JavaScript Coding Test Course, Understanding Dynamic Programming

Recently, programming interviews have been creating various types of problems to assess understanding of algorithms and data structures. Among them, Dynamic Programming has established itself as a powerful technique for efficiently solving many problems. In this article, we will explore the principles of Dynamic Programming and examine a specific problem-solving process using JavaScript.

What is Dynamic Programming?

Dynamic Programming is an algorithmic approach that breaks down large problems into smaller ones and finds the optimal solution to the problem. It is primarily used for optimization problems and improves performance by preventing duplicate calculations through the memoization technique.

Characteristics of Dynamic Programming

  • Divides the problem into smaller subproblems.
  • Saves the results of subproblems to avoid duplicate calculations.
  • Uses a state transition function to calculate the optimal solution.

Problem Presentation: Fibonacci Sequence

In this course, we will address the problem of calculating the Fibonacci sequence using Dynamic Programming. The Fibonacci sequence is defined as follows:

  • F(0) = 0
  • F(1) = 1
  • F(n) = F(n-1) + F(n-2) (n ≥ 2)

In other words, the goal is to efficiently calculate F(n) for a given integer n.

Problem Solving Process

1. Problem Analysis

The Fibonacci sequence is defined recursively. However, solving the problem with simple recursive calls is inefficient because the same value is calculated multiple times. To address this, we approach the solution using Dynamic Programming.

2. Definition of State Transition Function

The state transition function uses the results of already computed subproblems to solve the higher-level problem. In the case of the Fibonacci sequence:

F(n) = F(n-1) + F(n-2)

This function requires the values of F(n-1) and F(n-2) to calculate F(n).

3. Memoization Technique

Memoization is a method that caches the results of subproblems to prevent duplicate calculations. Below is an example code utilizing memoization in JavaScript:


function fib(n, memo = {}) {
    // Return if the value has already been calculated
    if (n in memo) return memo[n];
    // Base case
    if (n <= 1) return n;
    // Recursive call and memoization
    memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
    return memo[n];
}

// Test
console.log(fib(10)); // Output: 55

4. Tabulation Method

In addition to memoization, the tabulation method can also be used to implement Dynamic Programming. The tabulation method stores results of subproblems in an array and calculates them gradually. Below is the implementation of the Fibonacci sequence using the tabulation method:


function fibTable(n) {
    if (n <= 1) return n;

    const fib = new Array(n + 1);
    fib[0] = 0;
    fib[1] = 1;

    for (let i = 2; i <= n; i++) {
        fib[i] = fib[i - 1] + fib[i - 2];
    }
    return fib[n];
}

// Test
console.log(fibTable(10)); // Output: 55

Conclusion

In this article, we explored the basic principles of Dynamic Programming and the process of solving the Fibonacci sequence problem. Both memoization and tabulation methods have their advantages and disadvantages, so they should be chosen appropriately based on the characteristics of the problem. Dynamic Programming is useful for solving various algorithm problems, making it important to continuously practice and apply it to different problems.

References

  • CLRS, Algorithms
  • LeetCode, Dynamic Programming problems
  • GeeksforGeeks, Dynamic Programming Concepts

JavaScript Coding Test Course, Euclidean Algorithm

Hello! Today, we will learn about the algorithm for finding the Greatest Common Divisor (GCD) using the Euclidean algorithm. The Euclidean algorithm is a classical approach to find the GCD of two integers a and b based on the following principles. In this tutorial, we will explore the theory of the Euclidean algorithm and how to implement it in JavaScript step by step.

1. Overview of the Euclidean Algorithm

The Euclidean algorithm is a method proposed by the Greek mathematician Euclid around 300 BC to find the greatest common divisor of two natural numbers a and b. The algorithm works as follows:

  • If b is 0, then GCD(a, b) is a.
  • Otherwise, GCD(a, b) = GCD(b, a mod b).

Here, mod refers to the modulus operation, where a mod b returns the remainder when a is divided by b.

1.1 Example of the Algorithm

For example, if a = 48 and b = 18, we can find the GCD through the following steps:

GCD(48, 18)
= GCD(18, 48 mod 18)
= GCD(18, 12)
= GCD(12, 6)
= GCD(6, 0)
= 6

Therefore, GCD(48, 18) = 6.

2. Implementing the Euclidean Algorithm in JavaScript

Now let’s implement the Euclidean algorithm in JavaScript. Below is the code that implements a function to find the GCD:


function gcd(a, b) {
    if (b === 0) {
        return a;
    }
    return gcd(b, a % b);
}

// Example usage
const a = 48;
const b = 18;
console.log(`GCD of ${a} and ${b} is ${gcd(a, b)}`);

2.1 Explanation of the Code

  • function gcd(a, b): This is a function that takes two arguments a and b and calculates the GCD.
  • if (b === 0): If b is 0, it returns a. This is the basis of the Euclidean algorithm.
  • return gcd(b, a % b): This recursively calls the gcd function, swapping a with b and b with a mod b.

