Swift Coding Test Course, Finding Minimum Spanning Tree

Today, we will take a detailed look at one of the algorithm problems that frequently appear in coding tests: finding the Minimum Spanning Tree (MST). A minimum spanning tree is a tree that includes all the vertices of a given weighted graph, while having the minimum possible sum of weights. To solve this problem, we can use Kruskal’s Algorithm and Prim’s Algorithm. This article will focus on addressing the problem using Kruskal’s Algorithm.

Problem Description

Problem: Find the minimum spanning tree of the given graph.

The graph provided as input is presented in the following format:

8 11
0 1 4
0 7 8
1 2 8
1 7 11
2 3 7
2 5 4
2 6 2
3 4 9
3 5 14
4 5 10
5 6 2

The first line indicates the number of vertices and the number of edges, and from the second line onward, the information of the edges is provided. Here, each edge consists of two vertices and the corresponding weight. For example, “0 1 4” means that the edge connecting vertex 0 and vertex 1 has a weight of 4.

Problem Solving Strategy

To solve this problem, the following steps are necessary:

  1. Store the edge information of the graph.
  2. Apply Kruskal’s Algorithm to find the minimum spanning tree.
  3. Finally, sum up the weights of the selected edges and output the result.

1. Storing Edge Information of the Graph

First, read the input and store the edge information of the graph. For this, each edge should be stored as a tuple in the form of (weight, starting vertex, ending vertex). Each edge should be designed to be sortable based on its weight.


import Foundation

struct Edge {
    let weight: Int
    let start: Int
    let end: Int
}

func readGraph() -> [Edge] {
    let firstLine = readLine()!.split(separator: " ").map { Int($0)! }
    let vertexCount = firstLine[0]
    let edgeCount = firstLine[1]
    var edges = [Edge]()

    for _ in 0..

2. Implementing Kruskal's Algorithm

Kruskal's Algorithm sorts all the edges by weight, then adds them one by one from the lowest weight, ensuring that cycles are not formed. For this, a Union-Find data structure is used. The Union-Find structure allows for quick checks on whether two sets are connected and provides a method to unite two sets.


class DisjointSet {
    private var parent: [Int]

    init(size: Int) {
        self.parent = Array(0.. Int {
        if parent[node] != node {
            parent[node] = find(parent[node])
        }
        return parent[node]
    }

    func union(_ a: Int, _ b: Int) {
        let rootA = find(a)
        let rootB = find(b)
        if rootA != rootB {
            parent[rootB] = rootA
        }
    }
}

func kruskal(edges: [Edge], vertexCount: Int) -> Int {
    let sortedEdges = edges.sorted { $0.weight < $1.weight }
    let disjointSet = DisjointSet(size: vertexCount)
    var totalWeight = 0

    for edge in sortedEdges {
        if disjointSet.find(edge.start) != disjointSet.find(edge.end) {
            disjointSet.union(edge.start, edge.end)
            totalWeight += edge.weight
        }
    }

    return totalWeight
}

3. Main Function

Now that all parts are ready, let's write the main function to execute the entire process.


func main() {
    let edges = readGraph()
    let totalWeight = kruskal(edges: edges, vertexCount: 8)
    print("The sum of weights in the minimum spanning tree: \(totalWeight)")
}

main()

Complete Program

With all the steps completed, we can now write the overall program as follows:


import Foundation

struct Edge {
    let weight: Int
    let start: Int
    let end: Int
}

class DisjointSet {
    private var parent: [Int]

    init(size: Int) {
        self.parent = Array(0.. Int {
        if parent[node] != node {
            parent[node] = find(parent[node])
        }
        return parent[node]
    }

    func union(_ a: Int, _ b: Int) {
        let rootA = find(a)
        let rootB = find(b)
        if rootA != rootB {
            parent[rootB] = rootA
        }
    }
}

func readGraph() -> [Edge] {
    let firstLine = readLine()!.split(separator: " ").map { Int($0)! }
    let vertexCount = firstLine[0]
    let edgeCount = firstLine[1]
    var edges = [Edge]()

    for _ in 0.. Int {
    let sortedEdges = edges.sorted { $0.weight < $1.weight }
    let disjointSet = DisjointSet(size: vertexCount)
    var totalWeight = 0

    for edge in sortedEdges {
        if disjointSet.find(edge.start) != disjointSet.find(edge.end) {
            disjointSet.union(edge.start, edge.end)
            totalWeight += edge.weight
        }
    }
    return totalWeight
}

func main() {
    let edges = readGraph()
    let totalWeight = kruskal(edges: edges, vertexCount: 8)
    print("The sum of weights in the minimum spanning tree: \(totalWeight)")
}

main()

Conclusion

In this post, we explored how to solve the Minimum Spanning Tree problem using Swift. We learned the process of finding the MST using Kruskal's Algorithm based on edges and how to use Union-Find. When facing such problems in actual coding tests and interviews, I hope you can approach them with confidence based on your theories and algorithms. Next time, we will discuss finding the minimum spanning tree using Prim's Algorithm, so please look forward to it.

