Java Coding Test Course, Lowest Common Ancestor

In this article, we will take a detailed look at the definition and solutions of the Lowest Common Ancestor (LCA) problem.

Problem Definition

Problem: Lowest Common Ancestor (LCA)

Find the lowest common ancestor of two nodes in a binary tree. The lowest common ancestor is the ancestor of both nodes and is the closest node to both nodes.

For example, given the following binary tree:

                3
              /   \
             5     1
            / \   / \
           6   2 0   8
              / \
             7   4
        

The lowest common ancestor of node 5 and node 1 is 3, and the lowest common ancestor of node 5 and node 4 is 5.

Input Format

  • The root node of the binary tree and two nodes are given.

Output Format

  • Print the lowest common ancestor of the given two nodes.

Problem Solution

1. Problem Analysis

To find the lowest common ancestor of two nodes in the given binary tree, two conditions must be satisfied:

  • There must exist a node B that is an ancestor of node A.
  • There must also exist a node B that is an ancestor of node C.

To establish this relationship, we can traverse the binary tree while tracking the parent node for each node.

2. Algorithm Design

First, to find a specific node in the binary tree, we can use search techniques such as DFS (Depth First Search) or BFS (Breadth First Search).

The specific algorithm to find the lowest common ancestor is as follows:

  1. Start traversing the tree from the root node.
  2. Find the given two nodes in both subtrees.
  3. If one node is found in each subtree, the current node is the lowest common ancestor.
  4. If neither is found, return null.

3. Java Code Implementation

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}

public class LCA {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // Base case
        if (root == null || root == p || root == q) {
            return root;
        }

        // Find LCA in the left and right subtrees.
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        // If both are found, current node is the LCA.
        if (left != null && right != null) {
            return root;
        }

        // Return the result from the subtree where one was found.
        return left != null ? left : right;
    }
}
    

4. Code Explanation

The key points in the above code are as follows:

  • Recursively check if the current node is null or the node we are looking for.
  • Find the lowest common ancestor in the left and right subtrees, and return the current node if both are found.
  • Finally, if found only in one subtree, return that node; otherwise, return null.

5. Time Complexity

The time complexity of this algorithm is O(N), as each node in the collection is visited once. N is the number of nodes in the tree.

6. Space Complexity

The space complexity is O(H), where H is the height of the tree. In the worst case, if the binary tree is skewed to one side, the maximum height can be N.

Practical Code Test

To test the problem in practice, let’s create an example tree and write code to find the LCA.

public class Main {
    public static void main(String[] args) {
        // Creating the binary tree
        TreeNode root = new TreeNode(3);
        TreeNode node5 = new TreeNode(5);
        TreeNode node1 = new TreeNode(1);
        TreeNode node6 = new TreeNode(6);
        TreeNode node2 = new TreeNode(2);
        TreeNode node0 = new TreeNode(0);
        TreeNode node8 = new TreeNode(8);
        TreeNode node7 = new TreeNode(7);
        TreeNode node4 = new TreeNode(4);

        root.left = node5;
        root.right = node1;
        node5.left = node6;
        node5.right = node2;
        node1.left = node0;
        node1.right = node8;
        node2.left = node7;
        node2.right = node4;

        // Create an object to find the LCA
        LCA lcaFinder = new LCA();
        TreeNode lca = lcaFinder.lowestCommonAncestor(root, node5, node1);
        System.out.println("LCA of 5 and 1: " + lca.val); // Output: 3

        lca = lcaFinder.lowestCommonAncestor(root, node5, node4);
        System.out.println("LCA of 5 and 4: " + lca.val); // Output: 5
    }
}
    

Conclusion

In this article, we covered the problem of finding the lowest common ancestor in a binary tree, explaining the problem definition, algorithm design, Java code implementation, and code testing. This problem can be applied in various situations and is very helpful for understanding the basic concepts of algorithms and data structures.

Through this, you can learn useful algorithm patterns for preparing for coding tests in Java and contribute to improving your problem-solving skills.

Java Coding Test Course, Finding Least Common Multiple

In this article, we will explore in depth the method to calculate the Least Common Multiple (LCM) using Java, a problem that frequently appears in coding tests. The Least Common Multiple refers to the smallest multiple of two or more numbers and plays a very important role in various algorithmic problems.

1. Problem Definition

Here is a simple definition of the problem of finding the Least Common Multiple.

