Category: Java Coding Test

Hello! In this post, we will take some time to solve algorithm problems using Java. The topic is “Planning a Trip“. We will solve the problem through an algorithmic approach by selecting various places, just like planning a trip, and finding the optimal route.

Problem Description

The trip planning problem is given in the following form. A total of n travel destinations are provided, and the travel costs between each destination are specified. Our goal is to select some of the given destinations and find the minimum cost route to visit those destinations.

Problem Definition

Input:
- Number of destinations n (1 ≤ n ≤ 100)
- A n x n matrix cost representing the travel costs between each destination (cost[i][j]: cost of traveling from i to j)

Output:
- The minimum cost to visit the selected destinations

Problem Analysis

This problem can be viewed as one of the representative shortest path problems. The destinations can be represented as vertices, and the travel costs can be expressed as edge weights. This structure is typically solved using graph algorithms. In particular, since all vertices need to be visited, it can be solved similarly to the Traveling Salesman Problem (TSP).

Algorithm Design

There are various methods to solve the trip planning problem, but the basic idea is as follows:

• We need to consider all possible routes to visit the destinations.
• Calculate the costs for each possible route.
• Find the route with the lowest total cost.

To achieve this, the recursive backtracking method is effective and can guarantee an optimal solution. Additionally, using a memoization technique can reduce duplicate calculations.

Java Implementation

Below is the Java code to solve the problem:

import java.util.Arrays;

public class TravelPlan {

    private static int n; // Number of destinations
    private static int[][] cost; // Travel costs
    private static boolean[] visited; // Visit status
    private static int minCost = Integer.MAX_VALUE; // Minimum cost

    public static void main(String[] args) {
        n = 5; // Example number of destinations
        cost = new int[][] {
            {0, 10, 15, 20, 25},
            {10, 0, 35, 25, 30},
            {15, 35, 0, 30, 20},
            {20, 25, 30, 0, 15},
            {25, 30, 20, 15, 0}
        };
        visited = new boolean[n];
        visited[0] = true; // Mark starting point as visited
        findMinCost(0, 0, 1);
        System.out.println("Minimum cost: " + minCost);
    }

    private static void findMinCost(int currentCity, int currentCost, int count) {
        if (count == n) {
            // If all cities have been visited
            minCost = Math.min(minCost, currentCost + cost[currentCity][0]); // Add the cost to return to the starting point
            return;
        }

        for (int nextCity = 0; nextCity < n; nextCity++) {
            if (!visited[nextCity]) {
                visited[nextCity] = true; // Mark as visited
                findMinCost(nextCity, currentCost + cost[currentCity][nextCity], count + 1);
                visited[nextCity] = false; // Backtrack
            }
        }
    }
}

Code Explanation

The above code is written based on an example with 5 destinations. The findMinCost method takes the current city, current cost, and the number of visited cities as parameters and records the minimum cost when all cities have been visited.

• It recursively moves to the next city while accumulating the costs.
• It adds the cost of returning to the starting point after visiting all cities.

Result Analysis

When the above code is executed, it will output the minimum cost for the given example. This algorithm exhaustively explores all cities, so the time complexity is O(n!). As the number of cities increases, the execution time increases, so practical improvements are necessary:

• Using dynamic programming (DP) techniques to reduce duplicate calculations.
• Employing heuristic methods like the nearest neighbor algorithm to find approximate solutions.

Conclusion

Today, we learned about recursive backtracking and graph search techniques through the algorithm problem of planning a trip. This problem is frequently encountered in coding tests, so ample practice and understanding are necessary. If you desire a deeper understanding of algorithms, solving various modified problems is also a good approach.

I hope this article helps you prepare for your coding tests! In the next post, we will discuss methods using dynamic programming.

This is a course on algorithm problem-solving for employment. In this article, we will explain in detail which algorithm to use through actual problems.

1. Problem Description

The following is a problem of finding indices of two numbers in a given array that sum up to a specific value.

Problem: Given an array nums and an integer target, return the indices of the two numbers such that they add up to target. Each input has exactly one solution, and you may not use the same element twice. The returned indices should be zero-based and returned as an array.

Example:

• Input: nums = [2, 7, 11, 15], target = 9
• Output: [0, 1] (2 + 7 = 9)

2. Problem Analysis

This problem is a very common one that has been asked multiple times. As we need to find two numbers in the given array, the first method that comes to mind is using a nested loop. However, this method has a time complexity of O(n²) and is inefficient.

Using the optimal method, we can solve the problem with a time complexity of O(n). This method will approach the problem by using a hashmap for searching.

3. Algorithm Design

We will design the algorithm in the following steps:

1. Create an empty hashmap.
2. Iterate through all the numbers.
3. For each number, check if target - current number exists in the hashmap.
4. If it exists, return the two indices; if not, add the current number to the hashmap.

The key point of this algorithm is to find the two numbers in a single iteration.

