Category: Java Coding Test

Hello, everyone! In this post, we will cover a problem that frequently appears in coding tests, Identifying Friend Relationships. The problem we will introduce is a graph traversal problem that presents a task of analyzing friend relationships and calculating the number of friends satisfying certain conditions.

Problem Description

You have hosted a party with N friends. Each friend is connected to others through friendship. Friend relationships are bidirectional, and each friend A and friend B can only check friends among their friends. In other words, a friend of friend A is only a friend directly connected to A. Your mission is to count how many ‘2nd degree friends’ you can find among a specific friend’s friends.

Problem Definition

    Input:
    - N (1 ≤ N ≤ 100) : Number of friends
    - M (1 ≤ M ≤ 10^4) : Number of pairs of friend relationships
    - M pairs (A, B): Two friends A and B.

    Output:
    - X: Number of 2nd degree friends of a specific friend.

    You need to count the 2nd degree friends of a specific friend K.
    

Sample Input

    6 5
    1 2
    2 3
    1 4
    4 5
    5 6
    1
    

Sample Output

    2
    

Approach to the Problem

To solve this problem, we need to use a graph data structure to represent friend relationships. We will create an adjacency list for this purpose. After that, we need to find 2nd degree friends using either breadth-first search (BFS) or depth-first search (DFS). Each friend in the graph can be viewed as a vertex, while friend relationships are edges.

Step-by-Step Approach

1. Read the input values and create an adjacency list based on the connection information of the friend relationships.
2. Use BFS or DFS based on the specific friend K to explore the relationships of friends of friends.
3. During the exploration process, exclude friends directly connected to K, and use a set to prevent duplicates among friends of friends.
4. Finally, output the size of the set.

Implementation Code (Java)

    import java.util.*;

    public class FriendRelations {
        static List[] graph;
        static Set friendsOfFriends;

        public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            int N = scanner.nextInt();
            int M = scanner.nextInt();
            graph = new ArrayList[N + 1];
            for (int i = 1; i <= N; i++) {
                graph[i] = new ArrayList<>();
            }

            for (int i = 0; i < M; i++) {
                int A = scanner.nextInt();
                int B = scanner.nextInt();
                graph[A].add(B);
                graph[B].add(A);
            }

            int K = scanner.nextInt();
            friendsOfFriends = new HashSet<>();

            for (int friend : graph[K]) {
                for (int f2 : graph[friend]) {
                    if (f2 != K && !graph[K].contains(f2)) {
                        friendsOfFriends.add(f2);
                    }
                }
            }
            System.out.println(friendsOfFriends.size());
            scanner.close();
        }
    }
    

Code Explanation

The first part of the code defines the graph to store friend relationships and initializes the relationship of friends in the adjacency list format through input values. It then explores the friends of the specified friend K and adds them to a set to remove duplicates. Finally, it calculates the number of 2nd degree friends of K using the size of the set.

Code Execution

Now let’s execute the code to see the results. Using the above sample input, the outcome of the code will be ‘2’. This means that the specific friend K has a total of 2 second degree friends.

Conclusion

This problem serves as a good exercise in utilizing basic concepts of graph theory. The theory of friend relationships often appears in many coding interviews, and by solving such problems, you can further enhance your knowledge related to graph traversal.

In the next post, we will tackle another useful algorithm problem and its solution process. If you have any questions, please leave a comment!

In this course, we will discuss the problem of “Longest Common Subsequence (LCS)”, which is frequently encountered in coding tests. The LCS problem involves finding the longest common subsequence that can be formed by preserving the relative order of elements in both sequences when given two sequences.

Problem Description

Given two strings str1 and str2, find the length of the longest common subsequence of these two strings.

Input

• String str1 : “AGGTAB”
• String str2 : “GXTXAYB”

Output

Length of the longest common subsequence : 4

Problem Solving Process

To solve this problem, we will use the dynamic programming approach. Dynamic programming is a method for solving problems by breaking them down into smaller subproblems and storing the results to efficiently solve the problem.

Step 1: Initialize the Two-Dimensional Array

Step 2: Fill the Two-Dimensional Array

Step 3: Implement the Code

Now let’s implement this process in Java code.

public class LongestCommonSubsequence {
    public static int lcs(String str1, String str2) {
        int m = str1.length();
        int n = str2.length();
        int[][] dp = new int[m + 1][n + 1];

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[m][n];
    }

    public static void main(String[] args) {
        String str1 = "AGGTAB";
        String str2 = "GXTXAYB";
        System.out.println("Length of the longest common subsequence: " + lcs(str1, str2));
    }
}
    

Code Analysis

The code above operates in the following manner.

• It takes two strings as input and creates a dp array based on their lengths.
• Using a nested loop, it compares the strings and updates values using the dynamic programming method.
• The output provides the length of the longest common subsequence.

