Category: Swift Coding Test

Recently, many companies are conducting coding tests during the hiring process. In this article, I will cover a problem that can help you prepare for coding tests using the Swift programming language, and I will explain in detail how to solve that problem efficiently.

Problem: Sum of Two Numbers in an Array

Given an integer array nums and an integer target, you need to solve the problem of finding the two indices in the array such that their sum is equal to target. It is assumed that each input has exactly one solution, and you may not use the same element twice.

Input Format

• nums: [2, 7, 11, 15]
• target: 9

Output Format

[0, 1] (nums[0] + nums[1] = 2 + 7 = 9)

Problem Analysis

This problem is quite famous and can be solved in various ways. The most basic approach is to use a double loop, which has a time complexity of O(n^2). However, let’s explore a more efficient way.

Efficient Approach: Using Hash Map

You can use a method of iterating through the given array and storing each element in a hash map. By using a hash map, you can reduce the search time to O(1), thereby decreasing the overall time complexity to O(n).

Steps to Solve the Problem

1. Create an empty hash map.
2. Iterate through the array and calculate the difference between the current number current and target - current.
3. If target - current exists in the hash map, return the corresponding index and the current index.
4. Add the current number and its index to the hash map.

Swift Code

let nums = [2, 7, 11, 15]
let target = 9

func twoSum(nums: [Int], target: Int) -> [Int]? {
    var numDict = [Int: Int]()
    
    for (index, num) in nums.enumerated() {
        let complement = target - num
        
        if let complementIndex = numDict[complement] {
            // Return the two indices
            return [complementIndex, index]
        }
        
        // Add the current number and index to the hash map
        numDict[num] = index
    }
    
    // Return nil if no solution exists
    return nil
}

if let result = twoSum(nums: nums, target: target) {
    print(result) // [0, 1]
}
    

Result Verification

When you run the above code, it will print the result [0, 1]. This confirms that the sum of nums[0] and nums[1] is equal to target.

Optimization Considerations

This algorithm demonstrates an optimized approach to the given problem. Using the hash map allows us to solve the problem with an average time complexity of O(n). However, in the worst case, performance can degrade due to hash collisions, so it’s essential to use an appropriate hash function.

Conclusion

In this article, we explored how to solve coding test problems using Swift. The hash map approach can be applied in various situations and can significantly help in writing efficient code. I encourage you to continue solving various algorithm problems to improve your skills.

There will be more courses covering various algorithm problems and their solutions, so please stay tuned!

Problem Description

The meeting room management system is a process that allocates meeting rooms according to the schedule of meetings to use multiple meeting rooms efficiently. Given the start and end times of a meeting, propose a method to allocate meeting rooms as efficiently as possible.

When a specific meeting starts, it is necessary to find a meeting room that is not occupied by another meeting. The meeting room allocation problem includes the following inputs:

• The number of meetings n
• An array meetings containing the start and end times of the meetings, where each meeting consists of a start time and an end time.

Input

n = 3
meetings = [(0, 30), (5, 10), (15, 20)]

Output

Number of meeting rooms: 2

Problem Solving Approach

To solve this problem, I will follow these steps:

1. Sort the meetings based on their end times with respect to their start and end times.
2. Iterate through each meeting, comparing the start time of the current meeting with the end time of the previous meeting.
3. If a meeting room is needed, increase the number of meeting rooms, and release that room when the meeting ends.

Algorithm Implementation

I will implement an algorithm to solve the meeting room allocation problem using Swift. Below is the actual implementation code:

func minMeetingRooms(_ meetings: [[Int]]) -> Int {
    guard meetings.count > 0 else {
        return 0
    }
    
    var startTimes = meetings.map { $0[0] }
    var endTimes = meetings.map { $0[1] }
    
    startTimes.sort()
    endTimes.sort()
    
    var roomCount = 0
    var endPointer = 0
    
    for startTime in startTimes {
        if startTime < endTimes[endPointer] {
            roomCount += 1
        } else {
            endPointer += 1
        }
    }
    
    return roomCount
}

Code Explanation

The above code separates the start and end times of each meeting from the input array and sorts them. Then it uses two pointers to compare the start and end times of the meetings to calculate the number of meeting rooms needed.

