Category: Javascript Coding Test

In this course, we will address an algorithm problem that finds the Kth shortest path using JavaScript. This problem requires a solid understanding of graph theory and pathfinding algorithms and focuses on writing efficient code.

Problem Description

Essentially, you need to find the total length of the Kth shortest path among all possible paths from one vertex to another in the given graph.

Problem Input

• N: The number of vertices (1 ≤ N ≤ 100)
• M: The number of edges (1 ≤ M ≤ 1000)
• K: The rank of the desired path (1 ≤ K ≤ 10)
• Afterward, the edge information will be provided over M lines.
  Each edge information is composed of u, v, w, which means that the weight from vertex u to vertex v is w.

Output

Print the total length of the Kth shortest path. If the Kth shortest path does not exist, you should print -1.

Example

    4 5 2
    1 2 4
    1 3 2
    3 2 5
    2 4 1
    3 4 3
    

    5
    

Problem-Solving Approach

To find the Kth shortest path, one of the methods is to modify Dijkstra’s algorithm. The Dijkstra algorithm is primarily used to find the shortest path, but to find the Kth shortest path, it needs to allow for multiple explorations of paths.

Step-by-Step Solution Process

1. Graph Representation

The graph is represented in the form of an adjacency list, with each node storing information about its connected nodes and the weights of those paths. This allows Dijkstra’s algorithm to be applied efficiently.

2. Using a Priority Queue

Using JavaScript’s PriorityQueue to prioritize the exploration of the shortest path. It holds the information of each path, allowing for exploration starting from the minimum cost, just like Dijkstra’s algorithm.

3. Kth Path Exploration

While exploring paths, maintain a count to find the Kth shortest path. Store the lengths of paths in an array, returning the length when the Kth path is reached; otherwise, continue exploring.

4. Result Output

After exploring all paths, output -1 if the Kth path does not exist.

Implementation Code

The code below is a JavaScript version that finds the Kth shortest path based on the explanation above.

class MinHeap {
    constructor() {
        this.heap = [];
    }

    insert(node) {
        this.heap.push(node);
        this.bubbleUp();
    }

    bubbleUp() {
        let index = this.heap.length - 1;
        const element = this.heap[index];

        while (index > 0) {
            let parentIndex = Math.floor((index - 1) / 2);
            let parent = this.heap[parentIndex];

            if (element[1] >= parent[1]) break;

            this.heap[index] = parent;
            index = parentIndex;
        }
        this.heap[index] = element;
    }

    extractMin() {
        const min = this.heap[0];
        const end = this.heap.pop();
        if (this.heap.length > 0) {
            this.heap[0] = end;
            this.sinkDown();
        }
        return min;
    }

    sinkDown() {
        let index = 0;
        const length = this.heap.length;
        const element = this.heap[0];

        while (true) {
            let leftChildIndex = 2 * index + 1;
            let rightChildIndex = 2 * index + 2;
            let leftChild, rightChild;
            let swap = null;

            if (leftChildIndex < length) {
                leftChild = this.heap[leftChildIndex];
                if (leftChild[1] < element[1]) {
                    swap = leftChildIndex;
                }
            }

            if (rightChildIndex < length) {
                rightChild = this.heap[rightChildIndex];
                if ((swap === null && rightChild[1] < element[1]) || 
                    (swap !== null && rightChild[1] < leftChild[1])) {
                    swap = rightChildIndex;
                }
            }

            if (swap === null) break;

            this.heap[index] = this.heap[swap];
            index = swap;
        }
        this.heap[index] = element;
    }
}

function kthShortestPath(n, edges, k) {
    const graph = Array.from({ length: n + 1 }, () => []);
    edges.forEach(([u, v, w]) => {
        graph[u].push([v, w]);
    });

    const minHeap = new MinHeap();
    const paths = Array.from({ length: n + 1 }, () => []);

    minHeap.insert([1, 0]);
    paths[1].push(0);

    while (minHeap.heap.length > 0) {
        const [node, distance] = minHeap.extractMin();

        if (paths[node].length === k) continue;

        for (const [neighbor, weight] of graph[node]) {
            const newDistance = distance + weight;

            if (paths[neighbor].length < k) {
                paths[neighbor].push(newDistance);
                paths[neighbor].sort((a, b) => a - b);
                minHeap.insert([neighbor, newDistance]);
            } else if (newDistance < paths[neighbor][paths[neighbor].length - 1]) {
                paths[neighbor][paths[neighbor].length - 1] = newDistance;
                paths[neighbor].sort((a, b) => a - b);
                minHeap.insert([neighbor, newDistance]);
            }
        }
    }

    return paths[n][k - 1] || -1;
}

// Example usage
const n = 4;
const edges = [
    [1, 2, 4],
    [1, 3, 2],
    [3, 2, 5],
    [2, 4, 1],
    [3, 4, 3],
];
const k = 2;

console.log(kthShortestPath(n, edges, k)); // Output: 5

Conclusion

In this course, we discussed algorithms for finding the Kth shortest path. We learned techniques to solve problems based on shortest path algorithms. With this foundation, we can develop the ability to solve more complex graph problems.

