Category: Javascript Coding Test

Problem Description

This is a problem of determining the direction of the line segment AB defined by two given points A(x1, y1) and B(x2, y2). The result of the direction must be one of the following three:

• “UP”: If the line segment is directed upwards
• “DOWN”: If the line segment is directed downwards
• “HORIZONTAL”: If the line segment is horizontal

Input Format

The coordinates of the two points A(x1, y1) and B(x2, y2) are provided as input. Each coordinate is given as an integer.

Output Format

A string indicating the direction of the line segment AB is output.

Example

Solution Process

1. Understanding the Problem

To understand the problem, we need to determine the direction of line segment AB when two points A and B are given. The direction of the line segment is determined by the difference in y-coordinates (Δy) and the difference in x-coordinates (Δx). By evaluating these differences, we can deduce if the line segment is horizontal, upward, or downward.

2. Building the Logic to Determine the Direction of the Line Segment

The basic formulas needed to determine the direction of the line segment are as follows:

• Δy = y2 – y1
• Δx = x2 – x1

Now, we can determine the direction through three cases:

• If Δy > 0, the line segment is pointing upwards, so we return “UP”.
• If Δy < 0, the line segment is pointing downwards, so we return “DOWN”.
• If Δy == 0, the line segment is horizontal, so we return “HORIZONTAL”.

3. Implementing in JavaScript

The required JavaScript code to solve the problem is as follows:

function getDirection(x1, y1, x2, y2) {
    const deltaY = y2 - y1;
    
    if (deltaY > 0) {
        return "UP";
    } else if (deltaY < 0) {
        return "DOWN";
    } else {
        return "HORIZONTAL";
    }
}

// Example calls
console.log(getDirection(1, 2, 3, 5)); // "UP"
console.log(getDirection(1, 3, 4, 2)); // "DOWN"
console.log(getDirection(1, 3, 4, 3)); // "HORIZONTAL"

4. Complexity Analysis

The time complexity of this algorithm is O(1). Since it reads the coordinates of the two given points and determines the direction through simple operations, it is optimal for resolving within a fixed time.

5. Additional Improvements

The current algorithm is optimized for calculating the direction between two points in a 2D plane, but some improvements can be made:

• Exception handling can be added to handle various inputs. For example, we need to deal with cases where the input points are identical.
• A logic can be added to validate that the values of the input points are numeric to enhance stability.

Conclusion

In this tutorial, we implemented an algorithm to determine the direction of the line segment between two points using JavaScript. By utilizing simple mathematical principles and condition statements, we demonstrated that the problem can be effectively solved. We hope you will enhance your coding skills by solving more algorithmic problems in the future.

Let’s take a closer look at the Two Pointer technique, which is a common algorithm that appears in coding tests. This course will explain the basic concept of the Two Pointer method and solve actual problems using it.

1. What is Two Pointer?

The Two Pointer technique is an algorithmic method that efficiently processes data in data structures like arrays or lists by using two pointers. Generally, it is effective in reducing unnecessary loops and improving time complexity as the size of the problem increases.

• Efficiency: It often reduces the time complexity to O(N).
• Simplicity: The code becomes simpler and more readable.
• Application Scope: It can be used in various situations such as sorted arrays, strings, and subarray problems.

2. Basic Idea of Two Pointer

Two pointers are generally used in two ways:

• Left and Right Pointers: Start from both ends of an array and move towards the center to find elements that satisfy the condition.
• Moving in the Same Direction: Both pointers move in the same direction while examining surrounding elements until a specific condition is met.

3. Actual Problem: Two Sum

Here is an actual problem using the Two Pointer technique.

Problem Description

Given a sorted array numbers and a target sum target, write a function that returns the indices of the two numbers such that they add up to the target sum. Assume that each input has exactly one solution, and you cannot use the same element twice.

Input Format

numbers: [2, 7, 11, 15]
target: 9

Output Format

[0, 1]

Example Explanation

In the above example, since 2 + 7 = 9, the output is the indices 0 and 1.

4. Problem Solving Process

Let’s solve this problem using the Two Pointer method. Proceed with the following steps:

Step 1: Initialize Pointers

Initialize pointers at the start and end of the array. Name the left pointer as left and the right pointer as right.

