Category: Swift Coding Test

Hello! In this lecture, we will examine an algorithmic problem that finds the minimum value among prime numbers and palindrome numbers. Both prime numbers and palindrome numbers are basic mathematical concepts, but the combination of these two to solve specific requirements is common in coding tests. Therefore, develop your skills in efficient algorithm design and implementation through this problem.

Problem Description

Write a function that finds the minimum value of numbers that are both prime and palindrome among all integers within the given range.

For example, if we list the numbers that are both prime and palindrome in the range from 1 to 100, the minimum value among them is 2.

Input

An integer n (2 <= n <= 10,000) is provided. You need to find the numbers that are both prime and palindrome within this range.

Output

Output the minimum value of numbers that are both prime and palindrome. If there are no such numbers that meet the condition, output the message “There are no numbers that satisfy the condition.”

Problem Solving Process

To solve the problem, it can be divided into the following steps:

1. Write a function to determine if a number is prime
2. Write a function to determine if a number is a palindrome
3. Find prime and palindrome numbers within the given range
4. Return the minimum value

1. Write a function to determine if a number is prime

A prime number is a number that has only 1 and itself as divisors. To determine this, you can check if the number can be evenly divided by any number from 2 to sqrt(n).

func isPrime(_ number: Int) -> Bool {
        guard number > 1 else { return false }
        for i in 2...Int(sqrt(Double(number))) {
            if number % i == 0 {
                return false
            }
        }
        return true
    }

2. Write a function to determine if a number is a palindrome

A palindrome is a number that reads the same forwards and backwards. To check this, convert the number to a string and compare it with its reversed version.

func isPalindrome(_ number: Int) -> Bool {
        let string = String(number)
        return string == String(string.reversed())
    }

3. Find prime and palindrome numbers within the given range

Utilize the two previously written functions (isPrime, isPalindrome) to verify numbers from 2 to n that satisfy both conditions.

func findMinPrimePalindrome(upTo n: Int) -> Int? {
        var minPrimePalindrome: Int? = nil

        for i in 2...n {
            if isPrime(i) && isPalindrome(i) {
                if minPrimePalindrome == nil || i < minPrimePalindrome! {
                    minPrimePalindrome = i
                }
            }
        }
        return minPrimePalindrome
    }

4. Return the minimum value

After finding the minimum value, return it. If there is no minimum value, ensure an appropriate message is output.

let n = 100
if let minValue = findMinPrimePalindrome(upTo: n) {
    print("The minimum value of numbers that are both prime and palindrome is: \(minValue)")
} else {
    print("There are no numbers that satisfy the condition.")
}

Final Code

Integrating all parts, the final code is as follows:

func isPrime(_ number: Int) -> Bool {
        guard number > 1 else { return false }
        for i in 2...Int(sqrt(Double(number))) {
            if number % i == 0 {
                return false
            }
        }
        return true
    }

    func isPalindrome(_ number: Int) -> Bool {
        let string = String(number)
        return string == String(string.reversed())
    }

    func findMinPrimePalindrome(upTo n: Int) -> Int? {
        var minPrimePalindrome: Int? = nil
        
        for i in 2...n {
            if isPrime(i) && isPalindrome(i) {
                if minPrimePalindrome == nil || i < minPrimePalindrome! {
                    minPrimePalindrome = i
                }
            }
        }
        return minPrimePalindrome
    }

    let n = 100
    if let minValue = findMinPrimePalindrome(upTo: n) {
        print("The minimum value of numbers that are both prime and palindrome is: \(minValue)")
    } else {
        print("There are no numbers that satisfy the condition.")
    }

Conclusion

In this lecture, we covered the problem of finding numbers that are both prime and palindrome using the Swift language. I hope you were able to learn basic algorithm design and implementation skills through this problem. Additionally, you learned how to combine various functions to solve the problem. I look forward to encountering even more interesting and challenging problems in the next lecture!

Thank you!

Problem Definition

The ‘Salesman’s Dilemma’ problem seeks to find the minimum route for a salesman to visit all given cities
once and return to the starting city, based on the distance information between the cities. This is an
optimization problem known as the Traveling Salesman Problem (TSP), which is
a form of graph theory. This problem is known to be NP-complete, and there are various methods to solve it.

