Java Coding Test Course, Representation of Graphs

Hello! Today we will take a closer look at graph representation which is frequently asked in coding tests using Java, and solve related algorithm problems. Graphs are a crucial data structure for solving given problems, consisting of vertices and edges. There are various ways to represent graphs, and in this article, we will explain the Adjacency List and Adjacency Matrix methods.

Graph Representation Methods

1. Adjacency List

An adjacency list is a way of storing a list of adjacent vertices for each vertex. It is usually implemented using arrays or lists. This method is very memory efficient and is useful in operations that proceed relatively slowly.

class Graph {
    private int vertices; // number of vertices
    private LinkedList[] adjList; // adjacency list
    
    // graph constructor
    public Graph(int v) {
        vertices = v;
        adjList = new LinkedList[v];
        for (int i = 0; i < v; i++) {
            adjList[i] = new LinkedList<>();
        }
    }

    // method to add an edge
    public void addEdge(int source, int destination) {
        adjList[source].add(destination);
        // If it's an undirected graph, you should also add the next line
        // adjList[destination].add(source);
    }
}

2. Adjacency Matrix

An adjacency matrix is a way of representing a graph using a 2D array. This method allows you to check the existence of an edge in O(1) time complexity, but it has high space complexity and can be inefficient for graphs of variable sizes.

class Graph {
    private int vertices;
    private int[][] adjMatrix;

    // graph constructor
    public Graph(int v) {
        vertices = v;
        adjMatrix = new int[v][v];
    }

    // method to add an edge
    public void addEdge(int source, int destination) {
        adjMatrix[source][destination] = 1;
        // For undirected graphs
        // adjMatrix[destination][source] = 1;
    }
}

Problem: Building Bridges

Now let’s solve an actual problem. The given problem is Building Bridges.

Problem Description

Given N points on a 2D plane, we want to construct bridges so that every point can be connected to each other.
The length of the bridge is the Euclidean distance between two points.
Find the length of the shortest path. Note that two points cannot be connected again during the bridge construction process.

Input

  • The first line contains the number of points N (1 ≤ N ≤ 100).
  • The second line contains the coordinates of each point. (x1, y1), (x2, y2), …, (xN, yN)

Output

Print the length of the shortest path. (Up to 2 decimal places)

Problem Solving Approach

To solve the problem, I will approach it in the following order:

  1. Receive coordinates and store the points.
  2. Calculate the distances between the coordinates to find the bridge lengths.
  3. Use Dijkstra’s algorithm to find the shortest path.
  4. Print the result up to the second decimal place.

Java Code Implementation

import java.util.*;

class Point {
    int x, y;

    Point(int x, int y) {
        this.x = x;
        this.y = y;
    }

    double distance(Point p) {
        return Math.sqrt(Math.pow(this.x - p.x, 2) + Math.pow(this.y - p.y, 2));
    }
}

class Graph {
    private int vertices;
    private List points;
    private double[][] adjMatrix;

    public Graph(int v) {
        vertices = v;
        points = new ArrayList<>();
        adjMatrix = new double[v][v];
    }

    public void addPoint(Point p) {
        points.add(p);
        int idx = points.size() - 1;
        for (int i = 0; i < idx; i++) {
            double dist = points.get(i).distance(p);
            adjMatrix[i][idx] = dist;
            adjMatrix[idx][i] = dist; // undirected edge
        }
    }

    public double dijkstra() {
        double[] dist = new double[vertices];
        boolean[] visited = new boolean[vertices];

        Arrays.fill(dist, Double.MAX_VALUE);
        dist[0] = 0; // starting point

        for (int i = 0; i < vertices - 1; i++) {
            int minIndex = findMinIndex(dist, visited);
            visited[minIndex] = true;

            for (int j = 0; j < vertices; j++) {
                if (!visited[j] && adjMatrix[minIndex][j] != 0 &&
                    dist[minIndex] + adjMatrix[minIndex][j] < dist[j]) {
                    dist[j] = dist[minIndex] + adjMatrix[minIndex][j];
                }
            }
        }

        return dist[vertices - 1]; // distance to the endpoint
    }

    private int findMinIndex(double[] dist, boolean[] visited) {
        double min = Double.MAX_VALUE;
        int minIndex = -1;
        for (int i = 0; i < vertices; i++) {
            if (!visited[i] && dist[i] < min) {
                min = dist[i];
                minIndex = i;
            }
        }
        return minIndex;
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        Graph graph = new Graph(n);

        for (int i = 0; i < n; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            graph.addPoint(new Point(x, y));
        }

        double result = graph.dijkstra();
        System.out.printf("%.2f\n", result);
    }
}

Conclusion

Today, we explored graph representation methods and solved an algorithm problem.
Through this problem, we learned how to handle graphs and how to compute distances between points, as well as implement the shortest path algorithm using adjacency lists and adjacency matrices.
Graphs are a common concept in real life and are very useful in solving complex problems.