Step 2: Check Conditions

Use a while loop to repeat until the two pointers cross each other. In each iteration, calculate the sum of the two numbers pointed to by the current pointers and check if this sum equals target.

Step 3: Compare Sum

• If the sum is less than target, move the left pointer one step to the right.
• If the sum is greater than target, move the right pointer one step to the left.
• If the sum is equal to target, return the two indices.

Step 4: Code Implementation

function twoSum(numbers, target) {
        let left = 0; 
        let right = numbers.length - 1;

        while (left < right) {
            const sum = numbers[left] + numbers[right];

            if (sum === target) {
                return [left, right]; 
            } else if (sum < target) {
                left++; 
            } else {
                right--; 
            }
        }
        return []; // In case there is no answer
    }

Step 5: Code Analysis

The time complexity of this code is O(N), and the space complexity is O(1). That means it can solve the problem without storing the array additionally.

5. Additional Examples and Variations

Now, let’s look at other variation problems. This can be applied to finding all combinations of three numbers that add up to target from the given array.

Problem Description

Given an integer array numbers and an integer target, return the indices of all unique combinations of three numbers that sum up to target.

Input Format

numbers: [1, 2, 3, 4, 5]
target: 9

Output Format

[[0, 3, 4], [1, 2, 4], [2, 3, 4]]

Solution Approach

To solve this problem, we can add a second pointer to find combinations without duplicates. The modified algorithm is as follows:

function threeSum(numbers, target) {
        let result = [];
        numbers.sort((a, b) => a - b); // Sort the array

        for (let i = 0; i < numbers.length - 2; i++) {
            let left = i + 1;
            let right = numbers.length - 1;

            while (left < right) {
                const sum = numbers[i] + numbers[left] + numbers[right];

                if (sum === target) {
                    result.push([i, left, right]);
                    left++;
                    right--;
                    while (left < right && numbers[left] === numbers[left - 1]) left++; // Remove duplicates
                    while (left < right && numbers[right] === numbers[right + 1]) right--; // Remove duplicates
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }

6. Conclusion

The Two Pointer technique is a very useful method for processing arrays or lists. It can significantly improve performance, especially when dealing with sorted data. Through the content covered in this course, I hope you gain an understanding of the basic concepts and applications of Two Pointers, and help you solve real-world problems.

Continue practicing various situations where you can use Two Pointers in search and combination problems, and confidently apply them in actual coding interviews.

Practice Problems

Try the following problems:

• Given an integer array numbers and an integer target, return the combination of indices where the sum of the two closest numbers is equal to target.
• A problem that returns the length of a substring consisting of unique characters within a string.

By solving problems like these, you can enhance your understanding of the Two Pointer technique and gain experience in solving various problems.

1. What is Sliding Window?

The Sliding Window technique is an algorithmic approach used to find a subarray or substring that meets specific conditions in a given array or list. It is primarily suitable for problems that require consecutive elements and can use either a fixed-size or variable-size window depending on the problem.

1.1 Advantages of Sliding Window

The biggest advantage of the sliding window is that it can significantly reduce time complexity compared to the brute force method. It is often possible to reduce from O(n^2) to O(n). The sliding window uses two pointers to traverse the array, enabling efficient access.

2. Example Algorithm Problem

Problem: Maximum Length of Repeating Character Replacement

Given a string, write a function that returns the length of the longest substring that can be created by changing at most ‘k’ characters.

    Example:
    Input: s = "AABABBA", k = 1
    Output: 4
    Explanation: The characters that can be changed are 'A' or 'B'. You can change two 'B's to 'A' to make 'AAAA'.
    

Approach to the Problem

You can use a sliding window to solve this problem. Here, we will count the number of characters present in the current window and check if we can replace ‘k’ characters.

2.1 Steps of the Algorithm

1. Initialize the left pointer and the right pointer.
2. Use a HashMap to count the characters in the substring.
3. Check the validity of the current window.
4. If invalid, move the left pointer.
5. Record the current window size and move the right pointer.
6. Repeat this process to find the maximum length.

