Author: root

Hello! Today, we will take a detailed look at how to represent graphs, which frequently appear in coding tests, using Kotlin. Graphs are an important data structure in various fields and have essential fundamental functions for solving a range of problems. In this article, we will explain how to represent a graph and how to use that representation to solve problems.

What is a Graph?

A graph is a data structure composed of nodes (vertices) and edges. A node represents an object, and an edge represents the relationship between nodes. Graphs can be classified into directed and undirected graphs based on directionality, and into weighted and unweighted graphs based on the presence of weights.

Ways to Represent a Graph

Graphs can be represented in various ways, with the two most common methods being the Adjacency Matrix and the Adjacency List.

1. Adjacency Matrix

An adjacency matrix is a method of representing a graph using a 2D array of size N x N. N is the number of nodes in the graph, and each element of the matrix indicates whether nodes are connected. For example, if node A is connected to node B, then matrix[A][B] will be 1.

    fun createAdjacencyMatrix(vertices: Int, edges: List>): Array> {
        val matrix = Array(vertices) { Array(vertices) { 0 } }
        for (edge in edges) {
            val (from, to) = edge
            matrix[from][to] = 1
            matrix[to][from] = 1  // For undirected graphs
        }
        return matrix
    }

2. Adjacency List

An adjacency list is a method that uses a list to store connected nodes for each node. This method is memory efficient and advantageous for graphs with fewer nodes or edges.

    fun createAdjacencyList(vertices: Int, edges: List>): List> {
        val list = MutableList(vertices) { mutableListOf() }
        for (edge in edges) {
            val (from, to) = edge
            list[from].add(to)
            list[to].add(from)  // For undirected graphs
        }
        return list
    }

Problem: Friend Relationship Network

The following is a problem that represents a graph of friendships.

Problem: Given N people and their friendships, write a program to check whether a friendship exists between two people. Friend relationships are represented as undirected graphs. The numbers of the two people are given sequentially from 0 to N-1.

Problem-Solving Process

1. Understand the Input Format

First, we need to understand the given data. The input is as follows:

• The first line contains the number of people N.
• The second line contains the number of friendships M.
• Subsequently, M lines contain two integers a and b representing each friendship.

2. Choose a Data Structure

To store the given friendships, we will use an adjacency list. This allows for easy exploration of friendships.

3. Create the Graph

    fun main() {
        val reader = BufferedReader(InputStreamReader(System.`in`))
        val n = reader.readLine().toInt()
        val m = reader.readLine().toInt()

        val edges = mutableListOf>()
        for (i in 0 until m) {
            val (a, b) = reader.readLine().split(" ").map { it.toInt() }
            edges.add(Pair(a, b))
        }

        val adjacencyList = createAdjacencyList(n, edges)
    }

4. Check Friendship

The method to check whether two specific people are friends is to use DFS (Depth First Search) or BFS (Breadth First Search) to verify connectivity. Here, we will implement it using DFS.

    fun isConnected(adjList: List>, start: Int, target: Int, visited: BooleanArray): Boolean {
        if (start == target) return true
        visited[start] = true
        
        for (neighbor in adjList[start]) {
            if (!visited[neighbor]) {
                if (isConnected(adjList, neighbor, target, visited)) {
                    return true
                }
            }
        }
        return false
    }

    fun main() {
        // Continuing from the previous code
        val query = reader.readLine().split(" ").map { it.toInt() }
        val (x, y) = query

        val visited = BooleanArray(n) { false }
        val result = isConnected(adjacencyList, x, y, visited)
        println(if (result) "YES" else "NO")
    }

Conclusion

In this lesson, we learned about the basic concepts and methods of representing graphs, and how to solve the friend relationship network problem using these representations. Graphs are essential for solving various problems, and these graph representation and traversal techniques are very useful in algorithm tests.

Having implemented examples using Kotlin, I encourage you to practice by solving various graph problems yourself. Next time, we will discuss the differences between BFS and DFS, as well as various problem-solving strategies utilizing them. I hope this aids you in your learning journey!

Hello! Today, we will discuss the coding test problem “Finding the Sum of Intervals 3” using Kotlin. This problem requires efficiently calculating the sum of a specific interval of a given array. Such efficient algorithms are particularly important when needing to calculate interval sums repeatedly.

Problem Description

You are given an array A consisting of N integers and Q queries. Each query contains two integers L and R, which require calculating the sum from the L-th index to the R-th index of array A. Note that the array starts from index 1. You are to output the result for all Q queries.

