JavaScript Coding Test Course, Calculating Interval Sum 2

Problem Description

The interval sum problem is one of the common types encountered in algorithm problems. In this lecture, we will address the second problem of calculating the interval sum. We will learn how to efficiently calculate the interval sum between two points in a given array.

Problem:
Given an integer array A and a query array queries, compute the interval sum from the l-th to the r-th index of array A for each query (l, r).

The array indexing starts from 0, the size of the array is N, and the number of queries is Q.

Input Format

1. The first line contains the size of the array A, N (1 ≤ N ≤ 106).

2. The second line contains the elements of array A. (A[i] is an integer, -109A[i] ≤ 109)

3. The third line contains the number of queries Q (1 ≤ Q ≤ 105).

4. The next Q lines contain each query (l, r). (0 ≤ lr < N)

Output Format

For each query, output the interval sum on a new line.

Example

    Input:
    5
    1 2 3 4 5
    3
    0 2
    1 4
    2 2

    Output:
    6
    14
    3
    

Solution Process

The first step to efficiently obtain the interval sum is to create a prefix sum array. The prefix sum array stores the sum of the elements up to each index in the array, allowing us to calculate the interval sum in O(1) time.

Creating the Prefix Sum Array

The prefix sum array prefix is defined as follows:

prefix[0] = A[0],

prefix[i] = prefix[i-1] + A[i] (for 1 ≤ i < N)

With this prefix sum array, the interval sum A[l] + ... + A[r] can be calculated as follows:

sum(l, r) = prefix[r] - prefix[l - 1] (provided l > 0)

In the case where l = 0, we handle it as sum(0, r) = prefix[r].

Implementation

Now, let’s write the code to create the prefix sum array and calculate the interval sum for each query. Below is the implementation in JavaScript:

    
    function rangeSum(A, queries) {
        const N = A.length;
        const prefix = new Array(N);
        prefix[0] = A[0];
        
        // Create the prefix sum array
        for (let i = 1; i < N; i++) {
            prefix[i] = prefix[i - 1] + A[i];
        }

        const result = [];
        // Process queries
        for (const [l, r] of queries) {
            if (l === 0) {
                result.push(prefix[r]);
            } else {
                result.push(prefix[r] - prefix[l - 1]);
            }
        }

        return result;
    }

    // Example input
    const A = [1, 2, 3, 4, 5];
    const queries = [[0, 2], [1, 4], [2, 2]];
    console.log(rangeSum(A, queries)); // [6, 14, 3]
    
    

Time Complexity Analysis

Creating the prefix sum array takes O(N) time, and each query can be processed in O(1). Therefore, the overall time complexity of the algorithm is O(N + Q). This is an efficient solution that satisfies the input conditions of the given problem.

Conclusion

In this lecture, we learned how to calculate the interval sum. We learned to use the prefix sum array to compute the interval sum in O(1) time. This approach can also be applied to solving other similar problems, which will be very helpful in studying algorithms.

JavaScript Coding Test Course, Fast Forward with Time Machine

Coding tests are becoming mandatory for more and more companies, and JavaScript is one of the most popular languages in web development.
In this course, we will solve commonly occurring algorithm problems found in coding tests using JavaScript.
Today’s topic is the problem ‘Fast Travel with a Time Machine’.

Problem Description

You are a scientist who can operate a time machine. The time machine can move to a specific time, and
given two integers a and b, you need to calculate the minimum time (number of moves) it takes to move from a to b.
The time machine follows these rules:

  • You can add 1 or subtract 1 from your current position.
  • You can double your current position.

Write a function that calculates the minimum number of moves required to travel from a to b for given a and b.

Input Format

– Two integers a (0 ≤ a ≤ 105), b (0 ≤ b ≤ 105) are given.

Output Format

– Output the minimum number of moves required to travel from a to b as an integer.

Examples

        Input: a = 2, b = 9
        Output: 4
    
        Input: a = 5, b = 22
        Output: 7
    

Approach to Solve the Problem

To solve this problem, we will use BFS (Breadth-First Search). BFS is an algorithm suitable for finding the shortest path,
exploring all possible states from the given state to find the shortest move sequence to reach the target state.
In this case, each state represents a time the time machine could occupy.

Step 1: Prepare to Use BFS Algorithm

To implement BFS, we will use a queue. First, we add the starting position a to the queue.
Then, we will repeat adding all possible moves to the queue until we reach the target position b.
The possible moves from each position are as follows:

  • Add 1 to the current position
  • Subtract 1 from the current position
  • Double the current position

We will keep track of the number of moves taken for each move until we reach the target position, and output the count when we reach the target.

