Category: Kotlin coding test

Hello, everyone! Today, we will explore binary trees and take some time to solve related problems. A binary tree is one of the fundamental structures in computer science and algorithms, and it is a topic that frequently appears in job coding tests. In this article, we will introduce the basic concepts of binary trees, present a problem using these concepts, and explain the process of solving the problem with Kotlin in detail.

Basic Concepts of Binary Trees

A binary tree is a tree structure where each node can have at most two children. Each node contains data and references to its child nodes. Binary trees are used for various purposes and play a significant role in tree traversal, sorting algorithms, and more.

Structure of Binary Trees

class TreeNode(val value: Int) {
    var left: TreeNode? = null
    var right: TreeNode? = null
}

In the code above, the TreeNode class represents each node in a binary tree. The value is the value of the node, and left and right represent the left and right children, respectively.

Problem Statement

Now, I will present the problem we will solve. It is a problem of counting the number of paths in a given binary tree where the sum of nodes along each path equals a specified value.

Problem Description

Write a function to calculate the sum of paths from the root node of a given binary tree to each leaf node, and print the number of paths where this sum equals a given value.

Input:

• The root representing the root node of the binary tree.
• An integer sum, the sum of the path you want to find.

Output:

• The number of paths that match the given sum.

Example

Input:
        5
       / \
      4   8
     /   / \
    11  13  4
   /  \      \
  7    2      1
sum = 22

Output:
3

Problem-Solving Process

To solve this problem, we will use the DFS (Depth-First Search) algorithm. DFS is a useful method for traversing trees, visiting nodes through recursive calls.

Problem-Solving Approach

1. Explore each path using DFS.
2. Keep an accumulated sum as you visit each node.
3. When you reach a leaf node, check if the accumulated sum equals sum.
4. If they are equal, increase the count.

Kotlin Code

fun pathSum(root: TreeNode?, sum: Int): Int {
    return findPaths(root, sum, 0)
}

fun findPaths(node: TreeNode?, target: Int, currentSum: Int): Int {
    if (node == null) {
        return 0
    }

    val newSum = currentSum + node.value
    var pathCount = 0

    if (node.left == null && node.right == null && newSum == target) {
        pathCount += 1
    }

    pathCount += findPaths(node.left, target, newSum)
    pathCount += findPaths(node.right, target, newSum)

    return pathCount
}

Code Explanation

The code above is a function that calculates the sum of paths in a given binary tree. The pathSum function takes the provided root and sum as input and returns the number of paths. The findPaths function recursively explores nodes and calculates the sum of paths. When reaching each leaf node, it checks if the accumulated sum is equal to target, and if it is, it increases the path count.

Test Cases

Now let’s test the function we have written. By applying various test cases, we will validate the algorithm’s accuracy.

fun main() {
    // Sample test case
    val root = TreeNode(5).apply {
        left = TreeNode(4).apply {
            left = TreeNode(11).apply {
                left = TreeNode(7)
                right = TreeNode(2)
            }
        }
        right = TreeNode(8).apply {
            left = TreeNode(13)
            right = TreeNode(4).apply {
                right = TreeNode(1)
            }
        }
    }
    
    val sum = 22
    val result = pathSum(root, sum)
    println("Number of paths: $result") // Output: Number of paths: 3
}

Conclusion

In this article, we solved an algorithm problem using binary trees. We learned how to calculate the sum of each path using DFS and explored specific code writing methods in Kotlin. Since binary trees are widely used as a fundamental data structure to solve various problems, it is beneficial to apply them to multiple problem scenarios.

That concludes this tutorial. Let’s continue to improve our skills by solving more algorithm problems together!

Hello! In this course, we will explore the binary search algorithm, which commonly appears in job-related algorithm tests, and I will explain in detail how to solve problems using Kotlin. Binary search is an efficient algorithm for finding a specific value in a sorted array, with a time complexity of O(log n). In this course, we will solve problems using binary search and cover the problem-solving process in detail.