Problem Description

The following conditions are given:

• There are n cities.
• Each city is represented by an integer from 1 to n.
• The distances between the cities are given in a directed graph form, and the distance information is represented in a two-dimensional array.
• The salesman must visit all cities once and return to the starting city.

Input Format

The first line contains the number of cities, n. The following n lines contain the distance matrix between cities.
The distance matrix is constructed as follows:

        0 10 15 20
        10 0 35 25
        15 35 0 30
        20 25 30 0
        

In the example above, the first line indicates there are 4 cities, and it represents the distances between each city.
For example, the distance between city 1 and city 2 is 10, and the distance between city 1 and city 3 is 15.

Output Format

Print the total distance of the minimum route.

Problem Solving Approach

This problem can be approached in various ways, but here we will describe a method using backtracking and dynamic programming (DP).
While backtracking can be used to explore all routes and find the minimum path, the computational workload increases exponentially as the number of cities grows.
Therefore, combining dynamic programming with memoization can enhance efficiency.

Algorithm Implementation

Initially, we can implement the solution by generating all possible paths using backtracking with permutations, and then calculating the distance of each path to update the minimum value.
Below is an example code implemented in Swift.

        func tsp(graph: [[Int]], visited: inout [Bool], currentIndex: Int, count: Int, cost: Int, ans: inout Int) {
            // If all cities have been visited
            if count == graph.count && graph[currentIndex][0] > 0 {
                ans = min(ans, cost + graph[currentIndex][0])
                return
            }
            for i in 0.. 0 {
                    visited[i] = true
                    tsp(graph: graph, visited: &visited, currentIndex: i, count: count + 1, cost: cost + graph[currentIndex][i], ans: &ans)
                    visited[i] = false
                }
            }
        }

        func findMinCost(graph: [[Int]]) -> Int {
            var visited = [Bool](repeating: false, count: graph.count)
            var ans = Int.max
            visited[0] = true
            tsp(graph: graph, visited: &visited, currentIndex: 0, count: 1, cost: 0, ans: &ans)
            return ans
        }

        let graph = [
            [0, 10, 15, 20],
            [10, 0, 35, 25],
            [15, 35, 0, 30],
            [20, 25, 30, 0]
        ]

        let result = findMinCost(graph: graph)
        print("Distance of the minimum route: \(result)")
        

Code Explanation

The code above is structured as follows:

• tsp Function: Recursively explores all paths visiting the cities,
  traversing all unvisited cities from the current city. If the cost of visiting a city is minimal,
  it updates the minimum cost of the current path.
• findMinCost Function: Acts as an initializer and calls the tsp function
  while visiting the first city (0) to explore the entire path.

Complexity Analysis

The time complexity of this algorithm is O(n!). This is because it has to traverse each city and explore all paths.
It can be used when n is small, but it becomes inefficient as n increases.
Therefore, for more efficient methods, dynamic programming can be applied using bit masking to reduce complexity.
This method has a time complexity of O(n^2 * 2^n) and is practical for n below 20.

Conclusion

The ‘Salesman’s Dilemma’ problem is closely related to various optimization problems in reality and provides important insights when solving problems through effective algorithm design.
By tackling problems like this while preparing for coding tests, one can enhance their understanding of graph algorithms and dynamic programming, as well as develop systematic problem-solving skills.

Hello! In this lecture, we will take a closer look at the segment tree, which is one of the data structures. The segment tree is a powerful data structure that can efficiently handle interval queries, especially useful for calculating the range sum, range minimum, and range maximum of an array. In this text, we will explore the basic concepts of segment trees, how to implement them, and solve problems that are frequently asked in practice.

1. What is a Segment Tree?

A segment tree is a powerful tool for efficiently managing the interval information of an array. Typically, when the size of the array is N, a segment tree uses approximately 2 * 2^{⌈log₂(N)⌉} memory space. This is due to the use of a complete binary tree structure. Fundamentally, the segment tree can perform the following two operations efficiently:

• Interval Query: Quickly retrieve information about a specific interval.
• Update: Rapidly update the interval information affected after changing a specific element of the array.

