Course, Finding the K-th Number

Problem Description

This is a problem to find the K-th smallest number in a specific range when an array is given.
Given an integer array and two integers L and R, as well as K,
write an algorithm that returns the K-th smallest number among the numbers from L to R.

Input Format

    - The first line contains N (1 ≤ N ≤ 100,000) and Q (1 ≤ Q ≤ 100,000).
    - The second line contains N integers A₁, A₂, ..., Aₙ (-1,000,000,000 ≤ Aᵢ ≤ 1,000,000,000).
    - Following this, Q queries are given, each in the form of L, R, K.

Output Format

For each query, output the K-th smallest number.
Each output should be printed on a new line.

Example

    Input
    5 3
    1 5 2 6 3
    2 4 3
    1 5 2
    2 5 1

    Output
    5
    2
    3

Problem Solving Process

This problem fundamentally deals with interval queries.
While processing multiple queries, one must consider an optimal algorithm to find the K-th number each time in the interval.

Step 1: Problem Analysis

Given an array with N elements and Q queries,
each query must find the K-th number from the subarray of the array.
If the interval of the array is sorted every time to find the K-th number, it would take O(N * log(N)) time, which is
inefficient. The chances decrease as the number of queries increases.

Step 2: Designing a Solution Method

To solve this problem, the following methods can be used:
– **Peek Algorithm**: Sort the numbers in the interval to efficiently find the K-th number.
– **Counting Sort or Subarray Sorting**: Sort the subarray first and then find the K-th number.

Step 3: Algorithm Implementation

        func kthSmallest(arr: [Int], l: Int, r: Int, k: Int) -> Int {
            var subArray = Array(arr[l...r]) // Create subarray
            subArray.sort() // Sort
            return subArray[k - 1] // Return K-th smallest number
        }

        func solveProblem(arr: [Int], queries: [(Int, Int, Int)]) {
            for query in queries {
                let (l, r, k) = query
                let answer = kthSmallest(arr: arr, l: l, r: r, k: k)
                print(answer)
            }
        }

Step 4: Time Complexity Analysis

The above algorithm takes O(N log N) time for each query, so
when processing Q queries, the worst-case time complexity is O(Q * N log N).
This is very large, so we need to solve this problem more efficiently.

Efficient Solution Method: Segment Tree or Fenwick Tree

An efficient method could utilize a segment tree or a Fenwick tree to find the K-th number
in each interval. However, this part will not be covered here.

Step 5: Conclusion

We have examined the process of solving the problem of finding the K-th number.
Although approached by sorting the interval for each query,
using a more efficient method can lead to faster query processing.
This will be covered in the next lecture.

Conclusion

Understanding the problem accurately and designing an efficient algorithm is very important
in solving coding test problems.
Try various methods and establish your own approach.
The next lecture will cover more advanced topics such as the application of data structures.
Thank you!

Swift Coding Test Course, DNA Password

This article deals with algorithm problems that are frequently presented in coding tests, and explains in depth the process of solving those problems. Today’s topic is ‘DNA Password’, where we will solve a password problem based on DNA strings.

Problem Description

You need to design a system that generates passwords using DNA strings. The DNA string consists of uppercase letters A, C, G, and T, and each character becomes a component of the password in various combinations.

You are to find specific patterns in the given DNA string and generate a password based on these patterns. The following is a detailed problem definition:

Problem Definition

Given a string s and a natural number k, you should consider all substrings of length k from s and generate a password using the frequency of A, C, G, and T in these substrings.

The password is defined as the count of substrings where the frequency of A, C, G, and T is at least min_count.

Input Example

s = "ACGTACGTGACG", k = 4, min_count = 1

Output Example

3

Here, “ACGT”, “CGTA”, and “GTAC” are substrings with frequencies of at least 1.

Problem Solving Strategy

The strategy to solve this problem is as follows:

  1. Generate substrings of length k from the entire string s.
  2. Calculate the frequencies of A, C, G, and T for each substring.
  3. If the frequencies of A, C, G, and T are at least min_count, increase the count.
  4. Finally, output the count.

Implementation Steps

Now let’s look at how to implement the problem using Swift. Here is the step-by-step code to solve the problem.

1. Input

First, we take the string and variables as input. In Swift, values can be received through command-line arguments or direct input.


let inputString = "ACGTACGTGACG"
let k = 4
let minCount = 1

2. Generate Substrings

Generate all substrings of length k from the string.


var count = 0
let length = inputString.count

for i in 0...(length - k) {
    let startIndex = inputString.index(inputString.startIndex, offsetBy: i)
    let endIndex = inputString.index(startIndex, offsetBy: k)
    let substring = String(inputString[startIndex..