1. The guard statement returns 0 if there are no meetings.
2. Extract and sort the start and end times of the meetings.
3. The first pointer checks the start time while iterating through each meeting, and the second pointer tracks the end times.
4. If the start time is earlier than the end time, a new meeting room is needed, so roomCount is increased.
5. After all meetings are processed, the number of required meeting rooms is returned.

Complexity Analysis

This algorithm has the following complexities:

• Time Complexity: O(n log n) – it takes O(n log n) time to sort the start and end times.
• Space Complexity: O(n) – it uses additional space to store the meeting start and end times.

Conclusion

The meeting room allocation problem is an important issue of managing overlapping meetings. The given algorithm can enhance the efficiency of meeting room usage and is a common problem in coding tests. Understanding how to solve the problem with Swift and practicing actual code implementations can be beneficial.

Additional Practice Problems

• Determine how to handle cases where there are meetings with the same end time.
• Randomly generate meeting times to test meeting room allocation based on the given number of meetings.

I hope this post helps you. I encourage you to improve your algorithm problem-solving skills through this problem!

Hello! Today we will learn about one of the mathematical algorithms, the Extended Euclidean Algorithm. This course is designed to help you prepare for algorithm exams by explaining problem-solving methods in detail based on algorithmic thinking, and it can be implemented in Swift.

1. What is the Euclidean Algorithm?

The Euclidean algorithm is a classical algorithm for finding the greatest common divisor (GCD) of two numbers \(a\) and \(b\). The process is as follows:

GCD(a, b) = GCD(b, a % b)

This expression continues to repeat until \(b\) becomes 0, and \(GCD(a, 0) = a\) holds. In other words, the value remaining in the first argument is the GCD of the two numbers.

2. What is the Extended Euclidean Algorithm?

The Extended Euclidean Algorithm builds upon the Euclidean Algorithm to find the following expression for two integers \(a\) and \(b\):

ax + by = gcd(a, b)

Here, \(x\) and \(y\) are integers, commonly known as Bézout’s identity. Using this algorithm, you can find not only the GCD of the given two numbers but also the coefficients for the GCD.

3. Problem Description

Now, let’s look at a problem that can be solved using the Extended Euclidean Algorithm:

Problem: Given two integers a and b, find x and y that satisfy ax + by = gcd(a, b).

Input Format

• Two integers \(a\) and \(b\) are given. (1 ≤ a, b ≤ 1,000,000)

Output Format

• Output GCD(a, b), x, and y separated by spaces.

Example Input

30 12

Example Output

6 1 -2

4. Algorithm Solution Process

4.1. Problem Analysis

To solve this problem, we need to find the GCD of the given two numbers and the coefficients x and y to find that GCD. For this, we need to implement the Extended Euclidean Algorithm.

4.2. Algorithm Flowchart

1. Receive two numbers \(a\) and \(b\) as input.
2. Find the GCD using the Euclidean Algorithm.
3. Use the Extended Euclidean Algorithm to find the coefficients x and y for the linear combination.
4. Output the result.

4.3. Swift Implementation

Now let’s implement the above process in Swift code:

func extendedGCD(a: Int, b: Int) -> (gcd: Int, x: Int, y: Int) {
    if b == 0 {
        return (a, 1, 0)
    } else {
        let (gcd, x1, y1) = extendedGCD(a: b, b: a % b)
        let x = y1
        let y = x1 - (a / b) * y1
        return (gcd, x, y)
    }
}

func main() {
    let input = readLine()!.split(separator: " ").map { Int($0)! }
    let a = input[0]
    let b = input[1]
    let (gcd, x, y) = extendedGCD(a: a, b: b)
    print("\(gcd) \(x) \(y)")
}

// Program entry point
main()

5. Example Problem Solution

Let’s use the above Swift code to solve the given problem.

Input

30 12

Output

6 1 -2

For the above input, the GCD is 6, which can be confirmed with \(30x + 12y = 6\) where \(x = 1\) and \(y = -2\).