Input Format

    First line: N (1 ≤ N ≤ 100,000)
    Second line: N integers (each integer is -10,000,000 ≤ A[i] ≤ 10,000,000)
    Third line: Q (1 ≤ Q ≤ 100,000)
    The following Q lines contain each query in the form of L R.
    

Output Format

    Output the interval sum for each of the Q lines.
    

Approach

The basic method is to directly traverse the array to calculate the sum for the given L and R. However, this has a time complexity of O(Q * N), which could involve up to 1 billion operations in the worst case when both Q and N reach their maximum of 100,000. Therefore, this approach is inefficient. This problem requires a method to handle queries with a time complexity of O(1) through preprocessing.

One of the efficient methods to solve the interval sum problem is to use a cumulative sum array. By using the cumulative sum array, the sum for a specific interval L ~ R can be easily calculated.

Cumulative Sum Array

Define the cumulative sum array P. P[i] stores the sum from the first element of array A to the i-th element.

    P[i] = A[1] + A[2] + ... + A[i]
    

The sum for the interval L ~ R can then be calculated as follows:

    sum(L, R) = P[R] - P[L - 1]
    

This method allows us to compute the cumulative sum array in O(N), and subsequently obtain results for each query in O(1). The overall time complexity becomes O(N + Q).

Code Implementation

Kotlin Code

    fun main() {
        val reader = System.`in`.bufferedReader()
        val n = reader.readLine().toInt()
        val a = reader.readLine().split(" ").map { it.toInt() }
        val q = reader.readLine().toInt()
        
        // Initialize cumulative sum array
        val p = LongArray(n + 1)
        
        // Calculate cumulative sum
        for (i in 1..n) {
            p[i] = p[i - 1] + a[i - 1]
        }

        val result = StringBuilder()
        for (i in 0 until q) {
            val (l, r) = reader.readLine().split(" ").map { it.toInt() }
            // Calculate interval sum
            result.append(p[r] - p[l - 1]).append("\n")
        }

        // Print results
        println(result.toString())
    }
    

Code Explanation

The above code is a solution to the “Finding the Sum of Intervals 3” problem written in Kotlin. Let’s explain each part.

1. Input Handling

First, it receives the values for N, A, and Q. N signifies the size of the array, A is the integer array, and Q represents the number of given queries.

2. Cumulative Sum Array Creation

A cumulative sum array P is created. P[0] is initialized to 0 and the cumulative sums from P[1] to P[n] are calculated. This will store the results of adding each element of the array A.

3. Query Processing

For each query, the L and R values are received, allowing the interval sum to be computed within O(1) time using the cumulative sum array. This result is added to a StringBuilder and printed at the end.

Conclusion

Through the “Finding the Sum of Intervals 3” problem, we learned the importance of using cumulative sum arrays. This technique allows the efficient handling of queries even with large datasets. When solving algorithm problems, it is essential to find the optimal method by considering time complexity. I hope this helps you prepare for your coding tests.

Learn More

In the next lecture, we will cover a variety of algorithm problems. Keep preparing for coding tests and practice a lot. Thank you!

Among the various algorithms dealt with in coding tests, the ‘interval sum’ problem is frequently encountered.
In this post, we will solve a problem titled Calculating Interval Sums 2.
We will particularly utilize the Kotlin programming language for this.

Problem Description

This problem presents a method to efficiently calculate the sum of a specific interval given multiple integers.
Here is the definition of the problem:

Given integers N and M, 
N integers will be provided, 
and M queries will be given. 
Each query consists of two integers i and j, 
and the sum of integers between i and j needs to be calculated.

Input Format:
- The first line will contain N (1 ≤ N ≤ 100,000) and M (1 ≤ M ≤ 100,000).
- The second line contains N integers. Each integer does not exceed 1,000,000.
- Following M lines, i and j will be given. (1 ≤ i ≤ j ≤ N)

Output Format:
- For each query, output the sum from the i-th to the j-th number.

Problem Solving Process

To solve this problem, an efficient algorithm is needed. Simply looping through each query to calculate the sum would result in a time complexity of O(N*M), which may lead to a timeout failure if the input is large.

1. Using Cumulative Sum Array

Instead of calculating all sums each time from a checking perspective, there is the option of using a cumulative sum array. By utilizing the cumulative sum array, the sum of a specific interval can be determined with O(1) time complexity.

