Category: Javascript Coding Test

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.

Hello! In this session, we will learn about one of the algorithms frequently encountered in coding tests: Union-Find. This algorithm is useful for finding connected components in a graph or managing sets, and is utilized in various problem-solving scenarios. Before we begin, let’s understand the basic concept of Union-Find and learn specific applications through real problem-solving processes.

What is Union-Find?

Union-Find is a data structure that tracks how given elements are divided into connected sets. It provides the following two operations:

• Find: An operation to find which set a given element belongs to.
• Union: An operation to combine two sets.

This data structure is useful for finding connected components in a graph and is often used to determine the presence of cycles. Union-Find boasts very fast execution times by utilizing optimization techniques.

Problem Description

Now, let’s introduce a problem where we can apply Union-Find. We have the following problem:

Problem: Finding Friends

There are n people. Each person can make friends, and the friendship relationship is mutual. Write an algorithm that can determine whether two people are friends based on a given list of friendship relationships. The friendship relationships are given in the following format:

        [[1, 2], [2, 3], [4, 5]]
        

In the above example, 1 and 2 are friends, and 2 and 3 are friends, which means 1 and 3 are indirectly friends. 4 and 5 are separate friendships, so 1 and 4 are not friends. For each query, check if the two people are friends.

Implementing the Union-Find Algorithm

Now, let’s examine how to solve this problem using the Union-Find algorithm. First, we’ll define a Union-Find structure and implement the necessary functions.

    class UnionFind {
        constructor(size) {
            this.parent = new Array(size);
            this.rank = new Array(size).fill(1);
            for (let i = 0; i < size; i++) {
                this.parent[i] = i;
            }
        }

        find(x) {
            if (this.parent[x] !== x) {
                this.parent[x] = this.find(this.parent[x]); // Path compression
            }
            return this.parent[x];
        }

        union(x, y) {
            const rootX = this.find(x);
            const rootY = this.find(y);

            if (rootX !== rootY) {
                // Union by rank
                if (this.rank[rootX] > this.rank[rootY]) {
                    this.parent[rootY] = rootX;
                } else if (this.rank[rootX] < this.rank[rootY]) {
                    this.parent[rootX] = rootY;
                } else {
                    this.parent[rootY] = rootX;
                    this.rank[rootX]++;
                }
            }
        }

        areConnected(x, y) {
            return this.find(x) === this.find(y);
        }
    }
    

Problem-Solving Process

1. Process the input of the problem. Receive the friendship relationships and take the targets for performing the promised queries.

    function processFriendships(friendships, queries, numberOfPeople) {
        const uf = new UnionFind(numberOfPeople + 1); // +1 to accommodate 1-indexed people
        
        friendships.forEach(([a, b]) => {
            uf.union(a, b);
        });

        return queries.map(([x, y]) => uf.areConnected(x, y));
    }
    

2. Iterate through the list of friendships and perform the union operation for each pair.

3. For each query, determine whether the two people belong to the same set.

Final Code

    const friendships = [[1, 2], [2, 3], [4, 5]];
    const queries = [[1, 3], [1, 4], [4, 5]];
    const numberOfPeople = 5;

    const results = processFriendships(friendships, queries, numberOfPeople);
    console.log(results); // [true, false, true]
    

Interpreting Results

The final query results are as follows:

• 1 and 3 are friends: true
• 1 and 4 are not friends: false
• 4 and 5 are friends: true

The above code efficiently handles friendship relationships using Union-Find. Union-Find is particularly useful when there are many sets, and its time complexity is nearly constant time.

Conclusion

In this lecture, we solved the friend-finding problem using the Union-Find algorithm. Union-Find can be applied to various problems and is a very useful tool for solving algorithmic challenges. Continue to learn and practice various algorithms to achieve good results in coding tests!

Thank you!

Problem Description

The Binomial Coefficient is denoted as C(n, k), where n and k are two integers representing the number of ways to choose k elements from n elements. It is calculated using the following formula:

C(n, k) = n! / (k! * (n - k)!)

Here, n! (n factorial) is the product of all integers from n down to 1.

Example Input and Output

Input

• The first line contains two integers n and k (0 <= k <= n <= 30).

Output

• The value of C(n, k) is printed.

Problem Solving Process

1. Theoretical Background

To calculate the binomial coefficient, we first need to compute factorials. We need to calculate n!, k!, and (n-k)!, which will allow us to compute the binomial coefficient. Theoretically, we can compute it using the formula above, but to implement it efficiently using JavaScript, we can utilize both recursive functions and loops.

