Category: Swift Coding Test

Trees are one of the important data structures in computer science.
Having a deep understanding of trees can be advantageous when solving various problems.
In this lecture, we will explore the basic concepts of trees and how to traverse trees using Swift.
At the end, we will solve a problem that may appear in a real coding test.

1. Basic Concepts of Trees

A tree is a data structure composed of nodes.
Each node has a value and other nodes (child nodes) connected to it.
Trees have the following characteristics.

• Root Node: The topmost node of the tree. (A node with no parent)
• Leaf Node: A node that has no children.
• Parent Node: A node that has children.
• Subtree: A tree with its child node as the root.

2. Tree Traversal Methods

There are several ways to traverse a tree.
The most representative methods are Pre-order, In-order, and Post-order traversals.

2.1 Pre-order Traversal

• Order: Node => Left Subtree => Right Subtree
• Feature: The root node is visited first.

2.2 In-order Traversal

• Order: Left Subtree => Node => Right Subtree
• Feature: When traversing a binary search tree, values are output in ascending order.

2.3 Post-order Traversal

• Order: Left Subtree => Right Subtree => Node
• Feature: All child nodes are visited before visiting the parent node.

3. Coding Test Problem: Depth of a Binary Tree

Now we will implement a binary tree and apply each traversal method.
The given problem is to calculate the depth (maximum height) of a binary tree.

Problem Description

Write a function to find the maximum depth of the given binary tree, using the root node as the reference.
The depth is defined as the number of nodes in the longest path from the root node to a leaf node.

Example

Input: 
    1
   / \
  2   3
 / \
4   5

Output: 3 (Path from root 1 to 4 or 5)

3.1 Implementing the Binary Tree Node Class

First, we will implement a class (Node) to represent the nodes of a binary tree.

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

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

3.2 Implementing the Depth Calculation Function

The function to calculate the maximum depth of a binary tree can be structured as follows.

func maxDepth(_ root: Node?) -> Int {
    guard let node = root else { return 0 }

    let leftDepth = maxDepth(node.left)
    let rightDepth = maxDepth(node.right)

    return max(leftDepth, rightDepth) + 1
}

3.3 Complete Code

The complete code is as follows.

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

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

func maxDepth(_ root: Node?) -> Int {
    guard let node = root else { return 0 }

    let leftDepth = maxDepth(node.left)
    let rightDepth = maxDepth(node.right)

    return max(leftDepth, rightDepth) + 1
}

// Example tree creation
let root = Node(value: 1)
root.left = Node(value: 2)
root.right = Node(value: 3)
root.left?.left = Node(value: 4)
root.left?.right = Node(value: 5)

// Print maximum depth
let depth = maxDepth(root)
print("The maximum depth of the tree is: \(depth)")  // Output: The maximum depth of the tree is: 3

4. Reflections and Conclusion

In this lecture, we learned about the concepts of trees and how to traverse trees using Swift,
as well as how to calculate the depth of a tree through a coding test problem.
Since trees are utilized in various algorithms and data structures,
it is important to thoroughly understand and practice these fundamental concepts.
Implementing them in Swift while debugging and optimizing the code is also essential.
I hope you gain more experience by solving more algorithm problems in the future.

Problem Description

In this lecture, we will learn about the Trie data structure and solve algorithmic problems using the Trie.

The problem is as follows:

Problem: Phone Number List
Write a function to check if any phone number in a given list is a prefix of another phone number.
If any of the consecutive phone numbers have a prefix relationship, return false, 
otherwise return true.

Input:
- An array of phone number strings, numbers. (1 ≤ numbers.length ≤ 1000, 1 ≤ numbers[i].length ≤ 30)

Output:
- Return false if phone numbers have a prefix relationship, otherwise return true.

Problem Solving Process

1. Understanding the Trie Data Structure

A Trie is a tree structure optimized for storing and searching strings. Each node in the Trie represents a path to the next character, making it useful for storing strings like phone numbers.
A key feature of the Trie is that it can effectively save memory when there are strings that share a common prefix.

