Author: root

Hello! In this blog post, we will discuss an algorithm problem to prepare for the JavaScript coding test. The topic is ‘Creating Blu-rays’. The problem is to find the minimum number of Blu-rays needed to create the given movies. To solve this, we will need to utilize binary search and the greedy algorithm.

Problem Description

You want to create multiple movies on Blu-ray. The running time of each movie is given, and a Blu-ray can hold up to K amount of time. Your goal is to minimize the number of Blu-rays to store all the movies. However, each Blu-ray must contain movies such that their total running time does not exceed K.

Input

• The first line contains N and K. (1 ≤ N ≤ 1000, 1 ≤ K ≤ 10^6)
• The second line contains N natural numbers separated by spaces, representing the running times of each movie. (1 ≤ movie running time ≤ 10^6)

Output

Print the minimum number of Blu-rays.

Example

Input:
4 5
2 3 1 4

Output:
2

Problem Solving Process

To approach this problem, we can proceed in the following steps.

Step 1: Understand the Problem

Consider how to allocate the given movies into Blu-rays with a maximum running time of K. Since the running times of each movie are provided, we need to consider how to combine them without exceeding K.

Step 2: Derive Ideas

Since we cannot simply fit all movies into one Blu-ray, we will iteratively explore the movie list to check if they can fit into each Blu-ray. To do this, we will use binary search to find the minimum number of Blu-rays needed.

Step 3: Exception Handling

If the time to fit a movie exceeds K, we must place that movie on a new Blu-ray. It is important to be mindful of this condition to fit as many movies as possible into Blu-rays.

Step 4: Algorithm Implementation

Now, we will implement a JavaScript function based on the above ideas.

function minBluRays(N, K, movies) {
    let bluRays = 0;
    let currentTime = 0;

    for (let i = 0; i < N; i++) {
        if (currentTime + movies[i] > K) {
            bluRays++;
            currentTime = movies[i];
        } else {
            currentTime += movies[i];
        }
    }

    if (currentTime > 0) {
        bluRays++;
    }

    return bluRays;
}

// Example execution
const N = 4;
const K = 5;
const movies = [2, 3, 1, 4];
console.log(minBluRays(N, K, movies)); // Output: 2

Conclusion

In this article, we demonstrated the process of solving a commonly encountered algorithm problem in JavaScript coding tests through the ‘Creating Blu-rays’ problem. Understanding the essence of the problem and finding the necessary algorithms is very important. It is advisable to utilize the given time well to write efficient code.

I hope this problem helps you develop a basic understanding of JavaScript and an algorithmic mindset. Thank you!

Problem: Maze Exploration

This is a problem to determine whether there is a path from the starting point to the destination in the given 2D array. The values of the array are given as 0 (passable) and 1 (impassable). The starting point is (0, 0) and the destination is (N-1, M-1). Movement is allowed only up, down, left, and right, and if a path exists, it should return true; if not, it should return false.

Example Problem

Input

    [
        [0, 0, 1, 0],
        [1, 0, 1, 0],
        [0, 0, 0, 0],
        [0, 1, 1, 0]
    ]
    

Output

true

Explanation

In the above example, starting from the starting point (0, 0), the path goes through (1, 1), (2, 1), (2, 2), (2, 3), (3, 3) to reach the destination (3, 3).

Problem Solving Process

1. Overview of the Algorithm

This problem can be solved using the Depth First Search (DFS) algorithm. DFS is a method that goes as deep as possible into the nodes before backtracking to explore other paths. We will implement this using recursion. DFS is applied to explore the maze and check whether it is possible to reach the destination from the starting point.

2. Implementing the DFS Algorithm

To implement DFS, the following steps are necessary:

1. Check if the current position is within bounds.
2. Check if the current position is an impassable point.
3. Check if the current position is the destination.
4. Mark the current position as visited.
5. Recursively call DFS in all four directions (up, down, left, right).
6. After exploring all paths, unmark the visited position.