2. Structure of the Segment Tree

The segment tree is made up of nodes, each representing a specific interval. For example, there is a root node that manages the interval from index 0 to N-1 of the array, and this node divides the array into two halves through two child nodes. By continuously dividing the array in this manner, each node holds the information of a specific interval.

2.1 Node Definition

Each node contains the following information:

• Start Index (start): The starting point of the interval
• End Index (end): The ending point of the interval
• Value (value): A variable to store the information of the interval (e.g., sum, minimum value, etc.)

3. Creating and Querying the Segment Tree

To implement a segment tree, we first need to build the tree. For this, we use a method that recursively creates segment tree nodes based on the input array. As a simple example, let’s create a segment tree for range sums.

// Swift Code
class SegmentTree {
    var tree: [Int] // Array to store the segment tree
    var n: Int // Size of the array

    init(_ data: [Int]) {
        self.n = data.count
        self.tree = Array(repeating: 0, count: 4 * n) // Initialize the tree array
        buildTree(data: data, node: 1, start: 0, end: n - 1)
    }

    // Build the segment tree using the interval of the array
    func buildTree(data: [Int], node: Int, start: Int, end: Int) {
        if start == end {
            tree[node] = data[start] // Store value in leaf node
        } else {
            let mid = (start + end) / 2
            buildTree(data: data, node: 2 * node, start: start, end: mid) // Left subtree
            buildTree(data: data, node: 2 * node + 1, start: mid + 1, end: end) // Right subtree
            tree[node] = tree[2 * node] + tree[2 * node + 1] // Update parent node value
        }
    }
}

4. Processing Segment Tree Queries

The segment tree now needs to add a feature to calculate range sums. To handle range sum queries, we will add the following function:

// Function to calculate the sum of a given interval
func query(node: Int, start: Int, end: Int, l: Int, r: Int) -> Int {
    if r < start || end < l { // If intervals do not overlap
        return 0 // Default value
    }
    if l <= start && end <= r { // If the interval is completely contained
        return tree[node]
    }
    let mid = (start + end) / 2
    let leftSum = query(node: 2 * node, start: start, end: mid, l: l, r: r) // Query left subtree
    let rightSum = query(node: 2 * node + 1, start: mid + 1, end: end, l: l, r: r) // Query right subtree
    return leftSum + rightSum // Sum the results
}

5. Updating the Segment Tree

We will also add a function to update the elements of the array. When changing a specific index of the array, here’s how we can quickly update the information in the segment tree:

// Function to update a specific index of the array
func update(node: Int, start: Int, end: Int, idx: Int, value: Int) {
    if start == end { // Reached leaf node
        tree[node] = value // Update the node
    } else {
        let mid = (start + end) / 2
        if start <= idx && idx <= mid {
            update(node: 2 * node, start: start, end: mid, idx: idx, value: value) // Update left subtree
        } else {
            update(node: 2 * node + 1, start: mid + 1, end: end, idx: idx, value: value) // Update right subtree
        }
        tree[node] = tree[2 * node] + tree[2 * node + 1] // Update parent node value
    }
}

6. Example of a Practical Problem

Now that we have looked at the basic structure of the segment tree and how to perform queries and updates, let's tackle a problem that is frequently asked in practice. The problem is as follows:

Problem Description

Given an array, answer the following questions:

1. Calculate the sum of the interval [L, R].
2. Update the value of index I to V.

Input Format

N (array size)
arr[0], arr[1], ..., arr[N-1]
Q (number of queries)
Each query is given in the following format:
1 L R (range sum query)
2 I V (update query)

Output Format

Output for each range sum query

Example

5
1 2 3 4 5
3
1 1 3
2 2 10
1 1 3

Example Output
6

7. Problem-Solving Process

1. Build a segment tree for the input array.
2. Depending on the query type, call the query() function for range queries and the update() function for update queries.
3. Print the results.

// Main function to solve the problem
import Foundation

func main() {
    // Input processing
    let n = Int(readLine()!)!
    let arr = readLine()!.split(separator: " ").map { Int($0)! }
    let q = Int(readLine()!)!

    // Build the segment tree
    let segmentTree = SegmentTree(arr)

    // Process queries
    for _ in 0..