2. Implementing the Trie

To implement a Trie, we define a node class and design the Trie as a structure that includes node objects.

class TrieNode {
    var children: [Character: TrieNode] = [:]
    var isEndOfNumber: Bool = false
}

3. Implementing the insert Method

We implement the insert method to insert a new phone number into the Trie. This method adds each character of the phone number to the node, and sets isEndOfNumber to true at the end of the phone number.

class Trie {
    var root: TrieNode
    
    init() {
        root = TrieNode()
    }
    
    func insert(_ number: String) {
        var currentNode = root
        for char in number {
            if currentNode.children[char] == nil {
                currentNode.children[char] = TrieNode()
            }
            currentNode = currentNode.children[char]!
        }
        currentNode.isEndOfNumber = true
    }
}

4. Implementing the checkPrefix Method

Next, we implement the checkPrefix method to check if a phone number in the list is a prefix of another phone number.
This method traverses the Trie and checks if the current string is the end of a phone number before reaching the end of another phone number to determine the prefix relationship.

func checkPrefix(_ number: String) -> Bool {
    var currentNode = root
    for (index, char) in number.enumerated() {
        if let node = currentNode.children[char] {
            currentNode = node
            // If the current node is an ending phone number
            if currentNode.isEndOfNumber {
                return false
            }
        } else {
            break
        }
        // If there are nodes existing beyond the last character
        if index < number.count - 1 && currentNode.isEndOfNumber {
            return false
        }
    }
    return true
}

5. Overall Solution

Finally, we insert all phone numbers into the Trie for the given list of phone numbers and call checkPrefix for each phone number to return the result.

func solution(_ numbers: [String]) -> Bool {
    let trie = Trie()
    
    for number in numbers {
        // Check if it is a prefix of an already registered phone number
        if !trie.checkPrefix(number) {
            return false
        }
        // Register the current phone number
        trie.insert(number)
    }
    return true
}

6. Time Complexity and Space Complexity

The time complexity of this algorithm using a Trie is O(N * M), where N is the number of phone numbers and M is the maximum length of the phone numbers.
The space complexity is O(N * M), representing the space needed to store the phone numbers.

Conclusion

In this lecture, we explored how to solve the problem of determining prefix relationships among a list of phone numbers using the Trie data structure.
Tries are a powerful tool for string processing and can be applied to various string-related problems.
Especially when dealing with a large number of strings, considering space complexity allows for efficient design.

Problem Description

Given an integer array nums, find all unique pairs of indices i and j such that nums[i] + nums[j] == target.
Each number can be used only once, and index pairs (i, j) and (j, i) are considered the same combination, so they should not be counted multiple times.

For example, if the array is nums = [2, 7, 11, 15] and target = 9, the index pair to return should be [0, 1].
This is because nums[0] + nums[1] = 2 + 7 = 9.

Problem Analysis

An efficient method is required to solve this problem. Generally, using a nested loop results in a time complexity of O(n^2), which is inefficient for large arrays.
Therefore, we will introduce a way to reduce the time complexity to O(n) by using the two-pointer technique.

Understanding the Two-Pointer Technique

The two-pointer technique is a useful algorithmic method for processing arrays, where two pointers are used to solve the problem.
This technique allows us to simplify complex problems with arrays.
Let’s consider a simple example. Assume we want to find all combinations of two numbers in a sorted array that sum to a specific value (target).

Initially, place one pointer at the start of the array and the other at the end. If the sum of the values pointed to by the two pointers is greater than target,
move the right pointer one step to the left. On the other hand, if the sum is less than target, move the left pointer one step to the right.
By adjusting the pointers in this way, we can check all possible combinations.

Problem-Solving Process

I will explain the steps to solve the problem.

Step 1: Sort the Array

To apply the two-pointer technique, the array must be sorted. Sorting simplifies the comparison and movement of values at each pointer.
In a sorted array, each number can be found more quickly, and duplicate combinations can be prevented.

Step 2: Initialize Pointers

Initialize the pointers. One starts at the beginning (index 0), and the other starts at the end (length of the array – 1).

Step 3: Compare Sum and Move Pointers

– Repeat until the pointers cross each other.
– If the sum of the numbers pointed to by the current pointers is greater than target, move the right pointer one step to the left (i.e., choose a smaller value).
– If the sum is less than target, move the left pointer one step to the right (i.e., choose a larger value).
– If the sum equals target, record the indices and move both pointers.

Step 4: Return Results

After finding all possible combinations, return the list of index pairs.

