Swift Coding Test Course, Queueing

Hello! Today, as part of the Swift coding test course, we will address the queueing problem.
This is one of the frequently encountered problems in coding interviews and is an algorithm that is widely used in actual work environments.
In this post, I will explain the definition of the problem, the solution method, a Swift code example for it, and optimization strategies in detail.

Problem Definition

There are several students standing in line. Given each student’s number,
the problem is to calculate how many exchanges are needed for the students to be arranged in the correct order.
Each student knows their position in line, and each student’s
number is given as a natural number from 1 to N. Only two students can swap positions at a time, and
the goal is to solve the problem with the minimum number of exchanges.

Input Format

The first line contains the number of students N, and the next line contains N integers separated by spaces,
representing the current positions of the students in line.

Example Input

5
2 1 5 3 4
    

Output Format

Output the minimum number of exchanges needed to sort in the correct order.

Example Output

3
    

Solution Process

To solve this problem, we need to find a method to minimize position changes.
It is important to identify the positions where exchanges are needed while checking if each student is in the order they should be.
The basic approach is to use “cycle detection.”
We trace each student, tracking the process of getting them into the correct position.

Step 1: Cycle Analysis

Initially, we calculate the correct position for each student using their numbers.
For example, student 1 should be at index 0 (i=0).
Student 2 should be at index 1 (j=1), and student 5 should be at index 4 (k=4).
We check the given order to determine the direction in which swaps should occur to move to these correct positions.

Step 2: Calculate Number of Swaps

To ensure each student is in the correct position, we examine the length of the cycles.
If the length of the cycle is k, then (k-1) swaps are needed.
For example, if the cycle length is 3, 2 swaps are needed.
Thus, summing the number of swaps for each cycle gives us the minimum number of swaps required.

Step 3: Implement Swift Code

Now, let’s write Swift code according to the algorithm described above.


    func minimumSwaps(arr: [Int]) -> Int {
        let n = arr.count
        var indexedArr = Array(zip(arr, 0.. 0 {
                swaps += (cycle_size - 1)
            }
        }
        
        return swaps
    }
    

Example Execution

Let’s run the example using the above function:


    let studentPositions = [2, 1, 5, 3, 4]
    let result = minimumSwaps(arr: studentPositions)
    print("Minimum number of swaps: \(result)")
    

Code Explanation

– The minimumSwaps function takes the array of the students’ current positions as input.
– First, it pairs the students with their indices and sorts them based on their correct positions.
– It declares an array for visited checking, examining the cycles at each student’s position,
– counting the size of the cycles for students who are not in the correct positions.
– Finally, it counts the necessary swaps for each cycle ((cycle size – 1)) and returns the total.

Optimization and Additional Considerations

This algorithm has a complexity of O(N log N) due to the sorting process, which can lead to significant performance degradation.
However, if the number of students is not too large, this approach is sufficiently fast and efficient.
While there are other methods to solve this problem, the cycle-based analysis is the most intuitive and straightforward.

Conclusion

Today, through the queueing problem using Swift,
we had the opportunity to enhance our algorithm problem-solving skills.
The code is simple, but it combines multiple concepts, particularly the ideal process of cycle detection.
I hope to build skills through more algorithms and problem-solving and be helpful in interview preparation!

Swift Coding Test Course, Jumong’s Command

Hello! Today, as part of the Swift coding test course, we will cover a topic titled ‘The Command of Jumong’. Through this problem, we can learn various techniques necessary for solving algorithmic problems.

Problem Description

In an ancient kingdom, Jumong decided to give commands to his subordinates. The command is in the following format:

“Everyone, X go out from your places!”

According to this command, the subordinates move. However, the distances they move may vary, and some subordinates may ignore Jumong’s command and not go out. The problem is to find the number of subordinates who follow Jumong’s command.

Input

  • The number of subordinates N (1 ≤ N ≤ 105)
  • The distance K that the subordinates follow Jumong’s command (1 ≤ K ≤ 106)
  • An array A of distances that N subordinates follow (0 ≤ A[i] ≤ 106)

Output

Output the number of subordinates who follow Jumong’s command.

Example Input

   5
   10
   8 9 10 11 12
   

Example Output

   3
   

Problem Solving Process

The first step to solve this problem is to accurately understand the input. Here, we need to receive the number of subordinates, the command distance, and the distances each subordinate follows.

