Swift Coding Test Course, Calculating the Area of a Polygon

1. Problem Definition

In this session, we will handle the problem of calculating the area of a given polygon. When the vertices of the polygon are provided, we will implement an algorithm to calculate the area using them. The algorithm used for area calculation is based on the Strassen Algorithm.

2. Problem Input

The function has the following shape:

func polygonArea(vertices: [(Double, Double)]) -> Double

Here, vertices is an array of tuples representing each vertex of the polygon. Each tuple contains the x and y coordinates.

3. Algorithm Explanation

To calculate the area of the polygon, we will use the polygon area formula. This formula is as follows:

Area = 0.5 * | Σ (xi * yi+1 - yi * xi+1) |

Here, i represents the index from 0 to n-1, and the last vertex connects to the first vertex. To code this formula, we will follow these steps:

  1. Calculate the number of vertices n.
  2. Calculate the area contribution for each vertex.
  3. Sum all contributions.
  4. Convert the result to absolute value and multiply by 0.5.

4. Code Implementation

Now, let’s implement the above algorithm in Swift. Here is the complete code:

func polygonArea(vertices: [(Double, Double)]) -> Double {
    var area = 0.0
    let n = vertices.count

    for i in 0..

4.1 Code Explanation

The code above works as follows:

  • First, the area variable is initialized to prepare for area calculation.
  • n stores the number of vertices of the polygon.
  • For each vertex i, the next vertex j is calculated (where j is set to (i + 1) % n to connect the last vertex to the first vertex).
  • The area contribution is calculated and accumulated in area.
  • At the end of the loop, the absolute value of the area is divided by 2 to return the result.

5. Test Cases

Now we will validate this function with various test cases. Here are some examples:

let example1 = [(0.0, 0.0), (4.0, 0.0), (4.0, 3.0), (0.0, 4.0)]
let area1 = polygonArea(vertices: example1)
print(area1) // 12.0

let example2 = [(1.0, 1.0), (3.0, 1.0), (3.0, 3.0), (1.0, 3.0)]
let area2 = polygonArea(vertices: example2)
print(area2) // 4.0

let example3 = [(0.0, 0.0), (5.0, 0.0), (5.0, 5.0), (0.0, 5.0)]
let area3 = polygonArea(vertices: example3)
print(area3) // 25.0

6. Conclusion

In this tutorial, we implemented an algorithm in Swift to calculate the area of a polygon. We verified that the algorithm works correctly through various test cases. These types of problems can deepen our understanding of data structures and algorithms, which will be useful in future coding tests.

If more complex problems related to polygons or various types of area calculations are needed, we can consider additional optimizations and expansions based on this algorithm. In the next tutorial, we will cover these advanced techniques.

Swift Coding Test Course, Breadth-First Search

1. Introduction

Breadth-First Search (BFS) is an algorithm for traversing nodes in a graph or tree structure.
BFS explores adjacent nodes first from the starting node, then examines the adjacent nodes of those nodes,
visiting all levels of nodes in order. This method is useful for finding the shortest path since it explores
the graph or tree level by level and can be applied to many problems.

2. Problem Description

Problem: Finding the Shortest Path

The given 2D grid contains three types of elements:

  • 0: Walkable area
  • 1: Obstacle
  • 2: Starting point
  • 3: Destination point

The problem is to find the shortest path from the starting point (2) to the destination (3)
within the grid, and output the minimum number of moves required to travel along that path.
If the destination cannot be reached, output -1.

Input Format

        4 4
        0 0 1 0
        0 1 0 0
        2 0 1 0
        0 0 3 0
        

Output Format

        7
        

In the above example, the minimum number of moves from the starting point to the destination is 7.

3. Problem Analysis

BFS can be used to solve this problem.
BFS is suitable for finding the shortest path, starting from the starting point (2) in the given grid and
exploring the path to the destination (3) by moving to adjacent walkable areas (0).
Possible directions for movement are limited to four: up, down, left, and right.

One of the features of BFS is that it adds each node to the queue, allowing it to search in level order.
This makes it easy to count the number of moves needed to reach a specific node.
This method can be used to explore all 0s and 3s filled in the grid to check whether the destination can be reached.