3. Code Implementation

The following code implements the DFS algorithm using JavaScript:

function isPathExists(maze) {
    const rows = maze.length;
    const cols = maze[0].length;

    function dfs(x, y) {
        // Boundary conditions
        if (x < 0 || y < 0 || x >= rows || y >= cols) return false;
        // Impassable point
        if (maze[x][y] === 1) return false;
        // Destination
        if (x === rows - 1 && y === cols - 1) return true;

        // Mark current position as visited
        maze[x][y] = 1;

        // Explore DFS in four directions
        const found = dfs(x + 1, y) || dfs(x - 1, y) || dfs(x, y + 1) || dfs(x, y - 1);

        // Unmark visited position
        maze[x][y] = 0;

        return found;
    }

    return dfs(0, 0);
}

// Example Test
const maze = [
    [0, 0, 1, 0],
    [1, 0, 1, 0],
    [0, 0, 0, 0],
    [0, 1, 1, 0]
];

console.log(isPathExists(maze)); // Output: true

4. Code Explanation

Let me explain the key parts of the code:

• Boundary condition check: Ensures the current position does not exceed the array boundaries.
• Impassable point check: Verifies if the value at the current position is 1 to determine if it is passable.
• Destination reached: Returns true if the current position is the destination (N-1, M-1).
• Mark as visited: Marks the current position to prevent duplicate exploration.
• Recursive call: Calls DFS recursively in the four directions to explore the path.
• Unmarking: Unmarks the visited position after the exploration is complete.

5. Time Complexity Analysis

The time complexity of this algorithm is O(N * M), where N is the number of rows and M is the number of columns. This is because each cell can be visited once. However, there may be additional memory required in the worst-case scenario due to stack space overhead from recursion.

Conclusion

In this tutorial, we learned how to solve the maze exploration problem using Depth First Search (DFS). DFS can be effectively used even in complex graph structures, and it can be applied to various problems. Once you understand the characteristics and workings of DFS, it is advisable to apply it to different problems. In the next tutorial, we will explore the Breadth First Search (BFS) algorithm. I hope you found this tutorial helpful.

Problem Description

This problem is about finding the number closest to a given num value in the provided array arr. If there are multiple closest numbers, the one closer to num with a lower value takes precedence.

Input

• An integer array arr (-10⁶ ≤ arr[i] ≤ 10⁶, 1 ≤ arr.length ≤ 10⁵)
• An integer num (-10⁶ ≤ num ≤ 10⁶)

Output

Return the closest number.

Examples

Example 1:
Input: arr = [1, 2, 3, 4, 5], num = 3
Output: 3

Example 2:
Input: arr = [1, 2, 4, 5], num = 3
Output: 2

Example 3:
Input: arr = [5, 10, 15], num = 12
Output: 10

Solution Process

To solve this problem, we follow these steps.

Step 1: Understand the Problem

First, since the task is to find the number closest to num, the key is to calculate the absolute differences between each element and num and find the smallest value. If there are multiple numbers with the same difference, the smaller value should be returned.

Step 2: Choose an Approach

The simplest method is to iterate through the array and calculate the absolute differences for each element. However, this has a time complexity of O(n), so we need to consider a faster algorithm.

Step 3: Utilize Binary Search through Sorting

By sorting the input array, we can efficiently find the number closest to num using binary search. To find the index of a specific value num in the sorted array, we will use Array.prototype.sort() and then implement a binarySearch() function.

Step 4: Implement the Code

Below is the JavaScript code based on the above explanation.

function findClosest(arr, num) {
    // Sort the array
    arr.sort((a, b) => a - b);
    
    let left = 0;
    let right = arr.length - 1;
    let closest = arr[0];

    while (left <= right) {
        const mid = Math.floor((left + right) / 2);
        
        // Compare absolute differences
        if (Math.abs(arr[mid] - num) < Math.abs(closest - num) ||
            (Math.abs(arr[mid] - num) === Math.abs(closest - num) && arr[mid] < closest)) {
            closest = arr[mid];
        }
        
        // Determine the direction of binary search
        if (arr[mid] < num) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return closest;
}

// Example Tests
console.log(findClosest([1, 2, 3, 4, 5], 3)); // 3
console.log(findClosest([1, 2, 4, 5], 3));    // 2
console.log(findClosest([5, 10, 15], 12));      // 10

Step 5: Analyze Time Complexity

The above code first sorts the array, which has a time complexity of O(n log n), and then the searching process using binary search requires O(log n) time. Therefore, the overall time complexity can be evaluated as O(n log n).

Conclusion

This problem has taught us the importance of choosing an efficient algorithm considering time complexity. It is necessary to try various approaches for the given problem and select the appropriate algorithm based on the results. I hope you apply these principles well in future coding tests!

October 10, 2023

Problem Description

Efficiently assigning meeting rooms is a critical issue in modern office environments. Given a list of meetings with their start and end times, the problem is to write an algorithm to assign as many meetings as possible to meeting rooms without overlapping.

Input:

• meetings: A 2D array where each array consists of [startTime, endTime].

Output:

• The maximum number of meetings that can be assigned.