Next time, we will look into topics related to search algorithms. Thank you!

Java Coding Test Course, Finding Range Sum 3

Hello! Today, we will solve a Java coding test problem titled ‘Finding the Interval Sum 3’. The goal of this problem is to devise an algorithm that efficiently computes the sum of a specific interval in a given array. Let’s take a look at the problem.

Problem Description

Given an integer array A and m queries, each query is represented by two integers X and Y, and the task is to output the value of A[X] + A[X+1] + ... + A[Y]. Note that the queries are 1-indexed.

Input Format

  • The first line contains two integers N (the number of elements in the array) and M (the number of queries).
  • The second line contains an array A consisting of N integers.
  • From the third line onward, each of the M lines contains two integers X and Y representing each query.

Output Format

Print the result of each query on a new line.

Example

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

Output Example:
6
9
15

Approach to the Problem

To find the sum of a specific interval in a 1-indexed array, we can use the following method:

  1. First, create a sum array: Pre-calculate the sum of the original array and create an array that stores the sum at each index. Using this sum array allows us to calculate the interval sum in O(1) time complexity.
  2. Calculate the interval sum: The total sum of the interval can be computed as sum[Y] - sum[X-1]. Here, sum[i] is the sum from 1 to i in the input array.

Algorithm Implementation

Now, let’s implement the algorithm in Java. Below is the actual code:

import java.util.Scanner;

public class IntervalSum {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // Get the size of the array N and number of queries M
        int N = scanner.nextInt();
        int M = scanner.nextInt();

        // Declare array A and input values
        long[] A = new long[N + 1];
        for (int i = 1; i <= N; i++) {
            A[i] = scanner.nextLong();
        }

        // Declare sum array
        long[] sum = new long[N + 1];

        // Create sum array
        for (int i = 1; i <= N; i++) {
            sum[i] = sum[i - 1] + A[i];
        }

        // Process queries
        for (int j = 0; j < M; j++) {
            int X = scanner.nextInt();
            int Y = scanner.nextInt();
            // Calculate and output the interval sum
            System.out.println(sum[Y] - sum[X - 1]);
        }

        // Close the scanner
        scanner.close();
    }
}

Code Explanation

The above code is structured as follows:

  1. It uses Scanner to receive input. It reads the size of the array (N) and the number of queries (M).
  2. It creates the array A and assigns values in a 1-indexed manner.
  3. It generates the sum array sum. This array stores the sum from 1 to i of the array A.
  4. It reads each query’s X and Y, calculates the interval sum, and prints the result.

Time Complexity Analysis

The time complexity of this algorithm is as follows:

  • It takes O(N) time to create the sum array.
  • It takes O(1) time to process each query. Therefore, for M queries, it takes O(M) time.

In conclusion, the overall time complexity of the algorithm is O(N + M). This is an efficient way to solve the problem.

Conclusion

Today, we learned how to solve an algorithm problem in Java through the ‘Finding the Interval Sum 3’ problem. I hope you understand how to enhance query efficiency by utilizing the sum array. Try more practice problems for additional experience. We will meet again in the next lesson on a different topic. Thank you!

Java Coding Test Course, Calculating Prefix Sums 2

Author: [Your Name]

Creation Date: [Creation Date]

1. Problem Description

In this session, we will discuss the interval sum problem. This is one of the problems you often encounter in algorithms and coding tests, where we will learn how to efficiently calculate the sum of a specific range in a given array.

The problem is as follows:

Given an array A consisting of integers and various queries, find the sum from A[l] to A[r] for each range l and r specified in the queries.

Input format:

  1. Size N of the array A (1 ≤ N ≤ 100,000)
  2. Elements of the array A (1 ≤ A[i] ≤ 100,000)
  3. Number of queries Q (1 ≤ Q ≤ 100,000)
  4. Each query contains two integers l and r (1 ≤ l ≤ r ≤ N).

Output format:

Print the interval sum for each query.

2. Approach

The approach to the problem can be divided into two methods:

  1. A straightforward method using loops to calculate the sum (inefficient)
  2. A method using a prefix sum array to calculate the interval sum in O(1) time (efficient)

In the first approach, if we directly calculate the sum for each query, the time complexity becomes O(N * Q), as it is the product of the number of queries Q and the size of the array N. In this case, in the worst case, it requires 1010 operations, which is not efficient.