        Problem: Given two integers a and b, find the least common multiple of a and b.
        Example:
        Input: a = 4, b = 6
        Output: 12
    

2. What is Least Common Multiple (L.C.M)?

The Least Common Multiple refers to the smallest multiple among two numbers. For example, the multiples of 4 and 6 are as follows:

  • Multiples of 4: 4, 8, 12, 16, 20, …
  • Multiples of 6: 6, 12, 18, 24, 30, …

In the above example, the smallest number among the multiples of 4 and 6 is 12. Therefore, 12 is the Least Common Multiple of 4 and 6.

3. Common Divisors and Multiples

To understand the Least Common Multiple, it is essential to first understand the Greatest Common Divisor (GCD). The Greatest Common Divisor refers to the largest common divisor of two numbers. The Least Common Multiple can be calculated using the Greatest Common Divisor as follows:

        LCM(a, b) = (a * b) / GCD(a, b)
    

Using the above formula, we can quickly and efficiently find the Least Common Multiple even for large numbers.

4. Algorithm Design

The algorithm to find the Least Common Multiple can be designed as follows:

  1. Input the two integers a and b.
  2. Calculate GCD(a, b).
  3. Calculate LCM(a, b).
  4. Output the result.

5. Java Code Implementation

Now, let’s implement the above algorithm using Java code.

        public class LCMCalculator {
            // Method to calculate Greatest Common Divisor (GCD)
            public static int gcd(int a, int b) {
                while (b != 0) {
                    int temp = b;
                    b = a % b;
                    a = temp;
                }
                return a;
            }

            // Method to calculate Least Common Multiple (LCM)
            public static int lcm(int a, int b) {
                return (a * b) / gcd(a, b);
            }

            // Main method
            public static void main(String[] args) {
                int a = 4;
                int b = 6;
                
                int result = lcm(a, b);
                System.out.println("Least Common Multiple: " + result);
            }
        }
    

5.1 Code Explanation

Let’s explain how the above code works.

  1. gcd method: Accepts two integers a and b and calculates the Greatest Common Divisor. It efficiently computes the GCD using the Euclidean algorithm.
  2. lcm method: A method to find the Least Common Multiple of two numbers, applying the previously described GCD formula.
  3. main method: Calculates and outputs the Least Common Multiple for two numbers input by the user.

6. Additional Examples and Tests

Now, let’s test some other inputs. We can enhance the reliability of the code with various test cases.

Example 1

        Input: a = 15, b = 20
        Output: 60
    

Example 2

        Input: a = 9, b = 12
        Output: 36
    

Example 3

        Input: a = 7, b = 5
        Output: 35
    

7. Time Complexity Analysis

The time complexity of the above algorithm can be analyzed as follows:

  • The gcd method has a time complexity of O(log(min(a, b))).
  • Thus, the overall time complexity is O(log(min(a, b))), making it very efficient.

8. Conclusion

In this lecture, we introduced the algorithm for finding the Least Common Multiple using Java. We covered not only the design of the algorithm but also the implementation process and time complexity analysis. Since this problem frequently appears in coding tests, we encourage you to build your skills through sufficient practice. We plan to cover useful algorithms in the next lecture, so stay tuned!

Java Coding Test Course, Finding the Shortest Path

In this lecture, we will cover the problem of “finding the shortest path,” which often appears in algorithm tests for employment. The shortest path problem is an important topic in graph theory and is very useful when finding the optimal route in complex network environments. We will discuss how to solve this problem using Java.

Problem Description

This is the problem of finding the shortest path from a specific starting point to a specific destination in a given graph that satisfies the following conditions.

  • The graph is directed and has weighted edges.
  • The total number of nodes is V, and the total number of edges is E.
  • The nodes of the graph are numbered from 1 to V.

Example Problem

When given the following graph as input:

5 8          // 5 nodes, 8 edges
1 2 2       // weight 2 from node 1 to node 2
1 3 3       // weight 3 from node 1 to node 3
2 3 1       // weight 1 from node 2 to node 3
2 4 1       // weight 1 from node 2 to node 4
3 4 6       // weight 6 from node 3 to node 4
3 5 1       // weight 1 from node 3 to node 5
4 5 2       // weight 2 from node 4 to node 5
1 5 10      // weight 10 from node 1 to node 5

When the starting point is node 1 and the destination point is node 5, find the weight of the shortest path.

Problem Analysis

This problem can be solved using Dijkstra’s algorithm. Dijkstra’s algorithm is a method for finding the shortest path from one node to all other nodes in a given graph. This algorithm follows these steps:

  1. Set the starting node to 0 and all other nodes to infinity.
  2. Update the distances to adjacent nodes.
  3. Select the node with the shortest distance to determine the path.
  4. Repeat this process until all nodes have been selected.