Step 2: Implement the Code


function minimumMoves(a, b) {
    if (a >= b) return a - b; // If a is greater than or equal to b, determine moves by simply subtracting the difference
    const queue = [[a, 0]]; // [current position, number of moves]
    const visited = new Set(); // Record visited positions
    visited.add(a); // Mark the starting position as visited

    while (queue.length > 0) {
        const [current, moves] = queue.shift(); 

        // Possible moves
        const nextMoves = [current - 1, current + 1, current * 2]; 

        for (let next of nextMoves) {
            if (next === b) return moves + 1; // Return move count when the target position is reached
            if (next < 0 || next > 100000 || visited.has(next)) continue; // Only for valid range and unvisited positions
            visited.add(next); // Mark the next position as visited
            queue.push([next, moves + 1]); // Add the next position and move count to the queue
        }
    }
}
    

Step 3: Analyze Algorithm Complexity

The time complexity of the algorithm is O(n).
In the worst case, we need to explore all possible positions, which corresponds to O(n),
and the space complexity is also O(n). This complexity arises from the space needed for the queue and visited records.

Step 4: Optimization Possibilities

This algorithm is already based on BFS, which explores the shortest path
and does not require additional optimization. However, depending on the situation,
DFS (Depth-First Search) could also be applied, but BFS is more effective for this problem.

Step 5: Conclusion

Through the ‘Fast Travel with a Time Machine’ problem, we have briefly understood the principle of BFS and learned how to solve a problem using JavaScript.
In this way, various problems can be solved, and it is crucial to master basic algorithms to achieve good results in coding tests.

Additional Resources

– To further study the BFS algorithm and practice various problem-solving, we recommend the following resources.

JavaScript Coding Test Course, Implementing the Euler’s Phi Function

Hello, everyone! Today I will explain in detail how to implement the Euler’s totient function (𝜙(n)) using JavaScript. The Euler’s totient function represents the number of integers less than or equal to a given integer n that are coprime to n. This problem appears very frequently in algorithmic problems related to number theory and can be useful in various coding tests.

What is the Euler’s Totient Function?

The Euler’s totient function 𝜙(n) returns the count of positive integers from 1 to n that are coprime to n. In other words, two numbers a and b are said to be coprime if their greatest common divisor (GCD) is 1.

For example:

  • 𝜙(1) = 1 (The only number coprime to 1 is 1 itself)
  • 𝜙(2) = 1 (The number less than 2 and coprime to 2 is 1)
  • 𝜙(3) = 2 (The numbers less than 3 and coprime to 3 are 1, 2)
  • 𝜙(4) = 2 (The numbers less than 4 and coprime to 4 are 1, 3)
  • 𝜙(5) = 4 (The numbers less than 5 and coprime to 5 are 1, 2, 3, 4)

Problem Definition

Now, let’s define the coding test problem.

Problem: Write a function to calculate the Euler's totient function for a given integer n.
Input: Integer n (1 ≤ n ≤ 106)
Output: The value of the Euler's totient function 𝜙(n)

Problem Solving Approach

There are several methods to calculate the Euler’s totient function. The most efficient method is to modify the Sieve of Eratosthenes to compute 𝜙(n) in one go. The time complexity of this method is O(n log log n).

Properties of the Euler’s Totient Function

The key properties required to calculate 𝜙(n) are as follows:

  • 𝜙(p) = p – 1, where p is a prime number
  • 𝜙(p1 * p2 * … * pk) = p1 * p2 * … * pk * (1 – 1/p1) * (1 – 1/p2) * … * (1 – 1/pk)
  • 𝜙(n) = n * (1 – 1/p1) * (1 – 1/p2) * … * (1 – 1/pk), where pi are the prime factors of n.

JavaScript Code Implementation

Based on the above properties, let’s write the code to implement the Euler’s totient function.


function eulerPhi(n) {
    // Declare and initialize the array
    const phi = Array.from({ length: n + 1 }, (_, i) => i);
    
    // Find primes using Sieve of Eratosthenes
    for (let p = 2; p <= n; p++) {
        if (phi[p] === p) { // If p is prime
            for (let k = p; k <= n; k += p) {
                // Update the value of 𝜙(k)
                phi[k] = Math.floor(phi[k] * (1 - 1 / p));
            }
        }
    }
    
    // Return phi(n) for n
    return phi[n];
}

// Test cases
console.log(eulerPhi(1));  // 1
console.log(eulerPhi(2));  // 1
console.log(eulerPhi(3));  // 2
console.log(eulerPhi(4));  // 2
console.log(eulerPhi(5));  // 4

Code Explanation

The code works as follows:

  1. First, an array phi of size n + 1 is created, and each element is initialized to itself. phi[i] starts with the value of i.
  2. Loop from 2 to n, checking each number p to see if it is prime. If phi[p] equals p, it is considered prime.
  3. If p is prime, find its multiples and update the values of phi[k]. The update is performed as 𝜙(k) = 𝜙(k) * (1 - 1/p).
  4. Finally, return the value of phi[n] to compute the value of the Euler’s totient function for n.