Step 1: Get Input

We can use Swift’s basic input functions to receive the data.

   let n = Int(readLine()!)!
   let k = Int(readLine()!)!
   let distances = readLine()!.split(separator: " ").map { Int(String($0))! }
   

Step 2: Count Subordinates That Meet the Condition

To count the number of subordinates who follow Jumong’s command, we will iterate through the given distance array and check if each subordinate’s distance is greater than or equal to the command distance K. We will increase the count of subordinates that meet the condition through this process.

   var count = 0
   for distance in distances {
       if distance >= k {
           count += 1
       }
   }
   

Step 3: Output Result

Finally, we will output the number of subordinates who follow Jumong’s command.

   print(count)
   

Final Code

   import Foundation
   
   let n = Int(readLine()!)!
   let k = Int(readLine()!)!
   let distances = readLine()!.split(separator: " ").map { Int(String($0))! }
   
   var count = 0
   for distance in distances {
       if distance >= k {
           count += 1
       }
   }
   
   print(count)
   

Summary

In this lesson, we learned how to solve a simple algorithmic problem using the Swift language through the problem ‘The Command of Jumong’. We practiced how to approach the problem and write the necessary code for each step.

This problem-solving process is very similar to the thought process required in actual coding interviews, so I hope you improve your skills through practice!

Additional Practice Problems

If you want more practice, try solving the following additional problems.

  • Output the number of subordinates whose distances are less than Jumong’s command distance K.
  • Calculate and output the difference between the maximum and minimum distances of subordinates.
  • Randomly generate Jumong’s command distance K, and calculate the ratio of N subordinates who follow that command.

Thank you!

Swift Coding Test Course, Understanding Combinations

In this course, we will explore the concept of combinations and how to utilize it in preparation for Swift coding tests for employment. A combination refers to the way of selecting r elements from a given set of n elements. This can help solve various algorithmic problems.

Definition of Combinations

Combinations are used when the order of selection does not matter. For example, when selecting 2 elements from {A, B, C}, {A, B} and {B, A} are considered the same combination. Mathematically, combinations can be expressed as follows:

C(n, r) = n! / (r! * (n - r)!)

Here, ! denotes factorial, and n! represents the product of all positive integers up to n.

Problem Description

Now, let’s look at a problem that utilizes combinations:

Problem: Sum of Combinations

For given integers n and k, write a function countCombinations(n: Int, k: Int, target: Int) -> Int to find the number of combinations of selecting k elements from the numbers from 1 to n such that their sum equals target.

Input

  • n: Integer (1 ≤ n ≤ 20)
  • k: Integer (1 ≤ k ≤ n)
  • target: Integer (1 ≤ target ≤ 100)

Output

Print the number of combinations.

Problem-Solving Process

Now let’s explore the step-by-step process to analyze and solve the problem.

Step 1: Problem Analysis

This problem involves finding combinations of selecting k elements from the given n numbers such that their sum equals target. Since the selection does not depend on the order, combinations should be used. Therefore, it is essential to first understand and implement the concept of combinations.

Step 2: Combination Generation Algorithm

To generate combinations, a backtracking technique using recursion is commonly employed. This allows us to determine the current selected numbers and the range of numbers to be used next. In particular, each time a number is selected, it is necessary to check if that number has been chosen.

Step 3: Code Implementation

Below is an example code for a combination generator that includes the method to check if the sum of selected specific numbers equals target:

func countCombinations(n: Int, k: Int, target: Int) -> Int {
    var count = 0
    var combination = [Int]()
    
    func backtrack(start: Int, k: Int, sum: Int) {
        if k == 0 {
            if sum == target {
                count += 1
            }
            return
        }
        
        for i in start...n {
            combination.append(i)
            backtrack(start: i + 1, k: k - 1, sum: sum + i)
            combination.removeLast()
        }
    }
    
    backtrack(start: 1, k: k, sum: 0)
    return count
}

Step 4: Code Explanation

Let’s take a look at the above code:

  • countCombinations: This method counts the total number of combinations. It starts the combination generation by passing initial values to the backtrack method.
  • backtrack: Called recursively, it adds a specific number to the current combination and selects the next number.
  • Conditional Statement: If k is 0, it checks if the combination of the currently selected numbers matches target.

Step 5: Testing

Now let’s validate the function with several test cases:

print(countCombinations(n: 5, k: 2, target: 5))  // Output: 1 (Combination: [2, 3])
print(countCombinations(n: 7, k: 3, target: 10)) // Output: 4 (Combinations: [1, 2, 7], [1, 3, 6], [2, 3, 5], [1, 4, 5])
print(countCombinations(n: 10, k: 2, target: 8)) // Output: 3 (Combinations: [2, 6], [3, 5], [1, 7])

Conclusion

In this lecture, we explored the process of implementing combinations in Swift and solving the problem of finding the number of combinations that satisfy specific conditions. Understanding algorithms like combinations is crucial for coding tests in job interviews, so consistent practice is recommended.

Continue to tackle various algorithmic problems to enhance your skills. Thank you!