Swift Coding Test Course, Minimum Spanning Tree

In this article, we will learn about the Minimum Spanning Tree (MST) and implement an algorithm called Kruskal’s Algorithm in Swift to solve it. A minimum spanning tree is a tree that connects all the vertices in a given graph while minimizing the sum of the weights.

Problem Description

A company is trying to create a network connecting several cities. Each road between cities incurs a cost. To connect all cities while minimizing costs, what roads should the company choose?

Given a graph containing information about cities and roads, we need to solve the problem of finding the minimum spanning tree.

Input Format

            The first line contains the number of cities N (2 ≤ N ≤ 1000) and the number of roads M (1 ≤ M ≤ 10000).
            Following this, M lines provide information on each road and its weight.
            The road information is represented in the form (A, B, C), where A and B are the city numbers, and C is the road cost.
            

Output Format

            Print the total cost required to form the minimum spanning tree.
            

Problem-Solving Process

Step 1: Choosing the Algorithm

This problem can be solved using Kruskal’s Algorithm. Kruskal’s Algorithm sorts all the edges of the graph in ascending order of weights, then uses the Union-Find data structure to check for cycles while constructing the minimum spanning tree.

Step 2: Designing the Data Structure

To use the Union-Find data structure, we create two arrays: parent and rank. parent represents the parent of each node, and rank denotes the height of the tree. This data will be used to manage the connection status of the nodes.

            // Union-Find Structure
            struct UnionFind {
                var parent: [Int]
                var rank: [Int]

                init(size: Int) {
                    parent = Array(0.. Int {
                    if parent[u] != u {
                        parent[u] = find(parent[u])
                    }
                    return parent[u]
                }

                mutating func union(_ u: Int, _ v: Int) {
                    let rootU = find(u)
                    let rootV = find(v)

                    if rootU != rootV {
                        if rank[rootU] > rank[rootV] {
                            parent[rootV] = rootU
                        } else if rank[rootU] < rank[rootV] {
                            parent[rootU] = rootV
                        } else {
                            parent[rootV] = rootU
                            rank[rootU] += 1
                        }
                    }
                }
            }
            

Step 3: Implementing Kruskal's Algorithm

Sort all edges based on weights, and select edges in order only if they do not form a cycle. Through this process, we will construct the minimum spanning tree and calculate the sum of the weights at that time.

            func kruskal(n: Int, edges: [(Int, Int, Int)]) -> Int {
                var uf = UnionFind(size: n)
                var totalCost = 0
                let sortedEdges = edges.sorted { $0.2 < $1.2 }
                
                for edge in sortedEdges {
                    let (u, v, cost) = edge
                    if uf.find(u) != uf.find(v) {
                        uf.union(u, v)
                        totalCost += cost
                    }
                }
                return totalCost
            }
            

Step 4: Testing and Conclusion

Now let's take input, pass it to the kruskal function, and print the result.

            let n = 4
            let edges = [(0, 1, 10), (0, 2, 6), (0, 3, 5), (1, 3, 15), (2, 3, 4)]
            let minCost = kruskal(n: n, edges: edges)
            print("Cost of the minimum spanning tree: \(minCost)")
            

The above example prints the total cost of the minimum spanning tree.

In Conclusion

In this lecture, we learned how to solve the minimum spanning tree problem using Kruskal's Algorithm. Since this problem frequently appears in actual coding tests, it is important to practice repeatedly to improve your proficiency.

Swift Coding Test Course, Finding Minimum Cost

In this lecture, we will cover the ‘Minimum Cost Pathfinding’ problem, which is frequently presented in coding tests, and examine step-by-step how to solve it using Swift. We will guide you on understanding the algorithm as well as efficiently solving the problem using Swift’s features.

Problem Description

This problem entails finding a path that minimizes the cost of traveling from one city to another.
Each city is represented as a node in a graph, and cities are connected by roads.
Each road has a different cost associated with traveling on it.
The goal is to find the minimum cost to reach the destination city from the starting city.

Here is an example of the problem:

Input:

The number of cities N, and the number of roads M

Following this, M lines will provide information about each road.

The road information is given as (A, B, C), indicating that traveling from city A to city B requires a cost of C.

Lastly, the starting city and destination city are provided.