The ability to solve algorithm problems is very important in the process of preparing for coding tests. In this post, we will explore a JavaScript coding test problem on the topic of “Calculating the Sum of Consecutive Numbers.” I will explain the process of understanding and solving the problem step by step. This process will help you tackle complex algorithm problems.

Problem Description

You need to calculate the sum of consecutive numbers in a given array. Consecutive numbers refer to two adjacent elements in the array, and their sum can be calculated. However, this problem requires not just finding the sum of two elements, but calculating the sum of all possible consecutive subarrays and finding the maximum sum among them.

Input

• An integer array arr is given. (1 ≤ arr.length ≤ 10^5)
• Each element of the array falls within the range of -10^4 ≤ arr[i] ≤ 10^4.

Output

Return the maximum value among the sums of all consecutive subarrays.

Example

Input: arr = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: The sum of the continuous subarray [4,-1,2,1] is 6, which is the largest sum.

Problem Approach

To solve this problem, we need an algorithm that can effectively calculate the sum of consecutive subarrays. Here are the steps to follow to solve this problem:

Step 1: Understanding

Clearly understand the requirements of the problem and analyze the cases in which a large sum occurs. For example, if all elements of the array are negative, we need to realize that we should return the largest among them.

Step 2: Choosing an Algorithm

We will use the “Kadane’s Algorithm” to solve this problem. Kadane’s Algorithm is a very efficient algorithm that can find the maximum sum of consecutive subarrays with O(n) time complexity. This algorithm works by tracking the maximum sum so far and deciding whether to include the current element or not.

Step 3: Implementing the Algorithm

Now, let’s implement Kadane’s Algorithm in JavaScript.

function maxSubArray(arr) {
    let maxSoFar = arr[0]; // Maximum value among the sum of all elements
    let maxEndingHere = arr[0]; // Maximum value at the current position

    for (let i = 1; i < arr.length; i++) {
        maxEndingHere = Math.max(arr[i], maxEndingHere + arr[i]); // Compare current value and previous sum
        maxSoFar = Math.max(maxSoFar, maxEndingHere); // Update maximum sum
    }

    return maxSoFar;
}

Step 4: Testing

After implementing the function, it should be tested with various input values. Below are some test cases:

console.log(maxSubArray([-2, 1, -3, 4, -1, 2, 1, -5, 4])); // 6
console.log(maxSubArray([1])); // 1
console.log(maxSubArray([5, 4, -1, 7, 8])); // 23
console.log(maxSubArray([-1, -2, -3, -4])); // -1
console.log(maxSubArray([-5, -2, -3, -4, -1])); // -1

Conclusion

In this post, we solved the “Calculating the Sum of Consecutive Numbers” problem using Kadane’s Algorithm. Problems like this frequently appear in coding tests, so it is essential to practice enough. Understanding the algorithm and validating accuracy with various test cases is important. Continuous effort is needed to improve your coding skills by tackling various algorithm problems in the future.

In this course, we will cover coding test problems using JavaScript.
In particular, we will solve various algorithm problems with the theme
of making cocktails. The problem we will discuss today is “Making Cocktails from Subset Combinations.”

Problem Description

We need to find all possible combinations of cocktails using the given ingredients.
Each cocktail must consist of two or more ingredients, and
the combinations should be presented without duplicates.

Input

• Number of ingredients N (1 ≤ N ≤ 20)
• Array of strings ingredients containing the names of each ingredient

Output

Should return an array of all possible cocktail combinations. Each combination should
be represented in the form ["ingredient1", "ingredient2", ...],
and the ingredients in each combination must be sorted in alphabetical order.

Example

Input

const ingredients = ["vodka", "orange juice", "grenadine"];
    

Output

[["grenadine", "orange juice"], ["grenadine", "vodka"], ["orange juice", "vodka"], ["grenadine", "orange juice", "vodka"]]
    

Problem Solving Process

To solve this problem, we first need to think about how to create all possible combinations.
We can solve the problem using recursive functions or bit manipulation for combination generation.

Algorithm Design

We will use a recursive approach to generate cocktail combinations.
The main steps of the algorithm are as follows:

1. Sort the ingredient array. (This is to produce the output in sorted order)
2. Use a recursive function to generate combinations of ingredients.
3. If the combination consists of two or more ingredients, add it to the result array.

Code Implementation

Now we will implement the algorithm we designed in JavaScript.

function cocktailCombinations(ingredients) {
    const result = [];
    
    ingredients.sort();

    const generateCombinations = (start, currentCombination) => {
        if (currentCombination.length > 1) {
            result.push([...currentCombination]);
        }

        for (let i = start; i < ingredients.length; i++) {
            currentCombination.push(ingredients[i]);
            generateCombinations(i + 1, currentCombination);
            currentCombination.pop();
        }
    };

    generateCombinations(0, []);

    return result;
}

// Example usage
const ingredients = ["vodka", "orange juice", "grenadine"];
console.log(cocktailCombinations(ingredients));
    

Code Explanation

The above cocktailCombinations function takes an array of ingredients as input
and generates all possible cocktail combinations. It defines and calls an internal
recursive function called generateCombinations to generate the combinations.