Complexity Analysis

The time complexity of the above code is O(n log log n). This is due to the use of the Sieve of Eratosthenes method. The space complexity is O(n), as it requires space equivalent to the size of n to store the array phi.

Conclusion

We have learned how to implement the Euler's totient function in JavaScript. This method is very useful for algorithm testing and number theory, allowing for efficient computation of the Euler's totient value. Use this code to solve various problems!

Going Further

If you want to practice solving similar problems, working on problems involving the greatest common divisor or least common multiple would also be good practice. Additionally, exploring other concepts in number theory such as primality testing and prime generation is recommended. Enhance your understanding of algorithms through mathematical reasoning!

References

I hope this post was helpful. If you have any questions or additional inquiries, please leave a comment! Thank you!

JavaScript Coding Test Course, Finding Minimum Spanning Tree

Today, we will learn about a method to find the “Minimum Spanning Tree (MST)” which is a common problem in algorithm tests. In particular, we will explain how to solve this problem step by step using JavaScript. Through this tutorial, you will understand all the concepts and develop the ability to confidently solve this problem in actual coding tests.

1. What is a Minimum Spanning Tree?

A Minimum Spanning Tree is a subgraph that includes all vertices of a connected graph and has the minimum sum of edge weights. In other words, it refers to a tree that connects all vertices with the least cost. MST is used in various fields such as network design, transportation systems, and clustering.

2. Problem Description

When given the vertices and edges information of a graph, please write a function that finds the Minimum Spanning Tree and returns its total weight.

Input Format

  • The number of vertices n (1 ≤ n ≤ 1000)
  • The number of edges m (1 ≤ m ≤ 10000)
  • Each edge is given in the form of (a, b, c), where a and b are the vertices and c is the weight of the edge.

Output Format

Print the total weight of the Minimum Spanning Tree.

Example

    Input:
    4 5
    1 2 1
    1 3 4
    2 3 2
    1 4 3
    3 4 5

    Output:
    6
    

3. Algorithm Selection

There are several methods to find the Minimum Spanning Tree. Among these, Kruskal’s Algorithm and Prim’s Algorithm are widely used. We will use Kruskal’s Algorithm here.

Kruskal’s Algorithm

Kruskal’s Algorithm sorts the edges based on their weights and selects the edge with the lowest weight first, ensuring that no cycles are formed so that it can create the Minimum Spanning Tree. This method first sorts the given list of edges and then adds the lightest edges one by one.

4. Algorithm Implementation

Now, let’s write the JavaScript code to solve the problem using Kruskal’s Algorithm. The overall steps are as follows:

  1. After receiving edge information, sort them based on weights.
  2. Use the Union-Find data structure to include edges without forming cycles.
  3. After processing all edges, calculate and return the total weight of the Minimum Spanning Tree.

Code Implementation

    
    function find(parent, i) {
        if (parent[i] === -1) {
            return i;
        }
        return find(parent, parent[i]);
    }

    function union(parent, x, y) {
        const xset = find(parent, x);
        const yset = find(parent, y);
        parent[xset] = yset;
    }

    function kruskal(n, edges) {
        edges.sort((a, b) => a[2] - b[2]); // Sort by edge weight
        let parent = Array(n + 1).fill(-1);
        let minWeight = 0;
        const mst = [];

        for (let i = 0; i < edges.length; i++) {
            const [u, v, weight] = edges[i];

            if (find(parent, u) !== find(parent, v)) {
                union(parent, u, v);
                minWeight += weight;
                mst.push([u, v, weight]);
            }
        }

        return { minWeight, mst };
    }

    // Example input data
    const n = 4;
    const edges = [
        [1, 2, 1],
        [1, 3, 4],
        [2, 3, 2],
        [1, 4, 3],
        [3, 4, 5]
    ];

    const result = kruskal(n, edges);
    console.log("Total weight of the Minimum Spanning Tree:", result.minWeight);
    
    

5. Code Explanation

The above code is a function that uses Kruskal's Algorithm to find the Minimum Spanning Tree of the given graph. It is divided into the following key parts:

5.1. Union-Find Function

The Union-Find data structure is used to track the connected components of the graph. Each node has its own parent. The find function finds the representative of the set that the node belongs to, and the union function merges two sets.