Problem Description

Problem: Given a sorted array and a specific value, find the index of that value. 
If the value does not exist in the array, it should return -1.

Input:
- arr: a sorted array of integers in ascending order
- target: the integer to be searched for

Output:
- The index of target (or -1 if it does not exist)

Input Example

arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
target = 5

Output Example

4

Overview of the Binary Search Algorithm

The binary search algorithm reduces the search range by splitting it into two branches based on conditions. The most common implementation of binary search includes the following steps.

1. Define the starting index and the ending index of the array.
2. Calculate the middle index.
3. Compare the middle value with the target value.
4. If the middle value is greater than the target, search the left half; if it is smaller, search the right half.
5. Repeat until the value is found or the search range becomes empty.

Kotlin Implementation

Now let’s implement binary search using Kotlin.

fun binarySearch(arr: IntArray, target: Int): Int {
    var left = 0
    var right = arr.size - 1

    while (left <= right) {
        val mid = left + (right - left) / 2

        if (arr[mid] == target) {
            return mid // return index if found
        } else if (arr[mid] < target) {
            left = mid + 1 // search right
        } else {
            right = mid - 1 // search left
        }
    }
    
    return -1 // return -1 if not found
}

Problem Solving Process

Now let’s apply the binary search algorithm using the example provided in the problem.

Example Analysis

arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
target = 5

The starting index for the binary search is 0, and the ending index is 8. Let’s calculate the middle index:

mid = left + (right - left) / 2
   = 0 + (8 - 0) / 2
   = 4

The value of arr[4] is 5. Since it matches the target value, we return index 4.

Performance Analysis

The time complexity of binary search is O(log n). This is because the search range is halved with each comparison; for example, if there are 8 elements in the array, it can be found in 3 comparisons. Binary search is very efficient for searching through a large amount of data.

Conclusion

In this course, we learned the concept of the binary search algorithm and how to implement it in Kotlin. This algorithm is useful for effectively finding specific values in a sorted array. When solving algorithm problems, it is important to understand the problem and choose the appropriate algorithm.

Practice solving more problems and improving your algorithm skills!

Hello! In this article, we will explore algorithm problem-solving methods using Kotlin. Specifically, I will explain the Euclidean algorithm in detail and describe the process of solving a specific problem using it. The Euclidean algorithm is an efficient way to find the greatest common divisor (GCD) of two numbers and is utilized in many algorithm problems.

1. What is the Euclidean Algorithm?

The Euclidean algorithm is a method proposed by the ancient Greek mathematician Euclid for finding the greatest common divisor of two integers. For two integers a and b, the GCD is defined as follows:

GCD(a, b) = GCD(b, a % b), and GCD(a, 0) = a.

In other words, we repeat the process of swapping a with b and b with a % b until the remainder when dividing a by b becomes 0. Ultimately, the value of b will be the GCD.

2. Algorithm Problem

3. Problem-Solving Process

3.1 Problem Analysis

To solve the problem, we need to understand how to find the greatest common divisor of the two input values A and B. Since we can easily find the GCD using the Euclidean algorithm, we will use this method.

3.2 Algorithm Design

We will utilize the Euclidean algorithm for this problem. The algorithm proceeds as follows:

1. Receive the input values.
2. Repeat until A is not 0.
3. While possible, rearrange A and B and calculate the remainder.
4. When the value of B becomes 0, print the value of A.

3.3 Kotlin Code

fun gcd(a: Int, b: Int): Int {
        var x = a
        var y = b
        while (y != 0) {
            val temp = y
            y = x % y
            x = temp
        }
        return x
    }

    fun main() {
        val input = readLine()!!
        val parts = input.split(" ")
        val A = parts[0].toInt()
        val B = parts[1].toInt()
        println(gcd(A, B))
    }

3.4 Code Execution

Example input:
48 18

When the code is executed, the following output appears:

6

4. Complexity Analysis

The time complexity of the Euclidean algorithm is O(log(min(A, B))), making it a very efficient algorithm. Therefore, it maintains a sufficiently fast processing speed even as input values increase.