Algorithm Implementation

Now, let’s implement Dijkstra’s algorithm in Java.

import java.util.*;

public class DijkstraAlgorithm {
    static class Edge {
        int node;
        int weight;

        Edge(int node, int weight) {
            this.node = node;
            this.weight = weight;
        }
    }

    static final int INF = Integer.MAX_VALUE;
    static List> graph = new ArrayList<>();
    static int[] dist;
    static boolean[] visited;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int V = scanner.nextInt(); // Number of nodes
        int E = scanner.nextInt(); // Number of edges

        // Initialize the graph
        for (int i = 0; i <= V; i++) {
            graph.add(new ArrayList<>());
        }

        // Input edges
        for (int i = 0; i < E; i++) {
            int from = scanner.nextInt();
            int to = scanner.nextInt();
            int weight = scanner.nextInt();
            graph.get(from).add(new Edge(to, weight));
        }

        int start = 1; // Starting node
        dist = new int[V + 1];
        visited = new boolean[V + 1];

        // Initialize distances
        Arrays.fill(dist, INF);
        dist[start] = 0;

        // Run Dijkstra's algorithm
        dijkstra(start);

        // Output shortest distance
        int end = 5; // Destination node
        System.out.println("Shortest distance: " + dist[end]);
    }

    private static void dijkstra(int start) {
        PriorityQueue pq = new PriorityQueue<>(Comparator.comparingInt(e -> e.weight));
        pq.offer(new Edge(start, 0));

        while (!pq.isEmpty()) {
            Edge current = pq.poll();
            int currentNode = current.node;

            if (visited[currentNode]) continue;
            visited[currentNode] = true;

            for (Edge edge : graph.get(currentNode)) {
                if (dist[currentNode] + edge.weight < dist[edge.node]) {
                    dist[edge.node] = dist[currentNode] + edge.weight;
                    pq.offer(new Edge(edge.node, dist[edge.node]));
                }
            }
        }
    }
}

Code Explanation

The DijkstraAlgorithm class above implements Dijkstra's algorithm. Here is an explanation of each part:

  • Edge class: A class that stores nodes and weights.
  • graph: Represents the graph in adjacency list format.
  • dist: An array representing the shortest distance to each node.
  • visited: Indicates whether a node has been visited.
  • dijkstra(): A method implementing Dijkstra's algorithm, which uses a priority queue to update the shortest distances.

Results and Execution

When you run the code above, you can calculate the shortest distance from node 1 to node 5 for the given input graph. The result will be as follows:

Shortest distance: 3

Conclusion

In this lecture, we explored how to solve the shortest path problem using Java. Dijkstra's algorithm is used in various applications and is very useful when dealing with complex graphs. Through this problem, we learned the basic concepts of graph theory and how to implement them in Java. In the future, we will also explore more challenging algorithm problems and various graph traversal techniques.

Note: When solving algorithm problems, it is always good to consider various test cases to validate the problem. In particular, when there are multiple paths or negative weights, the performance of the algorithm should be analyzed closely.

Java Coding Test Course, Finding the Greatest Common Divisor

Hello! In this article, we will explore “Finding the Greatest Common Divisor” (GCD), one of the problems frequently presented in coding tests using Java. I will explain in detail the basic knowledge needed for solving algorithm problems, step-by-step approaches to problem-solving, and implementation of Java code.

1. Problem Description

Given two integers a and b, the task is to find the greatest common divisor (GCD) of these two numbers. The GCD is defined as the largest divisor that is common to both numbers. For example, if a = 12 and b = 15, the divisors of both numbers are as follows:

  • Divisors of 12: 1, 2, 3, 4, 6, 12
  • Divisors of 15: 1, 3, 5, 15

Thus, the greatest common divisor of 12 and 15 is 3.

2. Problem Approach

There are several methods to find the greatest common divisor, but among them, the **Euclidean algorithm** is very efficient and widely used. The basic idea of the Euclidean algorithm is as follows:

  • Let there be two integers a and b, and define r as the remainder when a is divided by b.
  • Then, GCD(a, b) = GCD(b, r). In other words, the GCD of a and b is the same as the GCD of b and the remainder of a divided by b.
  • This process is repeated until r becomes 0. At this point, b is GCD(a, b).

This method is very efficient, with a time complexity of O(log(min(a, b))), allowing results to be derived in relatively short time.