Example

For example, let’s assume there is a list of meetings as follows.

            [[0, 30], [5, 10], [15, 20]]
        

The maximum number of meetings that can be assigned here is 2. (The meeting [0, 30] does not overlap with [5, 10] and [15, 20]).

Approach to the Problem

This problem can be solved using a greedy algorithm. We will sort the meetings based on their ending times and then assign meetings starting from the one that ends the earliest. This way, we can utilize the resources of the meeting room to the fullest.

1. First, sort the list of meetings based on their ending times.
2. Select the first meeting, and if the start time of the next meeting is greater than the ending time of the currently selected meeting, select the new meeting.
3. Repeat this process until the end of the list of meetings.

JavaScript Implementation

Now, let’s implement the above approach in JavaScript code.

function maximumMeetings(meetings) {
    // Sort by meeting ending times
    meetings.sort((a, b) => a[1] - b[1]);
    
    let count = 0;
    let lastEndTime = 0;

    for (let i = 0; i < meetings.length; i++) {
        // If the start time of the current meeting is greater than or equal to the ending time of the last selected meeting
        if (meetings[i][0] >= lastEndTime) {
            count++;
            lastEndTime = meetings[i][1]; // Update the ending time of the last selected meeting
        }
    }

    return count;
}

// Example for testing
const testMeetings = [[0, 30], [5, 10], [15, 20]];
console.log(maximumMeetings(testMeetings)); // Output: 2
        

Complexity Analysis

The time complexity of this algorithm is O(n log n). This is due to the time taken to sort the meetings. The process of counting the maximum number of meetings from the sorted list is O(n). Therefore, the overall time complexity is O(n log n). The space complexity is O(1), meaning that no additional space is needed even if using a sorted list.

Conclusion

The “Meeting Room Assignment” problem is a representative problem that can be efficiently solved using a greedy algorithm. This problem teaches how to avoid meeting overlaps and make the most of resources. The code implemented in JavaScript above can help in solving this problem. Based on this, it will be possible to lay the groundwork for solving various algorithmic problems.

Problem Definition

Today’s problem is “I Will Become the Residents’ Association President”. This problem involves calculating the number of people living in a specific unit on a given floor when information about the floor and unit number is provided.

The residents’ association president manages according to the following rules. Each unit has one household living in it, and there is always 1 person living in unit 1 of each floor. Then, the population of each unit is calculated as the sum of the population of the same unit on the floor below and the population of the unit immediately to the left on the floor below.

Problem Description

Input:
– The first input value is the number of test cases T (1 <= T <= 1000).
– Each test case consists of two integers K (0 <= K <= 14) and N (1 <= N <= 14).
– K represents the floor number, and N represents the unit number on that floor.

Output:
For each test case, you need to output the number of people in unit N on floor K.

Example Input and Output

Problem Solving Process

1. Understanding the Combination Function

The key to this problem is calculating the population using dynamic programming. Basically, you use the population of the units on the floor below to compute the current floor’s population. The calculation process is as follows.

Calculation Method

def count_people(K, N):
    if K == 0:
        return N
    if N == 1:
        return 1
    return count_people(K-1, N) + count_people(K, N-1)

2. Progressing Through Loops

Since recursive functions can make the code complex, we will use loops for efficient computation. First, we create a 2D array to store the populations of floor K in advance.

3. Implementation

function countPeople(T, cases) {
    const results = [];
    const dp = Array.from(Array(15), () => Array(15).fill(0));

    for (let i = 0; i <= 14; i++) {
        dp[0][i] = i;  // For the 0th floor, there are i people
    }
    
    for (let k = 1; k <= 14; k++) {
        dp[k][1] = 1;  // For unit 1, there is always 1 person
        for (let n = 2; n <= 14; n++) {
            dp[k][n] = dp[k-1][n] + dp[k][n-1];  // Sum of the number of people in the unit above and the unit on the left
        }
    }

    for (let i = 0; i < T; i++) {
        let k = cases[i][0];
        let n = cases[i][1];
        results.push(dp[k][n]);
    }
  
  return results;
}

const T = 2;
const cases = [[1, 3], [2, 3]];
console.log(countPeople(T, cases)); // Output: [3, 6]

Conclusion

The function we implemented efficiently calculates the number of people for each unit on the specified floor for T test cases. This problem is an excellent example demonstrating the basic concepts of dynamic programming. Proper use of dynamic programming can effectively solve similar types of problems.

Through this algorithm, you can strengthen various aspects of programming while preparing for coding tests. I hope this course enhances your understanding of JavaScript coding tests.