3. Code Implementation

Below is the JavaScript code based on the algorithm described above.

    function characterReplacement(s, k) {
        const countMap = {};
        let left = 0;
        let maxLength = 0;
        let maxCount = 0;

        for (let right = 0; right < s.length; right++) {
            countMap[s[right]] = (countMap[s[right]] || 0) + 1;
            maxCount = Math.max(maxCount, countMap[s[right]]);

            while (right - left + 1 - maxCount > k) {
                countMap[s[left]]--;
                left++;
            }

            maxLength = Math.max(maxLength, right - left + 1);
        }

        return maxLength;
    }

    // Example usage
    console.log(characterReplacement("AABABBA", 1)); // 4
    

4. Code Explanation

In the above code, we are processing the string ‘s’ in relation to the given ‘k’ repeatedly. We use countMap to count the frequency of each character, and we track the number of the most frequent character in the current window.

4.1 Explanation of Key Variables

• countMap: An object that counts the occurrences of each character
• left: The left boundary of the window
• maxLength: Stores the maximum length
• maxCount: The number of the most frequent character in the current window

4.2 Movement of the Sliding Window

The right pointer increases and moves to the end of the string while the left pointer only moves when the current window is invalid. The valid condition is that the number of most frequent characters subtracted from the current window size must be less than or equal to ‘k’. This checks whether specific characters can be replaced.

5. Time Complexity and Space Complexity

The time complexity of this algorithm is O(n). It traverses each character of the string only once, and the space complexity is O(1) because it only needs to store 26 letters due to considering only uppercase alphabet letters.

6. Various Problem-Solving Methods

The sliding window technique can be applied in many diverse ways. Thus, mastering this concept will help in solving many other algorithm problems. For example:

• Maximum number of consecutive 1’s problem
• Minimum-length subarray sum problem
• Finding all anagrams problem

6.1 Additional Example Problem

The following problem can also be efficiently solved using the sliding window:

    Problem: Shortest subarray sum case
    Find the minimum length of the subarray that sums up to a specific number in the given array.
    

7. Conclusion

In this lecture, we covered the basic concept of the sliding window technique and solved an algorithm problem using it. This technique is particularly useful for string processing and subarray-related problems, so practice and familiarize yourself to tackle various variations of problems effectively.

Mastering the sliding window pattern can significantly reduce the difficulty of algorithm problems, and it often appears in coding tests in companies. I hope you acquire this technique perfectly through ample practice in the future.

8. Additional Resources

If you want to solve more sliding window problems, you can find numerous problems on online platforms such as LeetCode and HackerRank.

These days, many companies are evaluating applicants’ algorithm and problem-solving skills through coding tests. This time, we will take a look at the process of solving a problem with the topic of sorting cards. Sorting cards is one of the very simple yet frequently asked questions, requiring a fundamental understanding of algorithms and data structures.

Problem Description

We are trying to sort cards that have numbers written on them for a card game. The number of cards is N, and each card has an integer number from 1 to N. After sorting the given card array in ascending order, we need to print the final state of the card array.

Input Format

The first line contains the number of cards N (1 ≤ N ≤ 1000), and the second line contains the card numbers separated by spaces.

Output Format

Print the sorted card numbers separated by spaces.

Example Input

5
3 2 1 4 5

Example Output

1 2 3 4 5

Approach to the Problem

We can use the following approach to solve the problem.

1. Data Collection: Collect the number of cards and the card numbers.
2. Select a Sorting Algorithm: Use JavaScript’s built-in methods to sort the array.
3. Output Data in the Required Format: Output the sorted array.

Code Writing

Now let’s write the code. Below is the JavaScript code to solve the card sorting problem.

function sortCards(cards) {
    return cards.sort((a, b) => a - b);
}

function main() {
    const n = parseInt(prompt("Enter the number of cards:")); // Prompt to enter the number of cards
    const cardsInput = prompt("Enter the card numbers (separated by spaces):"); // Enter card numbers
    const cards = cardsInput.split(" ").map(Number); // Convert the space-separated string to an integer array

    const sortedCards = sortCards(cards); // Call the sorting method
    console.log(sortedCards.join(" ")); // Output the sorted cards
}

main();

Code Explanation

Let’s explain each part of the code we wrote.

sortCards Function

The sortCards function takes an array of card numbers and returns it sorted in ascending order. It uses JavaScript’s built-in sort method, along with an arrow function to compare the sizes of the numbers.