The second approach involves creating a prefix sum array to calculate the interval sum in O(1) time. The overall approach of this method is as follows:

  1. First, create a prefix sum array of size N.
  2. The i-th index of the prefix sum array stores the sum from A[0] to A[i].
  3. The sum of each query (l, r) can be calculated using the prefix sum array as S[r] – S[l-1].

3. Code Implementation

      public class IntervalSum {
        public static void main(String[] args) {
            int N = 5; // Size of array
            int[] A = {1, 2, 3, 4, 5}; // Input values
            int Q = 3; // Number of queries
            int[][] queries = {{1, 3}, {2, 5}, {1, 5}}; // Sample queries
            
            // Create prefix sum array
            long[] prefixSum = new long[N + 1];
            for (int i = 1; i <= N; i++) {
                prefixSum[i] = prefixSum[i - 1] + A[i - 1];
            }
            
            // Process each query
            for (int i = 0; i < Q; i++) {
                int l = queries[i][0];
                int r = queries[i][1];
                long sum = prefixSum[r] - prefixSum[l - 1];
                System.out.println("Sum from " + l + " to " + r + ": " + sum);
            }
        }
      }
    

4. Code Explanation

The above code operates as follows:

  1. It receives the array A, the number of queries Q, and l and r for each query.
  2. First, it creates the prefix sum array. The i-th index of this array stores the sum from the 0-th index to the (i-1)-th index of array A.
  3. When processing each query, the corresponding interval sum is calculated and output as prefixSum[r] – prefixSum[l-1].

5. Time Complexity Analysis

The time complexity of this algorithm is as follows:

  1. It takes O(N) time to create the prefix sum array.
  2. It takes O(1) time to process each query.

Therefore, the total time complexity is O(N + Q). This is a very efficient method that performs well even when both the size of the array and the number of queries are at their maximum values.

6. Final Summary

The interval sum problem is an important concept in understanding algorithms and data structures. In this session, we learned an efficient way to calculate interval sums using the prefix sum array. This method can significantly maximize performance when processing large amounts of data.

Such types of problems may also appear in various modified forms, making it important to practice further. I hope you continue to solve various algorithm problems to deepen and broaden your understanding.

Likes and subscriptions are a great help! Please look forward to more algorithm lectures.

Java Coding Test Course, Calculating Interval Sum 1

Hello! Today, I would like to take an in-depth look at one of the algorithms frequently featured in Java coding tests, which is calculating the range sum. In particular, this course will select a problem related to computing the range sum and explain the solution process in detail.

Problem Definition

Given an integer array arr, several queries are provided. Each query includes two integers start and end, and the task is to calculate arr[start] + arr[start + 1] + ... + arr[end]. What is required in this problem is to efficiently compute the range sum for each query.

Example Problem

    Input:
    arr = [1, 2, 3, 4, 5]
    queries = [[0, 2], [1, 3], [2, 4]]
    
    Output:
    [6, 9, 12]
    

Solution Algorithm

To solve this problem, we must efficiently handle multiple range sums on the array. Essentially, repeating the sum of parts of the array for each query may be inefficient. To reduce this inefficiency, I will introduce an approach using the “prefix sum array.”

Explanation of Prefix Sum Array

A prefix sum array is a method of preprocessing the elements of a given array by accumulating them. This allows for the calculation of each range sum in O(1) time complexity. The definition of the prefix sum array is as follows:

    prefix[i] = arr[0] + arr[1] + ... + arr[i]
    

In other words, the sum of the range arr[i] ~ arr[j] can be computed as prefixSum[j] - prefixSum[i-1]. Here, prefixSum[-1] is assumed to be 0.

Implementation Steps

  1. Create the Prefix Sum Array: Generate the prefix sum array using the given array.
  2. Process Queries: Calculate the range sum for each query with O(1) complexity.