Swift Code Implementation

Below is the Swift code implementing the algorithm described above.

swift
import Foundation

func twoSum(nums: [Int], target: Int) -> [[Int]] {
    var result: [[Int]] = []
    var numDict: [Int: Int] = [:]

    for (index, num) in nums.enumerated() {
        let complement = target - num
        if let complementIndex = numDict[complement] {
            result.append([complementIndex, index])
        }
        numDict[num] = index
    }

    return result
}

// Example usage
let nums = [2, 7, 11, 15]
let target = 9
let indices = twoSum(nums: nums, target: target)
print(indices) // Output: [[0, 1]]

Conclusion

The two-pointer technique is a powerful tool that reduces time complexity and provides better performance. This technique can be useful in many algorithm problems,
especially in finding the sum of two numbers in an array.
Through this lesson, I hope you learn how to solve algorithm problems using the Swift language, and further develop your ability to tackle various problems.

Additional Practice Problems

Try solving the following problems.

• Find all index pairs in a sorted array whose sum equals a specific value.
• Count the number of unique pairs in a sorted array.
• Find all intervals with a cumulative sum equal to a specific value.
• Find all pairs within a given range that includes a specified number.

1. Problem Description

As you prepare for resignation, it is important to develop your ability to solve algorithm problems using the Swift language. The following is one of the frequently asked questions in Swift coding tests.

Problem: Sum of Two Numbers in an Array

Given an integer array nums and an integer target, write a function that returns the indices of the two numbers in nums that add up to target. It is assumed that there is exactly one solution for each input, and you cannot use the same element twice. The order of the returned indices does not matter.

Example

    Input: nums = [2, 7, 11, 15], target = 9
    Output: [0, 1]  // nums[0] + nums[1] == 9
    

2. Understanding the Problem and Approach

This problem can be approached as follows:

• A method using a double loop to check all possible two-number combinations
• A method using a hash map to check the requirements while traversing only once

To optimally meet the requirements, I will choose to implement the method using a hash map. This implementation has a time complexity of O(n) and a space complexity of O(n).

3. Swift Implementation

3.1. Setting Up Required Libraries and Basic Structure

    import Foundation

    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var numDict = [Int: Int]()
        // The contents of the function will go here.
    }
    

3.2. Implementation Using Hash Map

The following code implements the functionality of finding the indices of two numbers using a hash map:

    import Foundation

    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 [complementIndex, index]
            }
            numDict[num] = index
        }
        return []
    }
    

4. Function Explanation

The above twoSum function performs the following tasks:

1. Iterates through the given array nums and stores each integer in the hash map.
2. Calculates the value by subtracting the number from target for each integer. This is called the complement.
3. Checks if the complement exists in the hash map. If it does, adds that index to the result array.
4. If it does not exist, adds the current number and its index to the hash map.

5. Test Cases

Let’s write several cases to test the implemented function.

    let nums1 = [2, 7, 11, 15]
    let target1 = 9
    print(twoSum(nums1, target1))  // [0, 1]

    let nums2 = [3, 2, 4]
    let target2 = 6
    print(twoSum(nums2, target2))  // [1, 2]

    let nums3 = [3, 3]
    let target3 = 6
    print(twoSum(nums3, target3))  // [0, 1]
    

6. Complexity Analysis

The time complexity of the twoSum function is O(n) because it traverses the array once. The space complexity is O(n) because the hash map can hold up to n elements.

7. Conclusion and Further Learning

The problem of the sum of two numbers in an array is a very important problem in preparing for coding tests using Swift. This problem helps to understand and utilize the efficiency of hash maps. Let’s focus on enhancing our skills by solving various algorithm problems in the future.

8. References

• Apple Swift Documentation
• LeetCode – Two Sum Problem

Problem Description

Problem Solving Process

1. Problem Analysis

When given pairs of times, you need to compare the intervals to find the smallest value. One method to calculate time intervals is to convert the time into consumed minutes and then find the absolute difference between the two times.

2. Algorithm Design

First, convert the given list of times into values stored in minutes. Next, you can use a method to compare all pairs of times to find the shortest time interval. This process can be broken down into the following steps:

1. Convert time strings into time values
2. Compare all time combinations and store their differences
3. Output the minimum difference

3. Implementation

import Foundation

func timeToMinutes(time: String) -> Int {
    let components = time.split(separator: ":").map { Int($0)! }
    return components[0] * 60 + components[1]
}

func minTimeDifference(times: [String]) -> Int {
    var minutesList = times.map { timeToMinutes(time: $0) }.sorted()
    var minDiff = Int.max

    for i in 0..