4. Algorithm Design

1. **Initialize the grid and objects**: Initialize the queue for breadth-first search and a visited array based on the input grid.
2. **Add the starting point to the queue**: Add the starting point to the queue and update the visited array.
3. **Execute BFS**: For each node removed from the queue, check its four neighbors.
If the neighbor is walkable, add it to the queue and mark it as visited.
4. **Check for reaching the destination**: If the destination is reached, output the number of moves at that point.
5. **Handle cases where destination is unreachable**: If the queue is empty but the destination has not been reached, output -1.

5. Code Implementation

Now, we will implement BFS in Swift based on the above algorithm design.


import Foundation

func findShortestPath(grid: [[Int]], start: (Int, Int), end: (Int, Int)) -> Int {
    let directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
    var queue: [(Int, Int, Int)] = [(start.0, start.1, 0)] // (x, y, distance)
    var visited = grid
    visited[start.0][start.1] = 1 // Mark as visited

    while !queue.isEmpty {
        let (x, y, distance) = queue.removeFirst()
        
        // If reached the destination
        if (x, y) == end {
            return distance
        }

        // Explore neighboring nodes
        for direction in directions {
            let newX = x + direction.0
            let newY = y + direction.1
            
            // If within valid range and walkable
            if newX >= 0, newY >= 0, newX < grid.count, newY < grid[0].count, visited[newX][newY] == 0 {
                queue.append((newX, newY, distance + 1))
                visited[newX][newY] = 1 // Mark as visited
            }
        }
    }
    
    // If destination is not reached
    return -1
}

// Example input
let grid: [[Int]] = [
    [0, 0, 1, 0],
    [0, 1, 0, 0],
    [2, 0, 1, 0],
    [0, 0, 3, 0]
]

if let start = grid.enumerated().flatMap({ $0.element.enumerated().compactMap { $1 == 2 ? ($0.offset, $0.element) : nil } }).first,
   let end = grid.enumerated().flatMap({ $0.element.enumerated().compactMap { $1 == 3 ? ($0.offset, $0.element) : nil } }).first {
    let result = findShortestPath(grid: grid, start: start, end: end)
    print(result) // Output the minimum number of moves
}

6. Code Explanation

The code above finds the starting point and destination in the given grid and then
performs breadth-first search to find the shortest path.

  • The four directions for exploration are defined as up, down, left, and right.
  • A visited array is used to ensure that
    already visited nodes are not added to the queue again.
  • When the destination is reached, the current distance is returned.
  • If the destination cannot be reached, it returns -1.

7. Conclusion

Breadth-First Search (BFS) is a highly useful algorithm for solving shortest path problems.
It works efficiently in structures like 2D grids and frequently appears in various programming challenges.
It is important to practice different types of BFS problems to enhance understanding of the algorithm and problem-solving skills.
Through this tutorial, we hope you have gained a fundamental understanding of BFS and learned how to solve actual problems.

8. Additional Practice Problems

Try your hand at the following variant problems:

  • Reach the destination while avoiding obstacles (1) using only a specific path in the given grid.
  • Finding the shortest paths between multiple starting and ending points.
  • Finding the shortest path when diagonal movements are also allowed, instead of only grid movements.

Tackling these problems will further enhance your understanding of the breadth-first search algorithm and improve your problem-solving abilities.

Swift Coding Test Course, Sorting Digits in Descending Order

In this article, we will explore how to solve a specific problem using Swift. The topic is Sorting Digits in Descending Order. This problem allows us to practice string manipulation, sorting algorithms, and handling data types in Swift.

Problem Description

Given an array of integers, write a function that converts each integer to a string, sorts the strings in descending order by their digits, and returns the result. For example, if the array is [321, 123, 456], it should convert each integer to a string resulting in ["321", "123", "456"] and sort it in descending order, returning ["321", "456", "123"].

Problem Examples

  • Input: [321, 123, 456] → Output: ["321", "456", "123"]
  • Input: [12, 345, 7, 89] → Output: ["89", "7", "345", "12"]
  • Input: [10, 2, 1] → Output: ["2", "10", "1"]

Problem Approach

The process to solve this problem can be divided into the following steps:

  1. Convert the integers to strings.
  2. Sort the strings in descending order by their digits.
  3. Return the sorted result.

Step 1: Convert Integer Array to String Array

To convert an integer to a string, you can use Swift’s String() initializer. This allows you to convert all integers in the array to strings. This conversion can be easily performed using the map() function.

Step 2: Sort Strings in Descending Order

When sorting strings, you need to provide a custom sorting logic using the sorted(by:) method. Here, you can compare the lengths of the strings, or in case of equal lengths, compare them lexicographically.

