C# Coding Test Course, Building Bridges

In the process of preparing for coding tests, solving various algorithm problems is very important. In this post, we will examine the problem of building a bridge according to given conditions. Bridge construction is a type of optimization problem, where the goal is to develop an algorithm that builds a bridge as efficiently as possible while meeting specified conditions.

Problem Description

You want to build a bridge across a wide river. The bridge is made of wooden planks with a specified range, and each wooden plank can support a unique load. To support the bridge, the maximum load that each plank can sustain must not be exceeded. Additionally, the total length of the bridge must be at least a given value, and it must be built at the minimum cost.

Input

  • First line: Length of the bridge L (1 ≤ L ≤ 1000)
  • Second line: Number of wooden planks N (1 ≤ N ≤ 100)
  • Third line: An integer array W of length N representing the maximum load each wooden plank can support (1 ≤ W[i] ≤ 1000)

Output

Print the minimum number of wooden planks that can be used. If no possible combination exists, print -1.

Solution Process

To solve this problem, we need to explore all possible combinations of the given wooden planks based on their load to construct the bridge. To approach the problem more systematically, we aim to resolve it through the following steps.

1. Problem Analysis

To build the bridge, we need to decide how to select the planks to match the total bridge length L while also considering the load of each plank. This ultimately boils down to a combination problem.

2. Considering Combinations

Using combinations from the given N wooden planks, we need to generate all combinations that can create a bridge length of at least L, and check the load of these combinations to find valid options.

3. Implementation Method

To aid understanding, I will show the implementation in C#. Here, I will try all combinations using recursive calls.


    using System;
    using System.Linq;

    class Program
    {
        static int[] woodWeights;
        static int minPlanks = int.MaxValue;

        static void Main(string[] args)
        {
            int L = int.Parse(Console.ReadLine());
            int N = int.Parse(Console.ReadLine());
            woodWeights = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();

            MakeBridge(0, 0, 0);
            Console.WriteLine(minPlanks == int.MaxValue ? -1 : minPlanks);
        }

        static void MakeBridge(int index, int totalLength, int count)
        {
            if (totalLength >= L)
            {
                minPlanks = Math.Min(minPlanks, count);
                return;
            }

            for (int i = index; i < woodWeights.Length; i++)
            {
                MakeBridge(i + 1, totalLength + woodWeights[i], count + 1);
            }
        }
    }
    

4. Code Explanation

The above code recursively checks the combinations of each wooden plank to find meaningful ways to build the bridge. The MakeBridge function starts from a given index, selects each wooden plank, and explores all possible combinations through recursive calls. Ultimately, when the length of the bridge is at least L, it compares the number of planks and updates the minimum value.

5. Time Complexity

The worst-case time complexity of this algorithm is O(2^N). This is because it attempts all combinations from N wooden planks. As the number of wooden planks increases, the time complexity grows further.

6. Additional Optimization

This problem has a reasonable range for N, so we can enhance performance by adopting specific conditions to prune the search space in the bridge-building problem. For instance, if the load of the planks selected so far already exceeds L, further recursive calls are unnecessary, thus optimizing performance.

Conclusion

The bridge construction problem is an interesting algorithmic challenge where we must find a solution that satisfies given constraints through various combinations. Therefore, there is a high probability of encountering similar problems in actual job interviews. Optimizing time and constructing efficient algorithms are crucial during the problem-solving process. You should practice more problems based on the algorithms understood from this post.

References

  • Algorithm Problem Solving Strategy
  • Complete Analysis of Programming Interviews

C# Coding Test Course, Preparing for Resignation

Hello, everyone! In this post, we will solve an algorithm problem that helps prepare for resigning. Algorithm problems are one of the commonly asked types in coding tests and are essential for skill building. We will focus on solving problems using C#.

Problem: Find Two Sum in an Array

Problem Description: Given an integer array and a specific integer (target), find the indices of the two numbers in the array that sum up to the target. It is assumed that there is exactly one solution for each input, and you may not use the same element twice. The result should be returned as an array of indices, sorted in ascending order.

Example:

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

Solution Process:

This problem can be approached in various ways. However, the most efficient method is to use a hashmap. By using a hashmap, we can solve it with a time complexity of O(n). Below is a step-by-step explanation of the solution process.

Step 1: Understand the Problem

First, we need to find the case where the sum of two numbers in the given array equals the target. Since we need to return the indices, we also have to remember the index of each number while calculating the sum.

Step 2: Set Up Hashmap Structure

In C#, we can implement a hashmap using Dictionary. This structure allows us to quickly find values without allowing duplicate numbers.

Step 3: Check Using a Loop

Iterate through the array one by one, finding the value that corresponds to the current number subtracted from the target. If this value exists in the hashmap, we return the indices of the two numbers.