Output:

The minimum cost from the starting city to the destination city

Problem Solving Strategy

To solve this problem, we will use Dijkstra’s algorithm.
Dijkstra’s algorithm is a method for finding the shortest path in a weighted graph and is particularly effective for handling non-negative costs.
The main idea of this algorithm is to select the nearest unvisited node based on the shortest path found so far.

Here is how to solve the problem step by step using Dijkstra’s algorithm:

  1. Construct the graph in the form of an adjacency list.
  2. Initialize an array to store the minimum cost to each city.
  3. Begin at the starting city and update the information for the adjacent cities.
  4. Use a priority queue to determine the next city to visit.
  5. Repeat until the destination city is reached.

Swift Code Implementation

Now, let’s implement the code in Swift according to the above strategy.
Here is an example code implementing Dijkstra’s algorithm:


import Foundation

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

func dijkstra(start: Int, end: Int, graph: [[Edge]]) -> Int {
    var minCosts = Array(repeating: Int.max, count: graph.count)
    var priorityQueue = [(cost: Int, vertex: Int)]()
    minCosts[start] = 0
    priorityQueue.append((0, start))
    
    while !priorityQueue.isEmpty {
        let current = priorityQueue.removeFirst()
        
        if current.vertex == end {
            return current.cost
        }
        
        for edge in graph[current.vertex] {
            let newCost = current.cost + edge.cost
            if newCost < minCosts[edge.to] {
                minCosts[edge.to] = newCost
                priorityQueue.append((newCost, edge.to))
            }
        }
        
        priorityQueue.sort { $0.cost < $1.cost }
    }
    
    return minCosts[end] == Int.max ? -1 : minCosts[end]
}

// Example of the graph
let N = 5  // Number of cities
let M = 7  // Number of roads
var graph = [[Edge]](repeating: [], count: N)

let roads: [(Int, Int, Int)] = [
    (0, 1, 10),
    (0, 2, 5),
    (1, 2, 2),
    (1, 3, 1),
    (2, 1, 3),
    (2, 3, 9),
    (3, 4, 2)
]

for road in roads {
    graph[road.0].append(Edge(to: road.1, cost: road.2))
    graph[road.1].append(Edge(to: road.0, cost: road.2)) // Assuming bidirectional roads
}

// Starting city and destination city
let start = 0
let end = 4
let minCost = dijkstra(start: start, end: end, graph: graph)

print("Minimum cost: \(minCost)") // Output: Minimum cost: 12
        

Code Explanation

The Swift code above implements an algorithm that finds the minimum cost to travel from one city to another using several functions.
The dijkstra function, starting from each city, receives connection information about the cities through the graph parameter and returns the minimum cost via the starting and destination cities.

1. Edge Structure:
Stores connection information between cities. to is the index of the connected city, and cost is the cost of that movement.

2. dijkstra Function:
This is the main logic for calculating the minimum cost. It explores the cities adjacent to the starting city and continuously updates the minimum cost.
A priority queue is used to determine the next move to the closest city.

3. Graph Structure:
The graph is constructed in an adjacency list format and contains information about the roads connected to each city.

4. Result: Outputs the minimum cost from the given starting city to the destination city.

Result Analysis

In the code above, based on the input cities and road information, the minimum cost to travel from starting city 0 to destination city 4 is calculated as 12.
This indicates that an appropriate connection and cost management were achieved.
Therefore, we can conclude that the optimized path provided through Dijkstra's algorithm can indeed be found.

Conclusion

In this lecture, we applied Dijkstra's algorithm to solve the minimum cost problem using Swift.
We learned how to understand the algorithm and implement it in actual code.
Understanding and utilizing the algorithm well will provide a foundation that can be useful in various coding tests and real-world applications.

Additionally, to tackle more complex problems, it is advisable to learn various algorithms and continuously practice whenever encountering problems.
In the next lecture, we will cover other algorithms to solve more problems and further enhance our skills.

Swift Coding Test Course, Finding the Least Common Ancestor 2

Hello! Today, we will tackle the problem of finding the Lowest Common Ancestor (LCA) in Swift coding tests for those preparing for it. This problem involves finding the closest common ancestor of two nodes in a tree structure and has various applications. It is also one of the frequently appearing problems in coding tests.

Problem Description

When given a binary tree with two nodes, write a function to find the lowest common ancestor of these two nodes. The nodes of the binary tree are defined as follows:

    class TreeNode {
        var value: Int
        var left: TreeNode?
        var right: TreeNode?
        
        init(value: Int) {
            self.value = value
            self.left = nil
            self.right = nil
        }
    }
    

For example, consider the following binary tree:

          3
         / \
        5   1
       / \ / \
      6  2 0  8
        / \
       7   4
    

The lowest common ancestor of node 5 and node 1 is node 3. In the case of node 5 and node 4, the lowest common ancestor is node 5.