Main Function

The main function performs the following tasks:

• Uses prompt to input the number of cards.
• Uses prompt again to input the card numbers and passes them as a string.
• Uses the split method to convert the space-separated string into an array, and uses the map method to convert each element into a number.
• Calls the sortCards function to get the sorted array.
• Converts the result to a string using the join method and outputs it to the console.

Testing and Validation

We will test the code we wrote to verify that it works correctly with various inputs.

Test Case 1

Input:
5
3 2 1 4 5

Output:
1 2 3 4 5

Test Case 2

Input:
4
8 3 5 2

Output:
2 3 5 8

Conclusion

In this tutorial, we explored how to solve the card sorting problem using JavaScript’s array methods. Additionally, we had a chance to understand the approach and the role of each function through code writing. Problems like this are frequently asked in algorithm tests, so it is important to improve problem-solving skills through practice.

Additional Resources

For those who want to practice more advanced algorithm problems, please refer to the links below:

• LeetCode – Provides various algorithm problems.
• HackerRank – Provides algorithm and data structure problems.
• Codewars – Offers problems of various difficulties and allows interaction with the community.

With that, we will conclude the JavaScript coding test course. Always strive to prepare for successful coding tests!

One of the frequently encountered problems in coding tests is array sorting. In this tutorial, we will learn about the Quick Sort algorithm and explain in detail how to implement it in JavaScript. Quick sort is an efficient sorting algorithm that uses the divide and conquer technique.

Problem Description

Sort the given array in ascending order using the quick sort algorithm.

Example Input: [34, 7, 23, 32, 5, 62]
Example Output: [5, 7, 23, 32, 34, 62]

Overview of Quick Sort Algorithm

Quick sort proceeds through the following steps.

1. Select one element from the array as the pivot.
2. Divide the array into two subarrays based on the pivot. One consists of elements smaller than the pivot, while the other consists of elements larger than the pivot.
3. Apply the same method recursively to each subarray.
4. Repeat until the subarrays have a size of 0 or 1.

Example Explanation

If the input array is [34, 7, 23, 32, 5, 62], it undergoes the following process.

1. Selecting Pivot: Choose 62, the last element of the array, as the pivot.
2. Partitioning: Divide the array into elements smaller than pivot 62 ([34, 7, 23, 32, 5]) and larger elements ([]) based on the pivot.
3. Recursive Call: Repeat the same process for the subarray [34, 7, 23, 32, 5], which is smaller than the pivot.
4. By repeating this process, the array will ultimately be sorted.

JavaScript Implementation

Now, let’s implement the quick sort algorithm in JavaScript.

function quickSort(arr) {
    if (arr.length <= 1) {
        return arr; // Return as is if the size is 0 or 1
    }
    
    const pivot = arr[arr.length - 1]; // Choose the last element as pivot
    const left = []; // Array to store elements less than the pivot
    const right = []; // Array to store elements greater than the pivot

    // Iterate through the array and compare with the pivot
    for (let i = 0; i < arr.length - 1; i++) {
        if (arr[i] < pivot) {
            left.push(arr[i]);
        } else {
            right.push(arr[i]);
        }
    }

    // Recursive calls and return the result array
    return [...quickSort(left), pivot, ...quickSort(right)];
}

// Example array
const array = [34, 7, 23, 32, 5, 62];
const sortedArray = quickSort(array);
console.log(sortedArray); // [5, 7, 23, 32, 34, 62]

Time Complexity of the Algorithm

The average time complexity of the quick sort algorithm is O(n log n). However, in the worst case, it can have a time complexity of O(n²). This can occur when the pivot selection is imbalanced. For this reason, quick sort can be optimized through techniques such as randomly selecting the pivot to enhance performance.

Advantages and Disadvantages of Quick Sort

Advantages

• An efficient sorting algorithm suitable for sorting large datasets.
• It uses less memory, allowing for in-place sorting.

Disadvantages

• In the worst case, the time complexity of O(n²) is inefficient.
• It uses stack memory due to recursive calls.

Conclusion

In this tutorial, we learned about the quick sort algorithm and how to implement it in JavaScript. Quick sort is a simple and efficient sorting algorithm frequently used in coding tests and algorithm problem solving. Based on what you have learned, try solving various array sorting problems.

In the next tutorial, we will cover another sorting algorithm, Merge Sort. Please stay tuned!