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

Swift Coding Test Course, Lowest Common Ancestor

This lecture will cover the Lowest Common Ancestor (LCA) problem. This problem involves finding the lowest common ancestor of two given nodes, a concept that is very useful in tree structures. It is particularly helpful in enhancing understanding of data structures and algorithms.

Problem Description

Write a function to find the lowest common ancestor of two nodes in a given binary tree. The binary tree is represented in the following format:

    class TreeNode {
        var value: Int
        var left: TreeNode?
        var right: TreeNode?

        init(value: Int) {
            self.value = value
            self.left = nil
            self.right = nil
        }
    }
    

Input

  • The root node of the tree (root)
  • The first node (first)
  • The second node (second)

Output

Returns the lowest common ancestor of the given two nodes.

Example

    Input:
        root = [3, 5, 1, 6, 2, 0, 8, null, null, 7, 4]
        first = 5
        second = 1

    Output: 3

    Input:
        root = [1, 2, 3]
        first = 2
        second = 3

    Output: 1
    

Problem Approach

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

  1. Start the search from the root of the tree.
  2. If the current node is one of the first or second nodes, return that node.
  3. Recursively search in the left and right children to find the first and second nodes.
  4. If both results from the left and right children are not nil, the current node is the lowest common ancestor.
  5. If only one side finds either first or second, return that child node.
  6. If nothing is found, return nil.

Swift Code Implementation

Based on this approach, let’s implement the Swift code:

    class TreeNode {
        var value: Int
        var left: TreeNode?
        var right: TreeNode?

        init(value: Int) {
            self.value = value
            self.left = nil
            self.right = nil
        }
    }

    func lowestCommonAncestor(root: TreeNode?, first: Int, second: Int) -> TreeNode? {
        // Base case: if the root is nil
        guard let root = root else {
            return nil
        }

        // If the current node is first or second node
        if root.value == first || root.value == second {
            return root
        }

        // Recursive calls for left and right children
        let left = lowestCommonAncestor(root: root.left, first: first, second: second)
        let right = lowestCommonAncestor(root: root.right, first: first, second: second)

        // If both sides found the nodes
        if left != nil && right != nil {
            return root
        }

        // Return the found node from one side
        return left ?? right
    }
    

Code Explanation

The above code works as follows:

  • If the root node is nil, it returns nil.
  • If the current node is one of the nodes being searched for, it returns that node.
  • It recursively calls the lowestCommonAncestor function to find the lowest common ancestor in the left subtree. It does the same for the right subtree.
  • If both the left and right calls result in non-nil, it means the current node is the lowest common ancestor, so it returns the current node.
  • If not, it returns either the left or right result.

Time Complexity

The time complexity of this algorithm is O(N). N is the number of nodes in the tree, as we have to visit each node at least once, leading to a time complexity proportional to the maximum number of nodes.

Space Complexity

The space complexity of this algorithm is also O(H). H is the height of the tree, corresponding to the stack space used in recursive calls. In the worst case, if the tree is skewed to one side, it can be O(N).

Conclusion

In this article, we explored how to solve the lowest common ancestor problem in a binary tree and how to implement it in Swift. This problem is a common topic in various technical interviews, so it is important to practice with multiple examples to enhance proficiency.

Frequently Asked Questions (FAQ)

Question 1: Can we find LCA in a graph that is not a binary tree?

Yes, it is possible to find LCA in general graphs that are not binary trees, but it may require more complex algorithms depending on the case. For instance, methods like dynamic programming may be used.

Question 2: Are there other methods to solve the LCA problem?

There are several algorithms to find the lowest common ancestor. For example, using parent pointers or utilizing bitmasks, but one might choose methods with less preprocessing and space complexity.

Question 3: Can the LCA problem be extended?

Yes, there are problems that extend the LCA problem to find the lowest common ancestor for more than K nodes. These problems require more complex structures or algorithms, thus necessitating deeper understanding through practice.

In Conclusion

The lowest common ancestor problem requires a deep understanding of tree structures, making it very important for learning algorithms. It is necessary to practice solving various types of problems to enhance one’s understanding of this topic.

Swift Coding Test Course, Finding the Least Common Multiple

Problem Description

When two integers A and B are given, write a program to find the Lowest Common Multiple (LCM) of these two numbers.

Example

  • Input: A = 12, B = 18
  • Output: LCM = 36

Problem Solving Process

1. Definition of Lowest Common Multiple

The lowest common multiple refers to the smallest number among the common multiples of the given two numbers A and B.
The formula to calculate LCM is as follows:

LCM(A, B) = |A * B| / GCD(A, B)

Here, GCD is the Greatest Common Divisor, which is the largest number among the common divisors of A and B.

2. Algorithmic Approach

To efficiently find the lowest common multiple, first calculate the greatest common divisor, and then apply the above LCM formula.
The Euclidean algorithm is used to find GCD, which is quick and efficient.

Explanation of the Euclidean Algorithm

To find the GCD of two numbers A and B, repeat the following process:

  1. Let the smaller of the two numbers A and B be B, and divide A by B.
  2. Update B to the remainder of A divided by B.
  3. Repeat this process until B becomes 0; then, A is the GCD.

3. Swift Code Implementation

Now, let’s implement code to find the lowest common multiple using the Swift programming language.

import Foundation

// Find GCD
func gcd(_ a: Int, _ b: Int) -> Int {
    if b == 0 {
        return a
    }
    return gcd(b, a % b)
}

// Find LCM
func lcm(_ a: Int, _ b: Int) -> Int {
    return abs(a * b) / gcd(a, b)
}

// Test example
let A = 12
let B = 18
let result = lcm(A, B)
print("LCM(\(A), \(B)) = \(result)") // LCM(12, 18) = 36

4. Explanation of the Code

– The gcd function takes two integers A and B as parameters and finds the GCD using the Euclidean algorithm.
– The lcm function calculates the lowest common multiple by dividing the product of A and B by the GCD.

5. Additional Tests

Test with several different numbers to ensure the code works well.


let testCases = [(12, 18), (4, 5), (6, 8), (15, 25)]
for (num1, num2) in testCases {
    let result = lcm(num1, num2)
    print("LCM(\(num1), \(num2)) = \(result)")
}

Conclusion

In this tutorial, we learned how to calculate the Lowest Common Multiple (LCM). We learned to find the GCD using the Euclidean algorithm and then calculate the LCM.
It is important to test the function with various numbers and practice frequently since it is a commonly used mathematical concept.