6. Conclusion

Today, we learned about the Extended Euclidean Algorithm and how to solve a given problem using it. This algorithm plays an important role in computer science and cryptography. Remember that it can be effectively used in various applications!

I hope you continue to enhance your coding skills through various algorithm-related courses. If you have any questions or want to learn more, please leave a comment!

In this lecture, we will solve an algorithm problem that finds the minimum number of multiplication operations required for matrix multiplication using Swift. This problem is a good example of utilizing dynamic programming techniques and is a type of problem that can frequently appear in practical coding tests.

Problem Description

The problem is to calculate the minimum number of multiplication operations required to multiply the given matrices in order. Since the size of the matrices affects the multiplication operations, we need to find the optimal multiplication order.

Problem Specification

   Input: Integer array arr (size n+1)
   Each arr[i] represents the number of rows and columns of the ith matrix. 
   (In other words, matrix A has the size arr[i-1] * arr[i]. Here, arr[0] is the number of rows of the first matrix, and arr[n] is the number of columns of the last matrix)

   Output: Minimum number of multiplication operations required for matrix multiplication

Example

Input: arr = [10, 20, 30, 40]
Output: 3000
Description: To achieve the minimum number of operations, we need to perform multiplication in the optimal order: (A1(10×20) * A2(20×30)) * A3(30×40) = 10 * 20 * 40 + (A1 * A2) * A3.

Problem Solving Approach

This problem can be solved using dynamic programming.
The basis of the algorithm is as follows.

• Understand the optimal structure of matrix multiplication.
• Define the recurrence relation: Define dp[i][j] as the minimum multiplication cost of multiplying matrices from index i to j.
• Derive the recurrence relation for minimum operations and fill the dp array.
• Return the result.

Recurrence Relation

Considering the multiplication from matrix A to B, and setting K as the point that divides A and B, we can establish the following equation:

   dp[i][j] = min(dp[i][k] + dp[k+1][j] + arr[i]*arr[k+1]*arr[j+1])

This equation represents the process of partitioning into several parts to find the optimal number of multiplication operations, where k is any possible splitting point between i and j.

Implementation

Now let’s write Swift code to solve the problem.

func matrixChainOrder(arr: [Int]) -> Int {
    let n = arr.count - 1
    var dp = Array(repeating: Array(repeating: 0, count: n), count: n)

    for length in 2...n { // Loop through the length of the matrices
        for i in 0..< (n - length + 1) {
            let j = i + length - 1
            dp[i][j] = Int.max
            for k in i..< (j) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + arr[i] * arr[k + 1] * arr[j + 1])
            }
        }
    }
    
    return dp[0][n - 1]
}

// Example usage
let arr = [10, 20, 30, 40]
let result = matrixChainOrder(arr: arr)
print("Minimum number of multiplication operations: \(result)")

Code Explanation

The above code is a function that calculates the minimum number of operations for matrix multiplication.

• First, it initializes the dp array based on the number of matrices received as input.
• Next, it uses a double loop to calculate the number of operations for all possible combinations of matrices.
• To find the smallest value, it explores possible split points in the third loop.
• Finally, it returns dp[0][n - 1] to output the minimum value for all matrix multiplications.

Complexity Analysis

The time complexity of this algorithm is O(n^3). This indicates that the number of operations increases proportional to the number of matrices. The space complexity is O(n^2), which increases due to the dp array used following the recurrence relation.

Conclusion

In this lecture, we explained the problem of finding the minimum number of multiplication operations for matrices. We learned how to derive the optimal solution using dynamic programming techniques and how to implement it in Swift. This algorithm is very useful for effectively solving complex problems and is a frequently presented topic in future coding tests.

We hope this problem encourages you to research various algorithms and deterministic problem-solving methods, and to enhance your coding skills further.

Let’s learn about the Floyd-Warshall algorithm, which often appears in coding tests, and solve related algorithm problems. This article will detail the overview, working principle, and example problem-solving process of the Floyd-Warshall algorithm.

