Category: Swift Coding Test

Hello! In this blog post, we will take a deep dive into how to solve the Longest Common Subsequence (LCS) problem using the Swift programming language. The LCS problem involves finding the longest subsequence that appears in both of two strings. This can be solved through various algorithms and dynamic programming. In this article, I will guide you through understanding the problem and implementing the algorithm in detail.

1. Understanding the Problem

The Longest Common Subsequence (LCS) problem is to find the longest subsequence that is common to both sequences while maintaining the original order. For example, given the two strings “ABCBDAB” and “BDCAB”, their LCS is “BDAB”, and its length is 4.

2. Problem Description

Let’s write a function to determine the length of the LCS for the given two strings S1 and S2. Let’s assume the two strings are defined as follows:

S1 = "ABCBDAB"
S2 = "BDCAB"

Now, let’s explore a general approach to find the LCS.

3. Problem-Solving Methodology

The most well-known method to solve the LCS problem is Dynamic Programming. This approach involves breaking the problem down into subproblems, solving those, and reusing the results to derive a solution for the entire problem. I will explain this process step by step.

3.1 Initializing the Dynamic Programming Table

First, let’s denote the lengths of the two strings S1 and S2 as m and n, respectively. We create and initialize a 2D array (dp) of size m+1 x n+1. Each element dp[i][j] represents the length of LCS of the first i characters of S1 and the first j characters of S2. The initialization process is as follows:

for i in 0 to m: 
    dp[i][0] = 0
for j in 0 to n: 
    dp[0][j] = 0

3.2 Dynamic Programming Recurrence Relation

Now let’s define the recurrence relation to update the LCS values for each character. If the i-th character of S1 and the j-th character of S2 are the same, the length of LCS up to that character is the previous LCS length plus 1. That is:

if S1[i-1] == S2[j-1] then
    dp[i][j] = dp[i-1][j-1] + 1
else
    dp[i][j] = max(dp[i-1][j], dp[i][j-1])

This recurrence relation allows us to calculate all elements, and ultimately, we can obtain the length of LCS at dp[m][n].

4. Implementing the Swift Code

Now, based on the above process, let’s implement LCS in Swift. Below is the function that calculates the longest common subsequence.

func longestCommonSubsequence(_ S1: String, _ S2: String) -> Int {
    let s1Array = Array(S1)
    let s2Array = Array(S2)
    let m = s1Array.count
    let n = s2Array.count
    var dp = Array(repeating: Array(repeating: 0, count: n + 1), count: m + 1)

    for i in 1...m {
        for j in 1...n {
            if s1Array[i - 1] == s2Array[j - 1] {
                dp[i][j] = dp[i - 1][j - 1] + 1
            } else {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
            }
        }
    }

    return dp[m][n]
}

The above function takes two strings S1 and S2 as arguments and returns the length of the longest common subsequence.

5. Test Cases

Let’s create a few test cases to test the function.

let S1 = "ABCBDAB"
let S2 = "BDCAB"
let result = longestCommonSubsequence(S1, S2)
print("The length of the longest common subsequence is \(result).") // Output: 4

Through testing, we can verify that our algorithm returns the correct results.

6. Time Complexity Analysis

The time complexity of this algorithm using dynamic programming is O(m*n), and the space complexity is also O(m*n). Here, m and n represent the lengths of the two strings. This complexity can grow rapidly as the length of the strings increases, so optimization techniques such as memoization can be applied to reduce space complexity.

7. Conclusion

In this article, we explored using dynamic programming techniques to solve the longest common subsequence problem and how to implement it in Swift. The LCS problem is widely used in computer science and plays an important role in various applications. This algorithm can be utilized in many scenarios, enhancing our ability to solve programming problems.

I hope that by continuing to solve various algorithm problems, you will find it helpful in preparing for coding tests. Thank you!

Problem: Finding Parentheses Arrangement that Minimizes Value

Algorithm problems play an important role in coding tests, and especially problems related to parentheses are often presented. In this article, we will address the challenge of finding the minimum value by considering all possible parentheses arrangements for a given expression. We will explain the theories and algorithms necessary to solve this problem in detail.

Problem Description

There is a string composed of given numbers and operators. This string can be calculated by placing parentheses in various ways. Our goal is to find the minimum value that can be obtained by placing parentheses.

Input: “1+2*3-4*5”
Output: -1

Approach to the Problem

This problem can be solved using Dynamic Programming. We need to consider all possible placements of parentheses for the given expression, so the Divide and Conquer technique will also be utilized.

1. Parsing the Expression

First, we need to separate the numbers and operators from the expression and store them in arrays. For example, we will convert the string “1+2*3-4*5” into the following format.

    Number array: [1, 2, 3, 4, 5]
    Operator array: ['+', '*', '-', '*']
    

2. Defining Dynamic Programming

Dynamic programming is a method that stores the results of each subproblem through several auxiliary arrays for future use. By using memoization, we can reduce time complexity by reusing results that have already been calculated.

3. Implementing the Recursive Function

We will calculate all possible combinations through a recursive function. This function divides based on the given index and calculates by placing parentheses differently for each side.

Swift Implementation

Below is the complete code implemented in the Swift programming language.

    func minValue(expression: String) -> Int {
        var numbers: [Int] = []
        var operators: [Character] = []
        
        // Step 1: Parse the expression
        var currentNumber = ""
        for char in expression {
            if char.isNumber {
                currentNumber.append(char)
            } else {
                numbers.append(Int(currentNumber)!)
                operators.append(char)
                currentNumber = ""
            }
        }
        numbers.append(Int(currentNumber)!)
        
        // Step 2: Create a memoization table
        var memo: [[Int?]] = Array(repeating: Array(repeating: nil, count: numbers.count), count: numbers.count)
        
        // Step 3: Define a recursive function
        func compute(_ left: Int, _ right: Int, _ op: Character) -> Int {
            switch op {
            case '+': return left + right
            case '-': return left - right
            case '*': return left * right
            default: fatalError("Unknown operator")
            }
        }

        // Recursive function for minimum value
        func findMinValue(left: Int, right: Int) -> Int {
            if left == right {
                return numbers[left]
            }
            if let result = memo[left][right] {
                return result
            }

            var result = Int.max
            for i in left..