Defining the cumulative sum array prefixSum, prefixSum[k] represents the sum from the start of the array to the k-th number.
Therefore, it holds the following relationship:

prefixSum[k] = prefixSum[k-1] + array[k-1] (1 ≤ k ≤ N)

In this case, the sum of a specific interval can be calculated as follows:

sum(i, j) = prefixSum[j] - prefixSum[i-1]

2. Implementing the Algorithm

Now, let’s implement the code based on this idea.
The following code is an example of solving the problem using Kotlin:

fun main() {
    val (n, m) = readLine()!!.split(" ").map { it.toInt() }
    val numbers = readLine()!!.split(" ").map { it.toInt() }.toIntArray()
    val prefixSum = LongArray(n + 1)

    for (i in 1..n) {
        prefixSum[i] = prefixSum[i - 1] + numbers[i - 1].toLong()
    }

    for (j in 0 until m) {
        val (i, j) = readLine()!!.split(" ").map { it.toInt() }
        println(prefixSum[j] - prefixSum[i - 1])
    }
}

3. Code Explanation

Let me explain how the above code calculates the interval sum:

• val (n, m) = readLine()!!.split(" ").map { it.toInt() }:
  Reads N and M from the first line.
• val numbers = readLine()!!.split(" ").map { it.toInt() }.toIntArray():
  Receives N integers from the second line and converts them into an array.
• val prefixSum = LongArray(n + 1):
  Creates an array to store cumulative sums. (The 0th index is initialized to 0)
• for (i in 1..n):
  Uses a loop to calculate the cumulative sums. Saves the sum to prefixSum[i] up to the i-th number.
• for (j in 0 until m):
  Receives input for each query and calculates the interval sum.
• println(prefixSum[j] - prefixSum[i - 1]):
  Outputs the sum from the i-th index to the j-th index.

Complexity Analysis

The time complexity of the above code is as follows:

• Time taken to create the cumulative sum array: O(N)
• Time taken to calculate the interval sum for each of the M queries: O(M)

Therefore, the overall time complexity is O(N + M). This is a good method for effectively handling the interval sums of the given problem.

Conclusion

In this lecture, we efficiently solved the problem of calculating interval sums 2 using cumulative sums.
This approach can be effectively applied to many algorithm problems, so try to apply it to various challenges.

We dealt with the problem of quickly and efficiently calculating the sum of a specific interval from an array consisting of multiple integers.
We have learned that various algorithm problems can be solved using data structures such as arrays and lists in Kotlin.

Additional Practice Problems

Please practice the following additional problems:

• Finding the minimum and maximum value of intervals in various query forms
• Finding unusual interval sums: Summing only intervals that meet specific conditions
• Finding interval sums in multidimensional arrays

I hope this post helps you prepare for your coding tests.
Thank you!

Hello! In this post, we will take a closer look at how to solve the problem of calculating the sum of intervals using Kotlin.
The interval sum problem is a common type encountered in programming contests and coding tests, and knowing how to solve it efficiently can help you quickly solve various problems.

Problem Description

The task is to calculate the sum of a specific interval when given an array. The aim is to solve the problem of
calculating the sum for each interval given the array and several queries. For example, let’s assume the array A is as follows:

        A = [10, 20, 30, 40, 50]
    

In this case, when we receive a query to find the sum in the interval (i, j) specified by the user,
we need to return the corresponding sum.

Input Format

The first line contains the size of the array N and the number of queries M.
Then, N integers are provided to form the array A.
The next M lines each contain two integers i and j,
indicating a query to calculate the sum from A[i] to A[j].

Output Format

Print the sum for each query.

Example Input

    5 3
    10 20 30 40 50
    1 3
    2 4
    1 5
    

Example Output

    60
    90
    150
    

Approach to Solve the Problem

There are two ways to solve this problem.
The first method is to simply iterate and calculate the sum for each query,
and the second is to use the prefix sum array.
The second method is much more efficient, so I will focus on explaining that.

Prefix Sum Array

By using a prefix sum array, we can calculate the interval sum in O(1) time complexity for each query.
Since calculating the prefix sum array takes O(N) time,
the total time complexity is O(N + M).
This becomes more efficient as the number of queries increases.

Definition of Prefix Sum Array

A prefix sum array is an array that stores the cumulative sums of a given array.
Specifically, we define an array S that stores the sum up to the i-th index of array A.
It is represented as S[i] = A[0] + A[1] + ... + A[i-1].