3. Frequency Calculation

Calculate the frequencies of each character (A, C, G, T) for the generated substrings.


var frequencies = [Character: Int]()

for char in substring {
    frequencies[char, default: 0] += 1
}

// Condition Check
if frequencies["A"]! >= minCount && 
   frequencies["C"]! >= minCount && 
   frequencies["G"]! >= minCount && 
   frequencies["T"]! >= minCount {
    count += 1
}

4. Output Result

Print the final count.


print("Total valid passwords: \(count)")

Complete Code

Now let's take a look at the complete code that integrates all the steps.


let inputString = "ACGTACGTGACG"
let k = 4
let minCount = 1
var count = 0
let length = inputString.count

for i in 0...(length - k) {
    let startIndex = inputString.index(inputString.startIndex, offsetBy: i)
    let endIndex = inputString.index(startIndex, offsetBy: k)
    let substring = String(inputString[startIndex..= minCount && 
       frequencies["C"]! >= minCount && 
       frequencies["G"]! >= minCount && 
       frequencies["T"]! >= minCount {
        count += 1
    }
}

print("Total valid passwords: \(count)")

Result

Running this code with the given input will output the number of valid passwords. Understanding the basic algorithm patterns of handling strings and checking conditions is very useful in coding tests.

Furthermore, you can enhance your problem-solving skills by testing the code against various input cases and adding exception handling.

I hope this article helps you prepare for coding tests using Swift. Keep practicing and solving problems to improve your skills!

Swift Coding Test Course, DFS and BFS Programs

DFS (Depth-First Search) and BFS (Breadth-First Search), which frequently appear in coding tests, are fundamental methodologies for graph and tree traversal. In this course, we will implement these two algorithms in Swift and solve the problems using them.

1. Overview of DFS and BFS

DFS (Depth-First Search) explores one path to the end before exploring the next one. It uses a stack and can also be called recursively.

BFS (Breadth-First Search) explores adjacent nodes starting from the initial node and then explores the nodes at the next depth. It is mainly implemented using a queue.

2. Problem: Counting Connected Components

We will tackle the problem of counting the number of connected components in a given undirected graph. The graph consists of vertices and edges, and the goal is to find the number of connected vertices.

Problem Definition

Integer n (1 ≤ n ≤ 1000): Number of vertices
List edges: A tuple (a, b) representing the two vertices connected by each edge.

Return the number of connected components.

Input Example

n = 5
edges = [(1, 2), (2, 3), (4, 5)]

Output Example

2

3. Algorithm Explanation

Both DFS and BFS methods can be used to solve this problem. Here, we will solve it using DFS. The following steps will be taken:

  • Convert the graph into an adjacency list format.
  • Create an array to check whether a vertex has been visited.
  • For each vertex, if there are unvisited vertices, execute DFS and mark all connected vertices as visited within the called DFS.
  • Each time DFS is called, increment the number of connected components by one.

4. Swift Implementation

Let’s implement the problem in Swift.

import Foundation

var graph = [[Int]]()
var visited: [Bool] = []
var connectedComponents = 0

// Define the DFS function
func dfs(node: Int) {
    visited[node] = true
    
    for neighbor in graph[node] {
        if !visited[neighbor] {
            dfs(node: neighbor)
        }
    }
}

// Main function
func connectedComponentsCount(n: Int, edges: [(Int, Int)]) -> Int {
    graph = Array(repeating: [Int](), count: n + 1)
    visited = Array(repeating: false, count: n + 1)
    
    // Create the graph
    for edge in edges {
        let (a, b) = edge
        graph[a].append(b)
        graph[b].append(a)
    }
    
    for i in 1...n {
        if !visited[i] {
            dfs(node: i)
            connectedComponents += 1 // New connected component discovered
        }
    }
    
    return connectedComponents
}

// Example usage
let n = 5
let edges = [(1, 2), (2, 3), (4, 5)]
let result = connectedComponentsCount(n: n, edges: edges)
print("Number of connected components: \(result)") // Output: Number of connected components: 2

5. Algorithm Analysis

The algorithm above explores the graph using DFS, resulting in a time complexity of O(V + E). Here, V is the number of vertices and E is the number of edges. This is because the algorithm visits each vertex once and checks each edge once as well.

6. Conclusion

In this course, we introduced an algorithm to count the number of connected components using DFS. Both DFS and BFS are important techniques for graph traversal, so it is crucial to understand the characteristics and implementation methods of each algorithm thoroughly and practice them. In the next course, we will tackle a similar problem using BFS.

7. References