Execution Result

Conclusion

In this lecture, we learned how to find the longest common subsequence. We demonstrated how effective it can be to use dynamic programming techniques to solve algorithmic problems. This problem can be applied in various fields and is fundamental knowledge to have when solving algorithm problems.

Furthermore, you can experiment with other string combinations based on the code shared in this lecture, deepening your understanding of the theory. I hope this will greatly assist you in developing your algorithm problem-solving skills.

Problem Definition

You need to find a way to create the minimum value using the given numbers and parentheses. For example, when a string containing numbers and operators is given, the problem involves calculating the possible minimum value by placing parentheses appropriately. This problem aims to optimize an expression that can yield different results depending on the placement of parentheses.

Example Problem

Approach to the Problem

To create the minimum value, you need to try all possible arrangements of parentheses and compare the results. The algorithm that can be used here is “divide and conquer.” The expression is partially divided, and each result is compared to derive the final result.

Step 1: Parsing the Expression

First, parse the input string to separate the numbers and operators. Then, prioritize each operation and calculate the results accordingly.

Step 2: Recursive Calculation

Recursively split the expression to calculate the results of each part. Each operator branches into child nodes, allowing the calculation of values for all combinations.

Step 3: Minimum Value Comparison

After calculating all possible cases, compare them to find the minimum value. During this process, results can be stored to prevent duplicate calculations.

Python Code Example

def minValue(expression):
    import re

    # Separate the expression into numbers and operators
    tokens = re.findall(r'\d+|[+*/-]', expression)
    
    # Function to search all combinations
    def dfs(tokens):
        if len(tokens) == 1:
            return int(tokens[0])
        
        min_result = float('inf')
        
        for i in range(1, len(tokens), 2):
            operator = tokens[i]
            left = tokens[:i]
            right = tokens[i + 1:]

            for left_result in dfs(left):
                for right_result in dfs(right):
                    if operator == '+':
                        result = left_result + right_result
                    elif operator == '-':
                        result = left_result - right_result
                    elif operator == '*':
                        result = left_result * right_result
                    elif operator == '/':
                        result = left_result // right_result
                    
                    min_result = min(min_result, result)
        
        return min_result
    
    return dfs(tokens)

# Example usage
print(minValue("1+2*3-4/2"))

Conclusion

This problem serves as practice in understanding how the placement of parentheses affects results and in finding the minimum value through recursive thinking. The process of optimizing while considering various conditions is a commonly used approach in coding tests. Implement the code and test various cases in practice.

Tips for Coding Test Preparation

• Understand the problem and accurately grasp its requirements.
• Generate various cases through examples.
• Learn and practice recursive approaches.
• Experiment with multiple ideas to find the optimal solution.

Additional Resources

Find and study materials describing various algorithm problems and solutions. By solidifying the basics of algorithms, you will be able to achieve good results in coding tests.

Problem Description

There is a given integer array. You need to find the minimum value within a specific range in this array.

Moreover, this range is dynamically provided and can be requested for multiple queries.

In other words, given specific indices i and j of the array,

you need to find the minimum value between i and j.

The goal of this course is to design and implement an efficient algorithm to solve this problem.

Problem Format

Input: An integer array nums and a list of queries queries.
– nums: An array of n integers (0 ≤ n ≤ 10^5, -10^9 ≤ nums[i] ≤ 10^9)
– queries: A list containing multiple pairs (i, j) (0 ≤ queries.length ≤ 10^5, 0 ≤ i ≤ j < n)
Output: Return a list of minimum values for each query.

Example Problems

Example 1

Input:

            nums = [2, 0, 3, 5, 1]
            queries = [[0, 2], [1, 4], [0, 4]]
            

Output:

            [0, 1, 0]
            

Example 2

Input:

            nums = [1, 2, 3, 4, 5]
            queries = [[0, 0], [0, 4], [2, 3]]
            

Output:

            [1, 1, 3]
            

Solution Strategy

There are several approaches to solving this problem.

The simplest way is to use linear search for each query.

However, this method has a worst-case time complexity of O(m * n),

so a more efficient method is required.

We will utilize data structures such as Segment Tree or Sparse Table

to find the minimum value for each query in O(log n) or O(1) time complexity.

Using Segment Tree

A Segment Tree is a data structure that efficiently handles range queries on a given array.

With this, we can build the Segment Tree in O(n) and process each query in O(log n).

Here is how to construct the Segment Tree and handle queries.