How to Calculate Interval Sums

The sum from A[i] to A[j] can be calculated as follows:
sum = S[j+1] - S[i]

Here, S[j+1] is the sum from A[0] to A[j],
and S[i] is the sum from A[0] to A[i-1].
Therefore, by subtracting S[i] from S[j+1], we obtain the sum from A[i] to A[j].

Code Implementation

Now, let’s write the code to solve this problem using Kotlin.

Kotlin Code

    fun main() {
        val (n, m) = readLine()!!.split(" ").map { it.toInt() }
        val a = readLine()!!.split(" ").map { it.toInt() }
        
        val s = IntArray(n + 1)
        
        for (i in 1..n) {
            s[i] = s[i - 1] + a[i - 1]
        }
        
        repeat(m) {
            val (i, j) = readLine()!!.split(" ").map { it.toInt() }
            println(s[j] - s[i - 1])
        }
    }
    

Code Explanation

1. The first line reads N and M.
2. The second line reads the array A.
3. The prefix sum array S is initialized with a size of N + 1.
4. A loop calculates the values for the prefix sum array S.
5. For each query, we calculate and print the sum for the respective interval.

Conclusion

The interval sum problem is a useful technique that can be applied to various problems.
Using a prefix sum array in Kotlin allows us to solve the problem much faster.
I hope this tutorial helps you enhance your understanding of the interval sum problem and
allows you to effectively utilize it in coding tests.

In the next post, we will explore other types of problems extending the concept of interval sums.
Keep continuing your learning!

1. Introduction

Hello. In this session, we will learn about how to efficiently calculate the interval sum in Kotlin.
This is one of the problems frequently encountered in coding tests, where we need to find the sum over a specific interval of a given sequence.
Through this process, we will discuss various methods to solve the interval sum problem and their respective time complexities.

2. Problem Description

Given a sequence A and a query Q, each query is given by two integers L and R.
We need to quickly find the sum from L to R.

Input

• The first line contains the size of the sequence N and the number of queries M.
• The second line consists of N integers forming the sequence A.
• For the next M lines, L and R are given.

Output

Print the result for each query one line at a time.

Constraints

• 1 ≤ N ≤ 100,000
• 1 ≤ M ≤ 100,000
• -1,000,000,000 ≤ A[i] ≤ 1,000,000,000

3. Approach

There are several ways to calculate the interval sum, but the basic approaches are as follows.

3.1. Simple Iteration

The most intuitive method is to directly calculate the sum in the defined range for each query. However, this method has a time complexity of O(N), and in the worst case, it becomes O(N * M) = O(10^10) when the maximum number of queries is 100,000, making it very inefficient.

3.2. Cumulative Sum Array

By using the following methodology, we can efficiently calculate the interval sum.

A cumulative sum array is an array that pre-calculates the sum up to a specific position, which can be created in linear time O(N).
This allows each query to be processed in O(1). Thus, the overall time complexity is O(N + M).

4. Implementation

Let’s implement the interval sum calculation algorithm using the cumulative sum array in Kotlin.

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

    val array = reader.readLine().split(" ").map { it.toInt() }.toIntArray()
    val sumArray = IntArray(n + 1)

    // Create cumulative sum array
    for (i in 1..n) {
        sumArray[i] = sumArray[i - 1] + array[i - 1]
    }

    val result = StringBuilder()
    // Process queries
    repeat(m) {
        val (l, r) = reader.readLine().split(" ").map { it.toInt() }
        val sum = sumArray[r] - sumArray[l - 1]
        result.append(sum).append("\n")
    }

    println(result)
}

            

The key point in the above implementation is that we can process each query in O(1) through sumArray.
The sum of the interval [L, R] can be easily calculated as sumArray[R] - sumArray[L-1].

5. Time Complexity Analysis

Analyzing the time complexity of this algorithm, we have:

• Creating the cumulative sum array: O(N)
• Processing queries: O(M)
• Total time complexity: O(N + M)

Therefore, under the given constraints, we can efficiently calculate the interval sum.

6. Example

6.1. Input Example

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

            

6.2. Output Example

6
14
15

            

In the above example, the results of the queries given as input are that the sum from 1 to 3 is 6,
the sum from 2 to 5 is 14, and the total sum from 1 to 5 is 15.

7. Conclusion

In this lecture, we explored various ways to calculate the interval sum and its precise implementation.
In particular, we discussed in depth about how to efficiently compute the interval sum using the cumulative sum array.
This algorithm can be effectively used in many coding test problems and practical applications. In the next session, we will explore more complex data structures and algorithms.