Swift Coding Test Course, Picking Up Pebbles

Hello, everyone! Today we will take an in-depth look at the coding test problem ‘Stone Removal’ using Swift. This problem is useful for laying the foundation of algorithms and is frequently featured in practical exams. In this post, we will explain the problem description, solution approach, Swift code, and detail the problem-solving process step by step.

Problem Description

Let’s assume a game of removing stones from a given array. Each element of the array represents the number of stones. In each turn, the player can remove a required number of stones, with the goal of removing all the stones. However, the rules for removing stones are as follows:

  • The player must remove at least one stone in each turn.
  • The maximum number of stones that can be removed in each turn must be equal to or less than the number of stone piles.

The objective of this game is to determine the minimum number of turns required to remove all the stones. Through this problem, you can practice algorithmic thinking and Swift syntax.

Approach

To solve this problem, we need to traverse each element of the given array and calculate the total number of turns. The problem can be solved using a loop. The main steps can be summarized as follows.

  1. Sum the number of stones from the array.
  2. To calculate the total number of turns, divide the number of stones by the maximum number that can be removed per turn.
  3. If there’s a remainder, an additional turn must be added.

Now, let’s write the Swift code.

Swift Code


func minimumTurns(to collectStones stones: [Int]) -> Int {
    // Calculate the total number of stones
    let totalStones = stones.reduce(0, +)
    
    // Calculate available turns
    let turns = totalStones / stones.count
    let remainder = totalStones % stones.count
    
    // Add an extra turn if there’s a remainder
    return remainder == 0 ? turns : turns + 1
}

// Test example
let stones = [3, 5, 2, 6]
let result = minimumTurns(to: collectStones: stones)
print("Minimum number of turns: \(result)")

Problem-Solving Process

The above code first uses the `reduce` function on the stone array to calculate the total number of stones. Next, it divides by the size of the array to compute the average number of turns. If there is a remainder, it adds 1 to the base number of turns to return the final number of turns.

Example Analysis

Let’s examine the case where the stone array is [3, 5, 2, 6]. First, the total sum of the stones is:

3 + 5 + 2 + 6 = 16

Since the number of stones is 4, the average number of turns is:

16 / 4 = 4

Since there is no remainder, the minimum number of turns is 4.

Conclusion

In this lecture, we covered the process of solving the ‘Stone Removal’ problem using Swift. By understanding the rules of the problem and implementing an algorithm to calculate the optimal number of turns, you can improve your algorithmic thinking. Don’t forget to continue practicing basic algorithm problems like this while preparing for coding tests. In the next lecture, we will tackle more complex algorithm problems. Thank you!

Swift Coding Test Course, Finding Non-Square Numbers

Problem Description

We are trying to find the number of non-perfect square numbers among integers from 1 to N. A perfect square refers to a number obtained by squaring an integer. For example, 1, 4, 9, 16, 25, etc., are all perfect squares. On the other hand, 2, 3, 5, 6, 7, 8, 10, etc., are not perfect squares.

Input Format

The first line contains the integer N (1 ≤ N ≤ 106).

Output Format

Print the count of non-perfect square numbers among integers from 1 to N.

Sample Input

10

Sample Output

7

Problem Solving Process

To solve this problem, we need to subtract the count of perfect squares from the total count of numbers. The following steps outline the procedure.

Step 1: Identify the range of perfect squares

Perfect squares are generated in the form of 1, 4, 9, 16, … etc. If N is 10, the perfect squares are 1(12), 4(22), and 9(32). In this case, there are a total of 3 perfect squares.

Step 2: Calculate the count of perfect squares

For N, we need to find the maximum integer k such that k2 <= N. This k can be calculated as the integer part of √N.

Step 3: Derive the result

The total count of numbers is N, and the count of perfect squares is k. Therefore, the count of non-perfect square numbers can be calculated as N - k.

Implementation Code (Swift)

func countNonPerfectSquares(N: Int) -> Int {
        // Calculate k to find perfect squares among numbers from 1 to N.
        let k = Int(sqrt(Double(N)))
        
        // Calculate the count of non-perfect square numbers.
        return N - k
    }

// Example execution
let N = 10
print(countNonPerfectSquares(N: N)) // Result: 7
    

Complexity Analysis

This algorithm has a time complexity of O(1). The operation of calculating the square root for a given N is performed in constant time, making it very efficient. Memory usage is also limited to a constant, so this problem operates reliably even with large inputs.

Post Analysis

This problem allowed us to understand the concept of perfect squares along with the efficiency of square root calculations. We also learned how to solve complex problems through very simple calculations.

Conclusion

In the process of solving algorithm problems, it is important to understand the problem, devise a step-by-step solution, and implement it efficiently. This process will greatly help in dealing with various types of algorithm problems. In the next lecture, we will tackle more complex algorithm problems. Thank you!