3. Various Uses and Applications

The Euclidean algorithm has various applications in different fields. For example:

  • Solving mathematical problems: It is used not only to find the GCD of two numbers but also to find the GCD of multiple numbers.
  • Computer Science: It is used to compute the GCD required for expressing fractions in their simplest form.
  • Cryptography: The GCD is important in RSA encryption algorithms.

4. Problem Solving: Finding the GCD of Two Numbers

Let’s solve the following problem:


Problem: Write a function that takes two integers and outputs their greatest common divisor.
Input: Two integers a, b (1 <= a, b <= 10000)
Output: The greatest common divisor of a and b

4.1 Problem Solving Process

  1. Input two integers.
  2. Calculate the GCD using the Euclidean algorithm.
  3. Output the calculated GCD.

Now let’s write the complete code as follows:


function gcd(a, b) {
    if (b === 0) {
        return a;
    }
    return gcd(b, a % b);
}

// Getting input from the user
const a = parseInt(prompt("Please enter the first integer: "));
const b = parseInt(prompt("Please enter the second integer: "));

console.log(`GCD of ${a} and ${b} is ${gcd(a, b)}`);

5. Conclusion

In conclusion, we have learned about the JavaScript algorithm using the Euclidean algorithm. I hope this helps you gain a basic understanding of algorithm problems by understanding and implementing the algorithm. If you have any further questions, feel free to leave a comment!

6. Additional Resources

If you want to study the Euclidean algorithm in depth, please refer to the following resources:

Thank you!

JavaScript Coding Test Course, Calculate ATM Withdrawal Time

Common problems that appear in programming interviews or coding tests are time calculation problems in various situations. In this post, we will discuss an algorithm problem that calculates the withdrawal time at an ATM. Assuming multiple users are using the ATM at the same time, we will write an algorithm to calculate the expected waiting time for each user.

Problem Description

Let’s assume there is one ATM machine. There are several users waiting in line for this ATM. Each user can use the ATM at a specific time, and the time required to use the ATM can vary from person to person.

The given input is an array of users’ withdrawal times. The index of the array represents the user’s order, and the value stored at that index indicates the time taken by that user to use the ATM in seconds. For example, if an array of [5, 3, 8] is provided, the first user uses the ATM for 5 seconds, the second user for 3 seconds, and the third user for 8 seconds.

Input

[5, 3, 8]

Output

Array of waiting times for each user: [0, 5, 8]

In the above example, the first user has a waiting time of 0 seconds, so they can use it immediately. The second user can use the ATM after the first user’s withdrawal is finished, so their waiting time is 5 seconds. The third user can use the ATM after the second user’s waiting time is over, making their waiting time 8 seconds.

Problem Solving Process

Step 1: Understanding the Problem

To understand the problem, the following must be clarified:

  • Each user using the ATM must wait until the previous user’s withdrawal is completed.
  • To calculate waiting times, cumulative time must be tracked.
  • The waiting times should be calculated in the order of withdrawal, and the result should be returned as an array.

Step 2: Designing the Algorithm

To solve this problem, the following algorithm can be used:

  1. Create an empty array to store each user’s waiting times.
  2. Initialize a variable totalTime to 0. This will be used to store cumulative waiting time.
  3. Loop through each user’s withdrawal time to calculate the waiting times:
    • The current user’s waiting time is set to totalTime.
    • Add the current user’s withdrawal time to totalTime.
  4. Return the waiting time array.

Step 3: Code Implementation

Now, let’s implement the algorithm in JavaScript code.

function calculateWaitTimes(atmTimes) {
    let waitTimes = [];
    let totalTime = 0;

    for (let i = 0; i < atmTimes.length; i++) {
        waitTimes[i] = totalTime; // Current user's waiting time
        totalTime += atmTimes[i]; // Update cumulative time
    }

    return waitTimes;
}

// Example execution
const inputTimes = [5, 3, 8];
const outputWaitTimes = calculateWaitTimes(inputTimes);
console.log(outputWaitTimes); // [0, 5, 8]

Testing and Validation

After implementing the function, let’s validate it using various test cases to ensure it produces the correct results.

console.log(calculateWaitTimes([0, 0, 0]));       // [0, 0, 0]
console.log(calculateWaitTimes([1, 2, 3]));       // [0, 1, 3]
console.log(calculateWaitTimes([10, 20, 30]));    // [0, 10, 30]
console.log(calculateWaitTimes([5]));              // [0]
console.log(calculateWaitTimes([]));               // []

Complexity Analysis

The time complexity of this algorithm is O(n). It increases in proportion to the number of users since it traverses the array once. The space complexity is also O(n), as an array is needed to store the waiting times.

Conclusion

In this post, we examined the problem of calculating withdrawal times at an ATM. This problem can serve as a useful exercise for mastering the basics of algorithm design and time calculation. Through the implementation process in JavaScript, I hope you have learned a more accessible approach to similar problems encountered in actual coding tests. It is hoped that this problem helps you understand the fundamental approaches to algorithms and time complexity and enhances your problem-solving skills.

JavaScript Coding Test Course, Finding the Direction of a Line Segment

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

Input:
A(1, 2), B(3, 5)
Output:
“UP”
Input:
A(1, 3), B(4, 2)
Output:
“DOWN”

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.

JavaScript Coding Test Course, Two Pointers

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.