5.2. Edge Sorting

Sort the list of edges by weight to select the minimum weight edge first. The sort method in JavaScript can be used to sort easily.

5.3. Minimum Spanning Tree Construction

For each edge, check the parents of the two nodes and select the edge only if it does not create a cycle. The selected edges are stored in the mst array, and the sum of the weights is incremented in the minWeight variable.

6. Performance Analysis

The time complexity of Kruskal's Algorithm is O(E log E). Here, E is the number of edges. Under the constraints of the given problem, this algorithm is efficient. You can expect additional performance improvements with the path compression technique of Union-Find.

7. Conclusion

In this tutorial, we learned about Kruskal's Algorithm to find the Minimum Spanning Tree using JavaScript and explained in detail how to solve the problem. Graph problems are frequently posed in algorithm competitions and various coding tests, so mastering this content will be very helpful. Next, we will solve various problems using other algorithms or data structures.

8. Practice Problem

Try to solve the following problem. After solving it, review your code to check for any parts that can be optimized.

Write an algorithm to extract the edges of the Minimum Spanning Tree from the given graph and output this list of edge weights.

9. References

JavaScript Coding Test Course, Radix Sort

This course will introduce the Radix Sort algorithm implemented in JavaScript and detail how to use it to solve coding test problems. We will systematically learn the concept of the Radix Sort algorithm, its implementation method, time complexity, and example problems.

What is Radix Sort?

Radix Sort is one of the sorting algorithms and an efficient method for sorting numbers with a similar number of digits. The key to this method is to sort the numbers by dividing them into individual digits (units, tens, etc.) and then sequentially considering the digits to sort the entire number.

The Principle of Radix Sort

Radix Sort proceeds in the following order:

  1. Find the maximum number of digits in the input array. This determines how many passes are needed to perform the sorting.
  2. Perform a stable sort for each digit, starting from the least significant digit (units) to the most significant digit (maximum digit).
  3. Finally, after all digit sorting is complete, the original array will be sorted.

Time Complexity

The time complexity of Radix Sort mainly depends on the stable sorting algorithm used, but it is generally O(nk), where n is the number of numbers to be sorted and k is the digit length of the largest number. Radix Sort can only be used for integers by nature, but it can also be applied in a modified version for characters or strings.

Implementing Radix Sort in JavaScript

Basic Algorithm Implementation

Below is an example code of Radix Sort implemented in JavaScript:

function getMax(array) {
    let max = array[0];
    for (let i = 1; i < array.length; i++) {
        if (array[i] > max) {
            max = array[i];
        }
    }
    return max;
}

function countingSort(array, place) {
    const n = array.length;
    const output = new Array(n);
    const count = new Array(10).fill(0);

    for (let i = 0; i < n; i++) {
        const digit = Math.floor(array[i] / place) % 10;
        count[digit]++;
    }

    for (let i = 1; i < 10; i++) {
        count[i] += count[i - 1];
    }

    for (let i = n - 1; i >= 0; i--) {
        const digit = Math.floor(array[i] / place) % 10;
        output[count[digit] - 1] = array[i];
        count[digit]--;
    }

    for (let i = 0; i < n; i++) {
        array[i] = output[i];
    }
}

function radixSort(array) {
    const max = getMax(array);
    for (let place = 1; Math.floor(max / place) > 0; place *= 10) {
        countingSort(array, place);
    }
    return array;
}

// Usage example
const numbers = [170, 45, 75, 90, 802, 24, 2, 66];
console.log(radixSort(numbers)); // Output: [2, 24, 45, 66, 75, 90, 170, 802]

Example Problem: Sorting an Integer Array

Now, let’s apply Radix Sort to a practical problem. The problem is as follows:

Problem: Given an array of integers, write a function to sort this array in ascending order using the Radix Sort algorithm.

Problem Approach

  1. Receive the input array as a function argument.
  2. Use the Radix Sort algorithm to sort the array.
  3. Return the sorted array.

Implementation and Testing

Based on the Radix Sort algorithm explained above, we can implement a function to solve the problem as follows:

function sortIntegers(array) {
    return radixSort(array);
}

// Test
const testArray = [5, 3, 8, 1, 2, 7, 4, 6];
console.log(sortIntegers(testArray)); // Output: [1, 2, 3, 4, 5, 6, 7, 8]

Conclusion

In this course, we learned about the Radix Sort algorithm and how to implement it in JavaScript. Radix Sort is very efficient for sorting large integers, especially showing excellent performance for numbers with fewer digits. Utilizing Radix Sort to approach various coding test problems will be a very beneficial experience. I hope you make good use of the concepts and implementation methods of Radix Sort in your upcoming coding tests.