Step 4: Implement the Code

    
    using System;
    using System.Collections.Generic;

    public class Solution 
    {
        public int[] TwoSum(int[] nums, int target) 
        {
            Dictionary numDict = new Dictionary();
            for (int i = 0; i < nums.Length; i++)
            {
                int complement = target - nums[i];
                if (numDict.ContainsKey(complement))
                {
                    return new int[] { numDict[complement], i };
                }
                if (!numDict.ContainsKey(nums[i]))
                {
                    numDict[nums[i]] = i;
                }
            }
            throw new ArgumentException("No two sum solution");
        }
    }
    
    

Step 5: Explain the Code

The code above works in the following way:

  • First, it initializes the Dictionary.
  • It iterates through the array, calculating the complement for each number.
  • If the complement exists in the Dictionary, it returns the indices.
  • If it does not exist, it stores the current number and its index in the Dictionary.

Complexity Analysis

Time Complexity: O(n), as it only traverses the array once.

Space Complexity: O(n), in the worst case, as we may have to store all numbers in the Dictionary.

Conclusion

Through this problem, we implemented an efficient algorithm in C#. Since various problems appear in coding tests, it is recommended to practice consistently. It is important to maintain your skills even after resigning!

In the next post, we will cover different types of problems. Stay tuned!

C# Coding Test Course, Hacking Efficiently

Coding tests are an essential process that many software developers and engineers must undergo to pass the entry barrier. In this article, we will discuss how to solve algorithm problems commonly encountered in coding tests using C#. In particular, we will explore what an efficient hacking process is and the algorithm patterns required for it.

Problem: Sum of Two Numbers

Given an integer array and a target integer ‘target’, find the indices of the two numbers in the array that add up to the ‘target’. The indices start from 0, and each number must be used exactly once.

Example Input

nums = [2, 7, 11, 15]
target = 9

Example Output

[0, 1]

Problem Interpretation

The above problem is quite famous as the ‘Two Sum’ problem. You must choose two numbers from the given array such that their sum equals the provided target value. This problem can be approached in various ways, but here we will focus on an efficient approach.

Approach

1. **Brute Force**: This method checks all possible pairs in the array to see if their sum matches the ‘target’. However, this approach has a time complexity of O(n^2), making it very inefficient.

2. **Using a HashMap**: This method uses a hashmap in the form of a dictionary to store the complement of the current number (target – current number) that we are currently looking at. This approach can solve the problem with a time complexity of O(n).

Implementation

C# Code Implementation

using System;
using System.Collections.Generic;

public class Solution
{
    public int[] TwoSum(int[] nums, int target)
    {
        Dictionary map = new Dictionary();

        for (int i = 0; i < nums.Length; i++)
        {
            int complement = target - nums[i];
            if (map.ContainsKey(complement))
            {
                return new int[] { map[complement], i };
            }
            map[nums[i]] = i;
        }
        throw new ArgumentException("No two sum solution");
    }
}
    

Code Explanation

  • First, we create an empty hashmap (>). This map will store numbers as keys and their indices as values.
  • As we iterate through the array, we calculate the complement for each number and check if it already exists in the hashmap.
  • If it exists, we return the index of that complement and the current index.
  • If it doesn’t exist, we add the current number and index to the hashmap.
  • We repeat this process for all numbers in the array.

Execution Example

When executed, based on the given nums and target, the following result will be produced:

var solution = new Solution();
var result = solution.TwoSum(new int[] { 2, 7, 11, 15 }, 9);
Console.WriteLine(string.Join(", ", result)); // Output: 0, 1
    

Tips for Improving Efficiency

When selecting data structures and algorithms in coding tests, it is important to choose what is suitable for each case. Below are some tips to consider.

  • **Time Complexity**: Consider the execution time of the algorithm. It’s preferable to choose the fastest algorithm.
  • **Space Complexity**: Consider memory usage. Instead of using additional arrays or lists, it’s better to utilize existing ones.
  • **Vary Test Cases**: Test a variety of scenarios to enhance generality and stability.

Conclusion

Preparing for coding tests using C# revolves around understanding which algorithms solve problems most efficiently. In this lesson, we explained a basic technique using hashmaps through the ‘Two Sum’ problem. We hope you gain experience by solving various algorithm problems in the future.

Finally, remember that coding tests involve not just problem-solving skills but also the thought process of understanding the problem and finding effective solutions. These skills are very important in your journey as a developer.

References

C# Coding Test Course, Utilizing Time Complexity

Hello, everyone visiting the blog! Today, we will dive deep into the concept of time complexity by solving commonly encountered algorithm problems in C# coding tests. Coding tests have become an essential part of modern programming interviews, and understanding algorithms and data structures, as well as the ability to calculate time complexity, is crucial for solving problems. In this article, we will explain these topics in detail through actual algorithm problems.

Problem: Two Sum

Problem Description:
Given an integer array nums and an integer target, write a function that returns the indices of the two numbers in nums such that their sum is equal to target. It is assumed that there is exactly one solution for each input, and you may not use the same element twice.

public int[] TwoSum(int[] nums, int target) {
    // Code to implement
}

Example Input

nums = [2, 7, 11, 15]
target = 9

Example Output

[0, 1]

Solution

This problem is relatively simple. However, it is important to solve it efficiently, taking time complexity into account. There are several approaches, but here we will introduce an approach using a hashmap.

Solution Process

  1. Traverse the given array and calculate the difference with target for each number.
  2. Check if this difference exists in the hashmap. If it does, return the index of that number along with the current index.
  3. Add the current number and its index to the hashmap.