Java Code Implementation

    public class RangeSum {
        public static void main(String[] args) {
            int[] arr = {1, 2, 3, 4, 5};
            int[][] queries = {{0, 2}, {1, 3}, {2, 4}};
            int[] result = rangeSum(arr, queries);
            
            for (int sum : result) {
                System.out.println(sum);
            }
        }

        public static int[] rangeSum(int[] arr, int[][] queries) {
            int n = arr.length;
            int[] prefixSum = new int[n];
            prefixSum[0] = arr[0];

            // Create the cumulative sum array
            for (int i = 1; i < n; i++) {
                prefixSum[i] = prefixSum[i - 1] + arr[i];
            }

            int[] results = new int[queries.length];
            for (int i = 0; i < queries.length; i++) {
                int start = queries[i][0];
                int end = queries[i][1];

                // Calculate the range sum
                results[i] = (start > 0) ? (prefixSum[end] - prefixSum[start - 1]) : prefixSum[end];
            }

            return results;
        }
    }
    

Code Explanation

The Java code above creates a user-defined class RangeSum and defines the array and queries in the main method. The rangeSum method generates the prefix sum array from the given array and calculates the range sum for each query.

1. Creating the Cumulative Sum Array

First, we initialize the first element, then create the cumulative sum array through a loop. This process has a time complexity of O(n), where n is the length of the array.

2. Processing Queries

For each query, we use the cumulative sum array to compute the range sum in O(1). This method should perform well in terms of efficiency, especially when there is a large amount of data or many queries.

Complexity Analysis

The time complexity of this code is O(n + m), where n is the length of the array and m is the number of queries. Thus, it consumes O(n) time complexity during the initialization process, and O(1) for each query, making the total time complexity O(n + m). This ensures satisfactory performance.

Conclusion

Today, we dealt with the problem of calculating range sums in Java coding tests. We learned that utilizing a cumulative sum array allows for efficient calculations of range sums for multiple queries. This is a useful algorithm to remember for actual coding tests! I will prepare more algorithm problem-solving courses in the future. Thank you!

Java Coding Test Course, Range Sum

Hello! In this post, we will discuss the range sum problem for coding test preparation using Java. The range sum problem is one where we efficiently calculate the sum of elements in a specific range of a given array, and it is a common type of question in algorithm problems.

Problem Description

Given an array A, there are Q queries. Each query consists of two integers L and R, which represent the indices of the array A. For each query, compute the range sum from index L to index R.

Input

    The first line contains the size of the array N and the number of queries Q.
    The second line contains N integers A[1], A[2], ..., A[N].
    Following that, Q lines are given with L and R separated by space.
    

Output

    Print the range sum for each query, one per line.
    

Example Input

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

Example Output

    6
    9
    15
    

Problem Solving Strategy

There are various ways to handle the range sum problem, but an inefficient way is to directly loop through and calculate the range sum for each query. In this case, the worst-case time complexity could be O(N * Q). Therefore, we need to find a method to calculate the range sum quickly.

Preprocessing Approach

One efficient method is to store the range sums in a cumulative array through preprocessing. This allows us to calculate the range sum for each query in O(1).

  1. First, create a cumulative sum array sum. Initialize it with sum[0] = 0, and set sum[i] = sum[i - 1] + A[i - 1].
  2. For each query, to find the sum in the range [L, R], use the formula sum[R] - sum[L - 1].

Java Code Implementation

    import java.util.Scanner;

    public class RangeSum {
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            
            // Input the size of the array and the number of queries
            int N = sc.nextInt();
            int Q = sc.nextInt();
            
            // Input the array
            int[] A = new int[N];
            for (int i = 0; i < N; i++) {
                A[i] = sc.nextInt();
            }
            
            // Create cumulative sum array
            long[] sum = new long[N + 1];
            for (int i = 1; i <= N; i++) {
                sum[i] = sum[i - 1] + A[i - 1];
            }
            
            // Process each query
            for (int i = 0; i < Q; i++) {
                int L = sc.nextInt();
                int R = sc.nextInt();
                long rangeSum = sum[R] - sum[L - 1];
                System.out.println(rangeSum);
            }
            
            sc.close();
        }
    }
    

Code Explanation

The code above operates in the following manner:

  1. First, it reads the size of the array N and the number of queries Q from the user.
  2. Then, it initializes the elements of the array A.
  3. A sum array is created to store the cumulative sums of each element, initializing index 0 to 0 to facilitate easy access to sum[i].
  4. For each query, it calculates the range sum using the provided L and R and outputs the result.

Testing and Validation

After writing the code, it is essential to verify that each query returns the correct result through various inputs. In addition to the example input, we should experiment with additional cases and review the results. Here are a few examples:

  • Input: 1 1 → Output: 1
  • Input: 2 5 → Output: 14
  • Input: 3 3 → Output: 3

Conclusion

In this post, we explored the process of solving the range sum problem. It is important to learn how to efficiently calculate range sums through preprocessing, and practicing with arrays and loops in Java will help develop good coding skills. I hope to tackle more algorithm problems in the future to enhance your technical abilities.

Thank you!