Segment Tree Implementation

            // Segment Tree class
            class SegmentTree {
                private int[] tree;
                private int n;

                public SegmentTree(int[] nums) {
                    n = nums.length;
                    tree = new int[4 * n]; // Sufficiently sized tree array
                    build(nums, 0, 0, n - 1);
                }

                private void build(int[] nums, int node, int start, int end) {
                    if (start == end) {
                        tree[node] = nums[start]; // Store value at leaf node
                    } else {
                        int mid = (start + end) / 2;
                        build(nums, 2 * node + 1, start, mid);
                        build(nums, 2 * node + 2, mid + 1, end);
                        tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]); 
                    }
                }

                public int query(int L, int R) {
                    return query(0, 0, n - 1, L, R);
                }

                private int query(int node, int start, int end, int L, int R) {
                    if (R < start || end < L) { 
                        return Integer.MAX_VALUE; // Return infinity if not in range
                    }
                    if (L <= start && end <= R) { 
                        return tree[node]; // Return node value if in range
                    }
                    int mid = (start + end) / 2;
                    int leftMin = query(2 * node + 1, start, mid, L, R);
                    int rightMin = query(2 * node + 2, mid + 1, end, L, R);
                    return Math.min(leftMin, rightMin); // Return minimum of both children
                }
            }
            

Final Implementation and Experiments

Now we will implement the main function that processes each query and returns the result.

By allowing users to request queries, we can efficiently retrieve the minimum value.

The final implementation is as follows.

            public class MinValueFinder {
                public int[] minInRange(int[] nums, int[][] queries) {
                    SegmentTree segmentTree = new SegmentTree(nums); 
                    int[] results = new int[queries.length];

                    for (int i = 0; i < queries.length; i++) {
                        results[i] = segmentTree.query(queries[i][0], queries[i][1]); 
                    }
                    return results; 
                }
            }
            

Time Complexity Analysis

– Segment Tree Construction: O(n)
– Each Query Processing: O(log n)
The overall time complexity is O(n + m * log n).
This is very efficient and can be sufficiently performed under restricted input conditions.

Conclusion

In this course, we solved the problem of finding the minimum value in a specific range of a given array using a Segment Tree.

The Segment Tree can efficiently handle multiple queries and is frequently used data structure in large datasets.

I hope this enhances your problem-solving abilities.

Problem Definition

You have to solve the problem of finding the minimum value in a given integer array. This problem is very basic and fundamental in coding tests and algorithm problem solving, requiring basic thinking about handling arrays. Through this problem, you will practice how to use arrays and loops, as well as how to utilize conditional statements.

Problem Description

Given an integer array nums, write a function that finds and returns the minimum value in the array.

For example, if the array is [3, 1, 4, 1, 5, 9, 2, 6, 5], it should return the minimum value 1.

Input and Output Format

• Input: Integer array nums (1 ≤ |nums| ≤ 10⁵, -10⁹ ≤ nums[i] ≤ 10⁹)
• Output: Minimum value in the array

Approach

This problem can be solved using the following approach.

1. Check if the array is empty. If it is, handle the exception.
2. Initialize the first element of the array as the minimum value.
3. Iterate through the array, comparing each element with the current minimum value.
4. If the current element is less than the minimum value, update the minimum value.
5. After checking all elements, return the minimum value.

Detailed Solution Process

1. Check if the array is empty

First, you need to check if the provided array is empty. If it is empty, it is impossible to find the minimum value, so appropriate exception handling should be done in this case. For example, you can throw an IllegalArgumentException when the array is empty.

2. Initialize the minimum value

Set the first element of the array as the minimum value. This creates a reference point for comparing all the elements as you iterate through the array.

3. Explore the array

As you iterate through the array, compare each element with the minimum value. To find the minimum reliably, a for loop is generally used. All elements need to be checked.

4. Compare and update the minimum value

If the current element being checked is less than the minimum value, update the minimum value. This will ultimately result in having the smallest value in the array.

5. Return the minimum value

After checking all elements, return the minimum value.

Java Implementation

public class MinFinder {
    public static int findMin(int[] nums) {
        // Check for empty array
        if (nums.length == 0) {
            throw new IllegalArgumentException("The array is empty.");
        }

        // Initialize with the first element of the array
        int min = nums[0];

        // Find the minimum value while iterating through the array
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] < min) {
                min = nums[i]; // Update the minimum value
            }
        }

        // Return the minimum value
        return min;
    }

    public static void main(String[] args) {
        int[] nums = {3, 1, 4, 1, 5, 9, 2, 6, 5};
        int minValue = findMin(nums);
        System.out.println("Minimum value: " + minValue); // Output: Minimum value: 1
    }
}

Time Complexity Analysis

The time complexity of this algorithm is O(n). Since we need to examine each element in the array once, it maintains a linear property concerning the size of the input. This is optimal time complexity.

Space Complexity Analysis

The space complexity of this algorithm is O(1). It uses no additional data structures and only one integer variable, so it occupies very little space.

Conclusion

Through this tutorial, you have implemented an algorithm to find the minimum value in an array. Although this problem is basic, it lays a foundation that can be expanded into more complex problems in the future. Having learned fundamental reasoning using arrays and loops, it can be beneficial for future algorithmic problems.

In the next tutorial, we will tackle more complex algorithm problems. We appreciate your anticipation.