3. Java Code Implementation

Now, let’s implement Java code based on the above algorithm. Below is the Java code to find the greatest common divisor:


public class GCD {
    // Method to calculate GCD using the Euclidean algorithm
    public static int gcd(int a, int b) {
        // If b is 0, then a is the GCD
        if (b == 0) {
            return a;
        }
        // Recursively call to calculate GCD using the remainder
        return gcd(b, a % b);
    }

    public static void main(String[] args) {
        int a = 12; // First integer
        int b = 15; // Second integer
        int result = gcd(a, b); // Calculate GCD
        System.out.println("Greatest Common Divisor: " + result); // Print result
    }
}

        

4. Code Explanation

In the above code, the gcd method takes two integers a and b as parameters and calculates their GCD. Inside the method, it checks if b is 0, and if so, returns a. In other cases, it recursively calls the gcd method, passing the remainder of a divided by b as the argument. This process is repeated until the final GCD is derived.

5. Testing with Various Inputs

It’s important to test various input values to ensure that the code operates correctly. Below are some input examples along with their GCDs:

  • a = 48, b = 18 → GCD = 6
  • a = 101, b = 10 → GCD = 1
  • a = 56, b = 42 → GCD = 14
  • a = 24, b = 36 → GCD = 12

The above examples cover different scenarios, allowing verification of the GCD calculation process for various combinations. Don’t forget to check if the results from the code are accurate for each combination.

6. Similar Problem Solving Approach

A problem similar to the GCD problem is finding the “Lowest Common Multiple” (LCM). The LCM for two numbers a and b can be defined as follows:

LCM(a, b) = (a * b) / GCD(a, b)

Based on this relationship, one can simply find the LCM of the two given numbers. Therefore, understanding and applying the GCD problem allows for the seamless resolution of LCM problems.

7. Conclusion

In this tutorial, we learned how to solve the GCD problem. I hope that understanding the Euclidean algorithm and implementing it in Java will enhance your ability to solve algorithmic problems. By practicing various problems and learning additional algorithms and data structures, I wish you success in coding tests.

Thank you!

Java Coding Test Course, Representing Sets

Hello! In this tutorial, we will learn about representing collections in Java. Collections are one of the important data structures in algorithm problem-solving. Therefore, understanding and being able to use them can greatly help in Java coding tests. In this post, I will explain a basic problem of representing a collection and its solution process in detail.

Problem: Create a Set without Duplicates

Given an array of integers, write a function that converts it to a Set and returns it as a list after removing duplicate elements.

Input

  • Array: [1, 2, 2, 3, 4, 4, 5]

Output

  • List: [1, 2, 3, 4, 5]

Problem-Solving Process

To solve this problem, it is important to first understand the definition of a Set. A Set is a data structure that does not allow duplicate values, and in Java, it can be implemented using HashSet. Below is a summary of the main processes to solve the problem.

1. Confirming the Need for a Set

To remove duplicate elements from the given array, we need to use a Set. Using a Set naturally handles duplicates and maintains the uniqueness of each element. For example, in the array [1, 2, 2, 3, 4, 4, 5], since 2 and 4 are duplicates, converting it to a Set results in [1, 2, 3, 4, 5].

2. Using HashSet in Java

To implement a Set in Java, we can use the HashSet class. HashSet is an implementation of a Set that uses a hash table internally, allowing elements to be added and searched with a time complexity of O(1).

3. Implementing the Function

Now, let’s implement the function needed to solve the given problem. Take a look at the following code.

import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.stream.Collectors;

public class SetExample {
    public static List<Integer> uniqueElements(int[] arr) {
        // Removing duplicates using HashSet
        Set<Integer> set = new HashSet<>();
        for (int num : arr) {
            set.add(num);
        }
        // Converting the Set to a List and returning it
        return set.stream().collect(Collectors.toList());
    }

    public static void main(String[] args) {
        int[] inputArray = {1, 2, 2, 3, 4, 4, 5};
        List<Integer> result = uniqueElements(inputArray);
        System.out.println(result); // [1, 2, 3, 4, 5]
    }
}

4. Explaining the Code

The code above defines a function called uniqueElements that takes an array of integers as a parameter. This function uses HashSet to remove duplicate elements and then collects all elements in the Set into a List to return.

5. Entire Code Result

The main method above defines a sample array, then calls the uniqueElements function and prints the result. When this program is executed, you can see the following result:

[1, 2, 3, 4, 5]

Conclusion and Further Learning

In this tutorial, we learned how to use HashSet to represent collections in Java. Sets are very useful in various algorithm problems, so it is necessary to practice to use them well. Enhance your skills by solving various problems related to Sets.

Additionally, I recommend studying the characteristics of Sets and various methods. For example, try implementing various Set operations like intersection, union, etc., and understand their differences from other data structures in the Java Collection Framework.

This tutorial ends here. Next time, I will return with a more interesting topic! Thank you.