4. Java Code Implementation

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

import java.util.HashMap;

public class TwoSum {
    public static int[] findTwoSum(int[] nums, int target) {
        HashMap numMap = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            if (numMap.containsKey(complement)) {
                return new int[] { numMap.get(complement), i };
            }
            numMap.put(nums[i], i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }

    public static void main(String[] args) {
        int[] nums = {2, 7, 11, 15};
        int target = 9;
        int[] result = findTwoSum(nums, target);

        System.out.println("Indices: [" + result[0] + ", " + result[1] + "]");
    }
}

The above code solves the problem in a very simple yet efficient way. It uses a HashMap to store the indices of each number and calculates the difference with the target for searching.

5. Time Complexity Analysis

The time complexity of this algorithm is O(n). Here, n is the length of the input array. Both the operations of adding numbers to the hashmap and searching have an average time complexity of O(1), allowing the entire iteration process to be O(n). The space complexity is also O(n) because we store as many numbers as the input numbers in the hashmap.

6. Additional Considerations

When solving this problem, there are a few additional considerations:

• In case of duplicate numbers: the same number should not be used twice.
• Performance of the hashmap: Java's hashmap has an average time complexity of O(1). However, if the input follows a specific pattern, it can reach O(n) in the worst case.
• Exception handling: appropriate exceptions should be thrown in case there is no solution.

7. Conclusion

Today, we learned about using hashmaps as an algorithm to solve a specific programming problem. As the ability to apply such skills in practical situations is important, it is advisable to practice by solving various problems. It is essential to continuously strive to improve your algorithm problem-solving skills for employment.

Introduction

Coding tests are a process that modern programmers inevitably go through. Among them, understanding and utilizing time complexity is very important. In this article, we will explain in detail how to solve algorithm problems using Java, along with how to analyze time complexity.

Problem Description

Given an integer array nums and an integer target, the problem is to find a pair of indices such that the two numbers in the array sum to target. If there is none, return -1, -1.

Problem Requirements

• 1 ≤ nums.length ≤ 10⁴
• -10⁹ ≤ nums[i] ≤ 10⁹
• -10⁹ ≤ target ≤ 10⁹
• Assume that there is only one solution.

Example

Input: nums = [2, 7, 11, 15], target = 9
Output: [0, 1]
Explanation: nums[0] + nums[1] = 2 + 7 = 9, so return [0, 1].

Solution Approach

This problem can be solved in two ways. The first method is using a double loop, and the second is using a hashmap. We will analyze the time complexity of each method and implement the hashmap method first.

Method 1: Double Loop

This method checks all pairs in the array to see if their sum equals target. The code implementation is as follows:

public int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] == target) {
                return new int[]{i, j};
            }
        }
    }
    return new int[]{-1, -1};
}

Time Complexity Analysis

In the double loop, each element is checked against all other elements. Therefore, the time complexity is O(n^2). This can be inefficient in the worst case.

Method 2: Using Hashmap

To reduce time complexity, we can use a hashmap to solve this problem. This method stores the indices of each element and checks if the value of target - nums[i] exists as a key, returning the index if it does.

import java.util.HashMap;

public int[] twoSum(int[] nums, int target) {
    HashMap map = new HashMap<>();
    
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }
    return new int[]{-1, -1};
}

Time Complexity Analysis

The time complexity of this method is O(n). Since each element is checked only once, it is efficient. The insertion and retrieval from the hashmap are performed in constant time, making this approach superior.

Complexity Analysis and Conclusion

This problem highlighted the importance of time complexity. While the double loop approach is simple, it can lead to performance degradation, so it is necessary to use optimized data structures like hashmaps in practice.

Additional Tips

• When you first encounter a problem, consider various approaches. Transitioning from a double loop to a hashmap can significantly reduce time complexity.
• Always review whether the given data structure or algorithm is suitable for solving the problem.
• When calculating time complexity, always consider the worst case. It’s also beneficial to consider the best and average cases.

Conclusion

In this article, we emphasized the importance of time complexity through solving coding problems using Java. We saw how crucial it is to choose the optimal approach in solving algorithm problems. In the next session, we will explore more tips and tricks through another algorithm problem.

Many people preparing for coding tests struggle with the difficulty and complexity of algorithm problems. Today, through a problem titled “Finding Interesting Primes,” we will introduce various algorithms for finding primes (premium numbers) and see how to efficiently solve the problem using Java.

Problem Description

Write a function to find all primes less than or equal to a given number N. Here, a prime is defined as a natural number that has no divisors other than 1 and itself.

Input: A natural number N is given on the first line. (2 ≤ N ≤ 1000)

Output: Print all primes less than or equal to N on one line, separated by spaces.

Example

Input: 
10

Output: 
2 3 5 7

Understanding and Analyzing the Problem

To solve the above problem, it is essential to understand the definition of a prime and basic methods for finding primes. A prime is a number that cannot be divided by natural numbers other than 1 and itself, so we can efficiently find primes using division operations.