Detailed Functions

• Ingredient Sorting: Sorts the input ingredients to maintain consistency in the output.
• Recursive Calls: Uses recursion to select each ingredient and explore combination possibilities.
• Add Combination: Only adds to the result if the current combination length is 2 or more.

Complexity Analysis

The time complexity of this algorithm is
O(2^N).
This is because it is a binary choice problem, deciding whether to select each ingredient, which
is equal to the maximum number of physically possible combinations when exploring all combinations.
The space complexity is proportional to the depth of the array, and in the worst case, the
additional memory used will depend on the number of combinations.

Test Cases

Let’s test the function with various inputs to verify its accuracy.

// Test cases
const test1 = ["gin", "tonic", "lime"];
const test2 = ["whiskey", "soda", "angostura", "lemon"];
const test3 = [];

// Output results
console.log(cocktailCombinations(test1)); // [["gin", "lime"], ...]
console.log(cocktailCombinations(test2)); // ...
console.log(cocktailCombinations(test3)); // []

    

Conclusion

In this course, we have gone through the process of solving the cocktail combination problem using
JavaScript. Through a deep understanding of arrays, recursion, and combination possibilities,
we are able to more effectively learn strategies to deal with common
problem types encountered in coding tests.

Additional Information

In actual coding tests, there may often be additional exception handling or input constraint requirements.
Reflecting these conditions to improve the code will also be good practice.
Since the process of adding various features tailored to individual circumstances is important,
we hope this problem will create opportunities for you to advance further.

Coding tests are an important step in assessing your capabilities as a software developer. In particular, algorithm problems are a great way to test problem-solving skills and creativity. In this course, we will solve the problem of finding the most basic solution to a linear equation using JavaScript. We will find the solution to the given equation Ax + By = C.

Problem Definition

The problem is as follows:

Given integers A, B, and C, find all pairs of integers (x, y) that satisfy Ax + By = C. Note that x and y must be at least 0 and can be at most 10000.

Problem Analysis

This problem is about finding the solution to a simple linear equation, which must satisfy the following conditions for the two variables x and y:

• Ax + By = C
• 0 ≤ x, y ≤ 10000

To solve this problem, we can use a loop to substitute all possible x values, calculate the corresponding y values, and check whether the conditions are satisfied.

Solution Strategy

1. Iterate x from 0 to 10000.

2. For each x, calculate y:

y = (C - Ax) / B

3. Check if y is a non-negative integer.

4. Store all pairs (x, y) that satisfy the conditions.

JavaScript Code Implementation

Now, based on the above strategy, let’s implement the code using JavaScript. Below is an example of the code:

function findSolutions(A, B, C) {
    const solutions = [];
    
    for (let x = 0; x <= 10000; x++) {
        // Handle the case when B is 0
        if (B === 0) {
            if (A * x === C) {
                solutions.push([x, 0]);
            }
            continue;
        }
        
        const y = (C - A * x) / B;

        // Check if y is an integer and non-negative
        if (y >= 0 && Number.isInteger(y)) {
            solutions.push([x, y]);
        }
    }
    
    return solutions;
}

// Example execution
const A = 1, B = 2, C = 3;
const result = findSolutions(A, B, C);
console.log(result); // [ [ 0, 1 ], [ 1, 1 ], [ 3, 0 ] ]

Code Explanation

The above JavaScript function findSolutions works as follows:

1. Create an array solutions to store the results.
2. Iterate x from 0 to 10000.
3. For each x, calculate y. Check if B is 0, and handle the case where B is 0 as an exception.
4. After confirming that y is non-negative and an integer, add the (x, y) pair to the solutions array if the conditions are satisfied.
5. Return the solutions array after all iterations are complete.

Test Cases

Now let’s verify that the function works correctly with several test cases.

console.log(findSolutions(1, 2, 3)); // [ [ 0, 1 ], [ 1, 1 ], [ 3, 0 ] ]
console.log(findSolutions(2, 3, 6)); // [ [ 0, 2 ], [ 3, 0 ] ]
console.log(findSolutions(0, 1, 5)); // [ [ 0, 5 ] ]
console.log(findSolutions(1, 0, 5)); // [ [ 5, 0 ] ]
console.log(findSolutions(0, 0, 0)); // [ [ 0, 0 ] ]

Conclusion

In this article, we implemented an algorithm to find the solutions to the given linear equation Ax + By = C using JavaScript. We detailed the logic step by step to approach the problem and confirmed that the function works correctly through various test cases. Such problems are often presented in coding tests, and developing an algorithmic mindset through them will be greatly beneficial. We hope to elevate your coding skills further through a variety of algorithmic challenges.