2. Recursive Approach

Factorials can be defined recursively. For example, n! can be defined as follows:

function factorial(n) {
        if (n <= 1) return 1;
        return n * factorial(n - 1);
    }

Using this approach, we can compute the binomial coefficient. However, the downside of this method is that it may affect memory limits and execution time for large numbers.

3. Iterative Approach

Another efficient way to compute the binomial coefficient is by using loops. Instead of calculating factorials directly, we can calculate the binomial coefficient directly. We can use the following loop:

function binomialCoefficient(n, k) {
      if (k > n) return 0;
      if (k === 0 || k === n) return 1;
      k = Math.min(k, n - k); // k must be less than or equal to n-k
      let result = 1;

      for (let i = 0; i < k; i++) {
          result *= (n - i);
          result /= (i + 1);
      }
      return result;
    }

4. Complete Code

Below is an example that integrates the complete code:

    function factorial(n) {
        if (n <= 1) return 1;
        return n * factorial(n - 1);
    }

    function binomialCoefficient(n, k) {
        if (k > n) return 0;
        if (k === 0 || k === n) return 1;
        k = Math.min(k, n - k); // k must be less than or equal to n-k
        let result = 1;

        for (let i = 0; i < k; i++) {
            result *= (n - i);
            result /= (i + 1);
        }
        return result;
    }

    // Example usage
    const n = 5;
    const k = 2;
    console.log(`C(${n}, ${k}) = ${binomialCoefficient(n, k)}`); // Output: C(5, 2) = 10
    

5. Performance Analysis

The time complexity of the above algorithm is O(k), and the space complexity is O(1). In other words, it works efficiently for small input values, but it may not be suitable for more complex problems that require global operations. In fact, this method can handle cases where n ≤ 30 quickly and efficiently.

6. Conclusion

The problem of calculating the binomial coefficient is one of the frequently encountered problems in many programming contests and coding tests. Through this lesson, we have explored how to calculate the binomial coefficient and learned various approaches to solving this problem using JavaScript. Through such theoretical and practical problem-solving methods, we can cultivate deeper algorithmic thinking.

Problem Definition

Given an integer array arr and an integer m, write a function to calculate the sum of the remainders when the sum of the array elements is divided by m. However, the sum of the remainders should not be greater than m.

Examples

Input: arr = [1, 2, 3, 4, 5], m = 3
Output: 15 % 3 = 0

Input: arr = [10, 20, 30], m = 5
Output: (10 + 20 + 30) % 5 = 0

Input: arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], m = 7
Output: (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) % 7 = 3

Problem Solving Process

This problem can be solved in the following steps:

Step 1: Understanding the Problem

Calculating the remainder of the sum of all elements in the array divided by m is a basic mathematical problem.
Since it specifies that the sum of the remainders must not be greater than m,
we need to keep this condition in mind while implementing the solution.

Step 2: Designing the Algorithm

A simple algorithm can be used as follows:

1. Sum all elements of the array arr.
2. Divide the summed result by m to get the remainder.
3. Return the result.

Step 3: Coding

Now, let’s implement the algorithm in JavaScript.
First, I will write the basic structure:

function remainderSum(arr, m) {
    const sum = arr.reduce((accum, value) => accum + value, 0);
    return sum % m;
}

arr.reduce() is used to sum all elements in the array and return the remainder when divided by m.
Next, I will prepare several cases to test this function.

Step 4: Writing Test Cases

console.log(remainderSum([1, 2, 3, 4, 5], 3)); // 0
console.log(remainderSum([10, 20, 30], 5)); // 0
console.log(remainderSum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 7)); // 3
console.log(remainderSum([15, 25, 35, 45, 55], 10)); // 5
console.log(remainderSum([1, 1, 1, 1, 1], 2)); // 1

Running the above test cases will help verify whether the results for each case are correct. If all results match expectations,
the code is successfully implemented.

Step 5: Optimization and Additional Considerations

The above implementation is simply written to sum all elements of the given array.
However, if the size of the array can be very large, it may be necessary to consider performance factors.
In such cases, optimization can be achieved by calculating the remainders during the summation process itself.

function optimizedRemainderSum(arr, m) {
    let remainder = 0;
    for (const value of arr) {
        remainder = (remainder + value) % m;
    }
    return remainder;
}

Here, the optimizedRemainderSum function stores intermediate results by calculating the remainder at each step,
thus calculating the final result in a much more efficient manner.

Conclusion

In this lesson, we covered the “Calculating Remainder Sum” problem. It’s a common problem of summing the elements of an array and finding their remainder,
but we also considered ways to optimize the algorithm.
I hope you found useful tips for preparing for coding tests using JavaScript.