However, simply checking all numbers from 1 to N is inefficient, with a time complexity of O(N^2). Therefore, a more efficient algorithm is needed. We can efficiently find primes using the Sieve of Eratosthenes algorithm.

Sieve of Eratosthenes

This algorithm works as follows:

1. Prepare a list of all natural numbers from 2 to N.
2. Select the first number in the list (2) and remove all its multiples.
3. Select the next remaining number (3) and remove its multiples.
4. Repeat this process until all numbers less than or equal to N have been removed. The remaining numbers are all primes.

Java Code Implementation

Now we will implement the above algorithm in Java.

public class PrimeSieve {
    public static void main(String[] args) {
        int N = 10; // The number N to use
        boolean[] isPrime = new boolean[N + 1];
        
        // Initialization
        for (int i = 2; i <= N; i++) {
            isPrime[i] = true;
        }

        // Sieve of Eratosthenes
        for (int i = 2; i * i <= N; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= N; j += i) {
                    isPrime[j] = false;
                }
            }
        }

        // Output results
        for (int i = 2; i <= N; i++) {
            if (isPrime[i]) {
                System.out.print(i + " ");
            }
        }
    }
}

Explanation

The code above goes through the following steps:

1. First, it takes the input N and initializes a boolean array for numbers from 2 to N. Each index in the array indicates whether the corresponding number is prime.
2. Then, it applies the Sieve of Eratosthenes algorithm to mark the indexes of non-prime numbers as false.
3. Finally, it iterates through the isPrime array, printing indexes that are true to list the primes.

Time Complexity

The time complexity of this algorithm is O(N log(log(N))), which operates efficiently even for large N. This is significantly more efficient than simply dividing all numbers.

Conclusion

In this lesson, we implemented the Sieve of Eratosthenes algorithm in Java to find primes. Familiarity with such basic algorithms can help you score higher when solving various algorithm problems in coding tests. Furthermore, remember that in coding tests, both the readability and optimization of your code are also important scoring factors, so it is necessary to keep this in mind when coding.

Let's continue working on various algorithm problems together to improve our skills!

Hello! In this post, we will explore the time complexity notation commonly discussed in Java coding tests. Understanding time complexity is crucial for solving algorithm problems. To save time, it is essential to assess which algorithms are more efficient. Thus, we will approach this understanding through examples.

Problem Description

First, I will introduce an algorithm problem.

Problem: Two Sum

Given an integer array nums and an integer target, find the indices of the two numbers in the array that add up to target. We do not consider combinations of arbitrary integers. It is assumed that there is exactly one solution for each input, and the same element cannot be used twice. The function should return an array of the two indices. For example:

    Input: nums = [2, 7, 11, 15], target = 9
    Output: [0, 1]
    Explanation: Since nums[0] + nums[1] = 2 + 7 = 9, the output is [0, 1].
    

Problem Solving Process

Step 1: Problem Analysis

This problem requires finding two numbers that sum to target. There are various approaches to solving this problem depending on the length of the array, but the most basic method is to use nested for loops.

Step 2: Approach through Nested For Loops

To simply find two numbers, we will check every element against all the other elements. This will result in n(n-1)/2 possible cases, leading to a time complexity of O(n2).

    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[] {i, j};
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
    

Step 3: Time Complexity Analysis

The above algorithm requires O(n²) time complexity. Algorithms with such high time complexity become inefficient as the input size increases. For example, if the size of the array is 10,000, it would require approximately 100,000,000 (one hundred million) operations.

Step 4: Efficient Approach

An efficient method is to use HashMap. This approach can reduce the time complexity to O(n). By using HashMap, we can quickly check previously seen numbers.

    public int[] twoSum(int[] nums, int target) {
        Map map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            if (map.containsKey(complement)) {
                return new int[] {map.get(complement), i};
            }
            map.put(nums[i], i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
    

Step 5: Time Complexity Analysis of the New Algorithm

The time complexity of this method is O(n). Since we only traverse the array once, there are no duplicate operations. In other words, we can find the necessary values immediately by scanning all elements of the array just once. Such an improved approach can significantly enhance performance.

Time Complexity Notation

To fundamentally understand the performance of algorithms, we use the following time complexity notations:

• O(1): Constant time complexity
• O(log n): Logarithmic time complexity, commonly used in binary search
• O(n): Linear time complexity
• O(n log n): Linear logarithmic time complexity, often seen in sorting algorithms
• O(n²): Quadratic time complexity, occurring in nested loops
• O(2ⁿ): Exponential time complexity, used in calculating Fibonacci numbers, etc.

Conclusion

In this article, we examined time complexity notation through a simple algorithm problem. Evaluating the efficiency of algorithms is a crucial part of programming. By developing the ability to identify the limits and optimal solutions for given problems, you can achieve good results in coding tests. In the next steps, we will address more complex algorithms and data structures, exploring ways to maximize performance.

I hope this article has been helpful. If you’re curious about more rich content regarding Java coding tests and time complexity, please keep following along!