Step 3: Return the Sorted Result

After sorting, you simply return the resulting array of strings. The final implementation will look like this.

Swift Code Implementation

func sortDigitsDescending(_ numbers: [Int]) -> [String] {
        // Step 1: Convert integers to strings
        let stringNumbers = numbers.map { String($0) }
        
        // Step 2: Sort strings in descending order (by digits)
        let sortedNumbers = stringNumbers.sorted { 
            (lhs, rhs) in
            if lhs.count != rhs.count {
                return lhs.count > rhs.count // Compare number of digits
            } else {
                return lhs > rhs // Compare values if lengths are equal
            }
        }
        
        // Step 3: Return the sorted result
        return sortedNumbers
    }

Code Explanation

The Swift function above works as follows:

  • numbers.map { String($0) }: Converts each element of the given integer array to a string.
  • sorted { ... }: Applies custom sorting logic to sort in descending order. It first compares the number of digits, and if they are the same, it compares the string values.
  • Returns the sorted array of strings as the result.

Test Cases

You can run a few test cases to check if the function works correctly. Below are some example test codes using the function.

let testCase1 = [321, 123, 456]
let result1 = sortDigitsDescending(testCase1)
print(result1) // ["321", "456", "123"]

let testCase2 = [12, 345, 7, 89]
let result2 = sortDigitsDescending(testCase2)
print(result2) // ["89", "7", "345", "12"]

let testCase3 = [10, 2, 1]
let result3 = sortDigitsDescending(testCase3)
print(result3) // ["2", "10", "1"]

Conclusion

In this article, we covered the problem of converting integers to strings and sorting them in descending order by digits using Swift. This was a useful exercise to enhance understanding of string manipulation and sorting algorithms. Problems like this frequently appear in coding tests and can be beneficial in real-world applications.

The coding implementation process allows you to utilize various features of Swift and aims to help improve problem-solving skills. In the next course, we will tackle even more interesting problems!

Swift Coding Test Course, Calculate the Sum of Remainders

Recently, the importance of algorithm coding tests in the field of software development has been growing. This is because many companies evaluate candidates’ problem-solving abilities through coding tests during the interview process. In this article, we will solve the ‘Subarray Sum Remainder’ problem using the Swift language, covering everything from understanding the problem to the overall resolution process.

Problem Description

Given an integer array nums and an integer k, write a function to calculate the remainder of the sum of all subarrays of nums when divided by k, and return the total sum of these remainders.

Example Problem

  • Input: nums = [1, 2, 3, 4, 5], k = 3
  • Output: 10
  • Description: The sums of the subarrays are 1, 1+2=3, 1+2+3=6, 1+2+3+4=10, ..., and we need to find the remainders of each when divided by 3.

Approach to the Problem

To solve this problem, we first need to generate all subarrays, calculate their sums, and then find the remainder when divided by k. However, generating all subarrays directly can lead to a time complexity of more than O(n^3), so a more efficient approach is needed.

We can accumulate the sums of the subarrays and use this to solve the problem. In this process, we can think about how to store the remainders.

Swift Implementation


func subarraySumRemainder(nums: [Int], k: Int) -> Int {
    var totalRemainderSum = 0
    let n = nums.count
    
    for start in 0..

Code Explanation

The above function subarraySumRemainder takes nums and k as inputs and calculates the sum of remainders.

  • totalRemainderSum: A variable that stores the sum of all subarray remainders.
  • n: Stores the size of the array.
  • for start in 0..: Sets the starting index for each subarray.
  • currentSum: A variable that stores the sum of the current subarray.
  • for end in start..: Sets the ending index for the subarray and updates the current sum.
  • totalRemainderSum += currentSum % k: Calculates the remainder and adds it to the total sum.

Time Complexity Analysis

The time complexity of this algorithm is O(n^2). This is because, in the worst case, we need to traverse all subarrays of the array. Since calculating the remainder can be done in constant time, the total complexity is determined by the time spent generating the subarrays.

Other Approaches

This problem can be solved in various ways. For example, using the prefix sum technique can provide better time complexity. By using prefix sums, we can quickly find the sum over a specific range, thus reducing the time required to calculate the remainder.

Method Using Prefix Sum


func subarraySumRemainderUsingPrefixSum(nums: [Int], k: Int) -> Int {
    var prefixSums = [Int](repeating: 0, count: nums.count + 1)
    for i in 1...nums.count {
        prefixSums[i] = prefixSums[i - 1] + nums[i - 1]
    }
    
    var totalRemainderSum = 0
    for start in 0..