1. Overview of the Floyd-Warshall Algorithm

The Floyd-Warshall algorithm is an algorithm used to find the shortest paths between all pairs of vertices in a graph. This algorithm is based on dynamic programming and updates the shortest paths by considering paths that pass through vertex k at each step. The time complexity of the Floyd-Warshall algorithm is O(V³), where V represents the number of vertices.

1.1. Significance of the Algorithm

The Floyd-Warshall algorithm exhibits excellent efficiency in that it can identify the shortest paths between all pairs of vertices through a single process, not just finding the shortest path from a single starting point.

1.2. Applications

This algorithm is applied in various fields including calculating the shortest paths between all nodes in a network, optimizing road networks, and computing movement paths in games.

2. Working Principle of the Algorithm

The Floyd-Warshall algorithm operates in the following manner:

1. Initialize the shortest paths for all edges of the graph. The distance between two directly connected vertices is set to the weight of the edge, while the remaining pairs are set to infinity.
2. For each vertex k, update the shortest paths for all combinations of vertices i and j as follows: dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
3. Repeat this process for all pairs of vertices.

This algorithm allows us to find the shortest paths when there are paths that pass through vertex k.

3. Example Problem

Problem Description

Here is a problem that can be solved using the Floyd-Warshall algorithm:

Given a directed graph consisting of N vertices and M edges, write a program to find the shortest distance between each pair of vertices. The weights of the edges are given as positive integers, and if there is no edge between two vertices, it should be set to infinity.

Problem Input

The first line contains the number of vertices N (1 ≤ N ≤ 100) and the number of edges M (1 ≤ M ≤ 100,000). The next M lines provide the starting vertex, ending vertex, and weight of each edge. If there are multiple edges between two vertices, only the edge with the smallest weight should be considered.

Problem Output

Output the shortest distance between each pair of vertices. If the shortest distance does not exist, output INF.

Example

Input:

        4 7
        1 2 1
        1 3 3
        2 3 1
        2 4 2
        3 4 1
        4 1 2
        4 2 3
        

Output:

        0 1 2 3
        INF 0 1 2
        INF INF 0 1
        2 1 2 0
        

4. Problem Solving Process

To solve the problem, we will implement the Floyd-Warshall algorithm. Below is the step-by-step process of problem-solving.

4.1. Initial Setup

First, set up the initial distance array using N vertices and M edges. The weights between directly connected vertices are set to the given values, and the distances for unconnected vertices are set to infinity.

4.2. Initialize Distance Array

        var N = 4
        var M = 7
        var dist = [[Int]](repeating: [Int](repeating: Int.max, count: N + 1), count: N + 1)
        
        for i in 1...N {
            dist[i][i] = 0
        }
        

This code initializes the distance array dist to infinity and sets the distance to itself as 0.

4.3. Input Edge Information

        dist[1][2] = 1
        dist[1][3] = 3
        dist[2][3] = 1
        dist[2][4] = 2
        dist[3][4] = 1
        dist[4][1] = 2
        dist[4][2] = 3
        

Now, set the weights between directly connected vertices using the edge information. If there are multiple edges, update it to the minimum weight.

4.4. Apply the Floyd-Warshall Algorithm

        for k in 1...N {
            for i in 1...N {
                for j in 1...N {
                    if dist[i][j] > dist[i][k] + dist[k][j] {
                        dist[i][j] = dist[i][k] + dist[k][j]
                    }
                }
            }
        }
        

Here, k acts as an intermediate vertex, updating the shortest path from i to j.

4.5. Output Results

        for i in 1...N {
            for j in 1...N {
                if dist[i][j] == Int.max {
                    print("INF", terminator: " ")
                } else {
                    print(dist[i][j], terminator: " ")
                }
            }
            print("")
        }
        

Finally, output the distance array to check the shortest distances between each pair of vertices.

5. Conclusion

In this lecture, we learned the concepts and operating principles of the Floyd-Warshall algorithm and examined the process of solving actual problems. This algorithm is a powerful tool for efficiently finding the shortest paths between all pairs of vertices. It is important to solve various examples to enhance your understanding of the algorithm. You should also practice solving algorithm problems in Swift to improve your skills!