5. Conclusion

In this post, we discussed the problem of finding the greatest common divisor using the Euclidean algorithm. It can be implemented simply in Kotlin, and this algorithm can be applied to various problems. Understanding and applying such techniques is very important when solving algorithm problems. In the next post, we will cover more complex algorithm problems.

Problem Description

This is a problem to determine whether the given graph is a bipartite graph. A bipartite graph is a graph that can be divided into two sets of vertices such that there are no edges between vertices of the same set.
In other words, if two vertices u and v are connected, u and v must belong to different sets.

Input

• Integer n (1 ≤ n ≤ 1000): number of vertices
• Integer m (1 ≤ m ≤ 5000): number of edges
• m pairs (u, v): edge information between vertex u and vertex v

Output

Print YES if the graph is a bipartite graph, otherwise print NO.

Problem Solving Process

To determine if a graph is bipartite, we can use DFS (Depth First Search) or BFS (Breadth First Search) algorithms. We proceed by coloring the vertices of the graph with two colors, and if adjacent vertices are colored the same, it is not a bipartite graph.

Step 1: Graph Representation

The graph is represented using an adjacency list. We use ArrayList to store the connectivity between vertices.

Step 2: Graph Traversal

While traversing the graph using DFS or BFS, we determine the color of each vertex and check its adjacent vertices to determine if it is a bipartite graph.

Step 3: Implementation

Below is the implementation in Kotlin.

fun isBipartiteGraph(n: Int, edges: List>): String {
    val graph = Array(n + 1) { mutableListOf() }
    for (edge in edges) {
        graph[edge.first].add(edge.second)
        graph[edge.second].add(edge.first)
    }

    val color = IntArray(n + 1) { -1 } // -1: unvisited, 0: set0, 1: set1

    fun bfs(start: Int): Boolean {
        val queue: Queue = LinkedList()
        queue.add(start)
        color[start] = 0 // Coloring the starting vertex with set0

        while (queue.isNotEmpty()) {
            val node = queue.poll()
            for (neighbor in graph[node]) {
                if (color[neighbor] == -1) { // Unvisited vertex
                    color[neighbor] = 1 - color[node] // Color with the opposite color of the current vertex
                    queue.add(neighbor)
                } else if (color[neighbor] == color[node]) {
                    return false // If adjacent vertices have the same color
                }
            }
        }
        return true
    }

    for (i in 1..n) {
        if (color[i] == -1) { // If unvisited vertex, start BFS
            if (!bfs(i)) {
                return "NO"
            }
        }
    }
    return "YES"
}

// Example usage
fun main() {
    val n = 5 // Number of vertices
    val edges = listOf(Pair(1, 2), Pair(1, 3), Pair(2, 4), Pair(3, 5)) // Edge information
    println(isBipartiteGraph(n, edges)) // Output: YES
}

Step 4: Results and Validation

By running the above code, you can determine whether the graph is a bipartite graph. Test various graphs to validate that it produces the correct results.

Conclusion

Through this course, we understood the concept of bipartite graphs and how to determine them, and learned how to effectively solve this problem using Kotlin programming.

Hello! In this article, we will explore the Union-Find algorithm using Kotlin. The Union-Find is a method for efficiently managing a set data structure, primarily used to find and combine connected components. This algorithm is an important concept frequently employed in problems related to graph theory.

Problem Description

Here is a problem that can be solved using Union-Find. We will understand the significance of Union-Find through the problem of counting the number of connected components.

Problem: Counting Connected Components

You have a graph with N nodes and M edges. Each node is numbered from 1 to N. Given M edges as follows, write a program to count the number of connected components.