Conclusion

In this article, we explored the method of solving the 'Subarray Sum Remainder' problem using the Swift language. We provided a comprehensive explanation from understanding the problem and approach to implementation and time complexity analysis. There are multiple approaches to algorithm problems, and they require creative thinking to solve. I hope you continue to develop your coding skills through consistent practice.

References

  • LeetCode (https://leetcode.com)
  • GeeksforGeeks (https://www.geeksforgeeks.org)
  • Algorithm Visualizer (https://algorithm-visualizer.org)

Swift Coding Test Course, Depth First Search

This course explains how to solve algorithm problems using the Depth-First Search (DFS) algorithm. In this article, we will address a real algorithm problem and detail the process and code to solve that problem.

Problem Description

We want to calculate the sum of all paths in a given binary tree. Each path is from the root to a leaf node. The binary tree is given in the following form:

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

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

Here, each node has a value (val) and is connected to a left child (left) and a right child (right). If the following binary tree is given:

               1
              / \
             2   3
            / \
           4   5
        

The sum of the paths from the root node 1 to the leaf nodes 4 and 5 is 7 (1+2+4) and 8 (1+2+5), so the final answer will be 15.

Approach to the Problem

We will use the DFS algorithm to solve this problem. DFS works by starting from a specific node, exploring as deeply as possible, and returning when no further nodes can be reached. In this case, we will calculate the cumulative sum of each path and record that sum when we reach a leaf node.

The steps to solve the problem are as follows:

  1. Start from the root node and perform DFS.
  2. Add the value (val) of the current node to the cumulative sum.
  3. If the current node is a leaf node, add the cumulative sum to the result.
  4. If it is not a leaf node, proceed to the left child and the right child, recursively calling DFS.
  5. Repeat the above process for all paths to calculate the cumulative sum.

Swift Code Implementation

Now let’s implement the above algorithm in code using Swift.

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

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

        class Solution {
            func sumNumbers(_ root: TreeNode?) -> Int {
                return dfs(root, 0)
            }

            private func dfs(_ node: TreeNode?, _ currentSum: Int) -> Int {
                guard let node = node else { return 0 }
                let newSum = currentSum * 10 + node.val

                if node.left == nil && node.right == nil {
                    return newSum
                }

                return dfs(node.left, newSum) + dfs(node.right, newSum)
            }
        }
        

The above code recursively performs DFS to calculate the sum of the paths. The sumNumbers function takes the root node as an argument, while the dfs function takes the current node and cumulative sum as arguments and returns the final sum. The process is as follows:

  1. When the sumNumbers function is called, DFS starts with a current cumulative sum of 0.
  2. For each node, the current sum is multiplied by 10 and the node value is added to create a new sum.
  3. When a leaf node is reached, that sum is returned, and after adding the sums of the left and right children, the final result is returned.

Test Cases

Let’s create a few test cases to verify that the code works correctly. Here are examples of various binary tree structures.

        let root = TreeNode(1)
        root.left = TreeNode(2)
        root.right = TreeNode(3)
        root.left!.left = TreeNode(4)
        root.left!.right = TreeNode(5)

        let solution = Solution()
        print(solution.sumNumbers(root)) // Output: 15
        

In the above test case, we calculate the sum of all paths starting from 1, passing through 2, and reaching 4 and 5. As a result, the sum of 1-2-4 and 1-2-5 adds up to 15.

Performance and Optimization

This problem has a time complexity of O(N) using the DFS method, where N is the number of nodes. The space complexity is O(H), where H is the height of the tree. Since DFS uses a stack, in the worst case, we may need to visit all nodes, resulting in memory usage proportional to the height.

Alternatively, we could use the BFS (Breadth-First Search) method to solve the problem, but for this specific problem of calculating the sum of paths to leaf nodes, DFS is more intuitive and efficient.

Conclusion

In this lecture, we addressed the problem of calculating the path sum to leaf nodes in a binary tree using depth-first search. We understood the concept of the DFS algorithm and implemented it in Swift. Such problems frequently appear in many coding tests, so we encourage you to practice more with various variations of the problem.

Now that we have covered basic DFS, challenge yourself with more complex problems. For example, consider exploring graph traversal, finding connected components, or shortest path problems. We hope you develop more proficient coding skills through practice.

We hope this article has been helpful in your preparation for code testing. If you have any questions or additional comments, please leave them below.