Input Format

The function is defined in the following format:

    func lowestCommonAncestor(_ root: TreeNode?, _ p: TreeNode?, _ q: TreeNode?) -> TreeNode?
    

Here, root is the root node of the binary tree, and p and q represent the respective nodes. These nodes exist in the tree.

Solution Process

There are various approaches to solve this problem, but one of the most efficient methods is to use recursion. Below, I will explain the process of solving the problem using this method step-by-step.

Step 1: Basic Idea

Through tree traversal, we can consider the following three cases for each node:

  • If the current node matches p or q
  • If we are searching for p or q in the left subtree
  • If we are searching for p or q in the right subtree

Through these cases, we can deduce the lowest common ancestor. If we find one node in each of the two subtrees, the current node is the lowest common ancestor.

Step 2: Implementing the Recursive Function

Now, we implement a recursive function to find the two nodes at each node:

    func lowestCommonAncestor(_ root: TreeNode?, _ p: TreeNode?, _ q: TreeNode?) -> TreeNode? {
        // If the node to check is nil
        guard let current = root else { return nil }

        // If the current node is p or q
        if current === p || current === q {
            return current
        }

        // Search for the respective nodes in the left and right subtrees
        let left = lowestCommonAncestor(current.left, p, q)
        let right = lowestCommonAncestor(current.right, p, q)

        // If we found one in both the left and right subtrees
        if left != nil && right != nil {
            return current
        }

        // If we found one only in one subtree
        return left ?? right
    }
    

In the code above, we check whether the current node is nil using the guard let statement and test whether the current node matches p or q. After that, we recursively call the function on the left and right subtrees and store the results in left and right. If we find results in both subtrees, we return the current node as the lowest common ancestor.

Step 3: Writing Test Cases

Now, we will add some test cases to test the function we wrote:

    let root = TreeNode(value: 3)
    let node5 = TreeNode(value: 5)
    let node1 = TreeNode(value: 1)
    let node6 = TreeNode(value: 6)
    let node2 = TreeNode(value: 2)
    let node0 = TreeNode(value: 0)
    let node8 = TreeNode(value: 8)
    let node7 = TreeNode(value: 7)
    let node4 = TreeNode(value: 4)

    // Connecting the tree structure
    root.left = node5
    root.right = node1
    node5.left = node6
    node5.right = node2
    node1.left = node0
    node1.right = node8
    node2.left = node7
    node2.right = node4

    // Testing
    let lca1 = lowestCommonAncestor(root, node5, node1) // Result should be 3
    let lca2 = lowestCommonAncestor(root, node5, node4) // Result should be 5
    

We execute the test cases to check the results. In this case, we validate whether the expected results come out and whether each node’s lowest common ancestor is found correctly.

Time Complexity Analysis

The time complexity of this algorithm is O(N). This is because, in the worst case, we may need to traverse all nodes; thus, it takes up to O(N) time for a tree with N nodes. Additionally, the space complexity is O(H), where H represents the depth of the tree. This refers to the depth of the recursive call stack.

Conclusion

Today, we learned how to solve the problem of finding the lowest common ancestor in a binary tree using Swift. This problem frequently appears in coding tests, so it is important to develop the skill to solve problems quickly when they arise. By utilizing a recursive approach to solve the problem, we improved the efficiency of the code and emphasized the understanding. I will return with another interesting algorithmic problem next time!

Swift Coding Test Course, Finding the Lowest Common Ancestor 1

Coding tests are a very important process for software developers. In particular, algorithm problems provide a good opportunity to demonstrate your problem-solving skills. In this course, we will cover the problem of finding the Lowest Common Ancestor (LCA) in a tree structure.

Problem Description

In the given binary tree, the Lowest Common Ancestor (LCA) of two nodes A and B is the deepest node among the ancestors of nodes A and B. In other words, you need to find the nearest common ancestor along the paths of A and B.

Problem: Given a binary tree, write a function that takes the values of two nodes A and B and returns the Lowest Common Ancestor.

Input

  • The root node of the binary tree is given.
  • The values of nodes A and B are provided.

Output

  • Returns the value of the Lowest Common Ancestor node.

Constraints

  • The nodes in the binary tree are unique.
  • Nodes always exist, and A and B must be included in the tree.

Algorithm Approach

To find the Lowest Common Ancestor, it is necessary to traverse the tree. A depth-first search (DFS) approach is used to recursively explore until each node is reached. The method to find A and B at each node is as follows.

  • Examine the root node to see if it matches A or B.
  • If the current node is null, return null.
  • Recursively check the left child and the right child, and examine the results of both child calls.
  • If both child calls return non-null results, the current node is the Lowest Common Ancestor.
  • If only one child is non-null, that child is the Lowest Common Ancestor.

Swift Implementation