Input Format

N M
u1 v1
u2 v2
...
uM vM
    

Where N is the number of nodes, M is the number of edges, and u_i and v_i represent the endpoints of the i-th edge.

Output Format

Print the number of connected components in a single line.

Example

Input:
5 3
1 2
2 3
4 5

Output:
2
    

By examining the graph given in the example, nodes 1, 2, and 3 form one connected component, while nodes 4 and 5 form another, resulting in a total of 2 connected components.

Introduction to the Union-Find Algorithm

The Union-Find (Union-Find) or Disjoint Set Union data structure supports two main operations:

• Find: Finds the representative element of the set to which a specific element belongs.
• Union: Combines two sets into one.

Union-Find is primarily used to manage the connectivity of disconnected graphs, and it employs path compression and union by rank techniques for efficiency. These techniques help in quickly performing the ‘Find’ operation by maintaining a tree structure after merging.

Implementing in Kotlin

Now, let’s implement Union-Find in Kotlin. Below is the code to solve the above problem.

class UnionFind(val size: Int) {
    private val parent = IntArray(size + 1) { it } // Initialize each node's parent to itself
    private val rank = IntArray(size + 1) { 1 } // Initialize the rank of each set

    // Find operation: Path compression technique
    fun find(node: Int): Int {
        if (parent[node] != node) {
            parent[node] = find(parent[node]) // Recursively find the parent while compressing the path
        }
        return parent[node]
    }

    // Union operation: Merging sets based on rank
    fun union(node1: Int, node2: Int) {
        val root1 = find(node1)
        val root2 = find(node2)

        if (root1 != root2) {
            if (rank[root1] > rank[root2]) {
                parent[root2] = root1
            } else if (rank[root1] < rank[root2]) {
                parent[root1] = root2
            } else {
                parent[root2] = root1
                rank[root1]++
            }
        }
    }

    // Counting the number of connected components
    fun countComponents(): Int {
        val rootSet = mutableSetOf()
        for (i in 1..size) {
            rootSet.add(find(i)) // Find the root of each node and add to the set
        }
        return rootSet.size // The number of unique roots is the number of connected components
    }
}

fun main() {
    val reader = System.`in`.bufferedReader()
    val (n, m) = reader.readLine().split(" ").map { it.toInt() }
    val unionFind = UnionFind(n)

    for (i in 0 until m) {
        val (u, v) = reader.readLine().split(" ").map { it.toInt() }
        unionFind.union(u, v) // Combine the two nodes based on edge information
    }

    println(unionFind.countComponents()) // Output the number of connected components
}
    

Code Explanation

The code above implements the Union-Find data structure. To store each node’s parent, it uses the parent array, and to store the rank (height) of each set, it uses the rank array.

1. The Find method recursively finds the root of the node through path compression. In this process, all nodes directly point to the root, speeding up future searches.

2. The Union method finds the roots of two nodes and connects the lower rank set to the higher rank set. If their ranks are equal, it connects one set to the root of the other and increases the rank.

3. The countComponents method returns the number of connected components by finding the root for all nodes and counting the unique roots.

Time Complexity Analysis

Each operation of Union-Find is very fast and can be performed in nearly constant time. The average time complexity of this algorithm is as follows:

• Find: O(α(N))
• Union: O(α(N))

α(N) is the inverse of the Ackermann function, which grows extremely slowly. Therefore, Union-Find operates efficiently even with large datasets.

Conclusion

In this article, we learned about the basic concepts and applications of the Union-Find algorithm, as well as how to solve problems involving connected graphs. Implementing it in Kotlin not only allowed us to understand the theory but also provided practical experience, which will be beneficial for preparing for coding tests.

Since the Union-Find algorithm is an important algorithm that can be used in various fields, it is highly recommended that you familiarize yourself with it. We will also have upcoming courses on other algorithms and data structures, so please stay tuned!