Time Complexity Analysis

The time complexity of this approach using a hashmap is O(n). Since each element is checked only once, it is efficient. The space complexity is O(n) due to the numbers stored in the hashmap.

C# Code Implementation

public int[] TwoSum(int[] nums, int target) {
    Dictionary numDict = new Dictionary();

    for (int i = 0; i < nums.Length; i++) {
        int complement = target - nums[i];
        if (numDict.ContainsKey(complement)) {
            return new int[] { numDict[complement], i };
        }
        numDict[nums[i]] = i;
    }
    throw new ArgumentException("No two sum solution");
}

Result

This means that executing the above code will allow you to find the indices of the desired two numbers in the given array. In this example, for nums = [2, 7, 11, 15] with target = 9, the output will be [0, 1].

The Significance and Application of Time Complexity

There are several reasons why time is important when solving algorithm problems. In coding tests or systems that need to be operated within a finite time, execution time is a very critical factor.

When analyzing time complexity, the following methods are used:

  • Constant Time Complexity (O(1)): Algorithms that reflect results within a fixed time regardless of input size.
  • Logarithmic Time Complexity (O(log n)): If the input size doubles, the algorithm's running time increases by a constant ratio. The binary search algorithm is a typical example.
  • Linear Time Complexity (O(n)): Algorithms whose execution time increases proportionally with the input size. Checking each element of a given array exactly once falls under this.
  • Linear Logarithmic Time Complexity (O(n log n)): Complexity in the form of input size multiplied by a logarithm. Merge sort or quicksort algorithms fall into this category.
  • Polynomial Time Complexity (O(n^k)): Performance degrades in proportion to the k-th power of input size. This often happens in cases with nested loops.
  • Exponential Time Complexity (O(2^n)): Algorithms whose running time increases rapidly even for small input sizes. This is common in recursive calculations of the Fibonacci sequence.

Considerations When Solving Algorithm Problems

Here are some things to keep in mind while solving problems:

  • Understand the range and characteristics of the input.
  • Consider ways to break the problem down into manageable steps.
  • Implement in a way that minimizes time and space complexity as much as possible.
  • Where possible, validate the accuracy of the algorithm by considering various test cases.

Final Review

In conclusion, we have looked at how to utilize time complexity in C# coding tests and the problem-solving process. Practice using time complexity well by solving various algorithm problems, including the two sum problem. I hope you achieve good results in future coding tests!

Conclusion

I hope this article serves as a useful guide in preparing for C# coding tests, and I recommend continually practicing problem-solving and writing more effective code by considering time complexity. If you have any additional questions or feedback, please feel free to leave a comment!

Thank you!

C# Coding Test Course, Finding the Sum of Remainders

Problem Description

Given an integer array numbers and an integer target,
you need to determine if any two elements from the numbers array can be added together to yield the remainder of target.
If possible, return the indices of those two elements, otherwise return -1.
Each element must be used exactly once, and the same element cannot be chosen twice.

Input Format

  • numbers: Integer array (0 ≤ numbers.length ≤ 105, -109 ≤ numbers[i] ≤ 109)
  • target: Integer (0 ≤ target ≤ 109)

Output Format

Indices of the two elements or -1

Solution Process

1. Problem Analysis

This problem uniquely asks to obtain target through the addition of two elements followed by a modulo operation.
In other words, if the two numbers we are looking for are x and y, it must hold that
(x + y) % target == target. It is essential to understand the implications of the addition and modulo conditions.

2. Algorithm Design

To solve this problem, we will traverse the array only once and calculate the remainder for each element to match the target condition.
One method involves using a hash set to store previous values and check the condition by calculating the remainder with the current value.
The reason for using a hash set is to maintain an average time complexity of O(1).

3. Code Implementation

Now, let’s implement this algorithm in C#. Below is the solution code.

using System;
using System.Collections.Generic;

public class Solution
{
    public int[] FindTwoElementsWithTargetModulo(int[] numbers, int target)
    {
        // Dictionary to store the index of elements
        Dictionary valueToIndex = new Dictionary();
        
        for (int i = 0; i < numbers.Length; i++)
        {
            // Calculate the remainder for the current element
            int requiredValue = (target - numbers[i]) % target;

            // Check if the remainder exists in the dictionary
            if (valueToIndex.ContainsKey(requiredValue))
            {
                return new int[] { valueToIndex[requiredValue], i };
            }

            // Record the current element in the dictionary
            if (!valueToIndex.ContainsKey(numbers[i]))
            {
                valueToIndex[numbers[i]] = i;
            }
        }
        
        // If no two elements satisfy the condition
        return new int[] { -1 };
    }
}
        

4. Time Complexity

The time complexity of this algorithm is O(n). We can find the result with a single traversal of the array.
Additionally, the extra space complexity is O(n) due to the storage of values in the hash set.

5. Test Cases

Let’s write various test cases to verify the functionality.

public class Program
{
    public static void Main()
    {
        Solution solution = new Solution();

        // Test case 1
        int[] numbers1 = { 1, 2, 3, 4, 5 };
        int target1 = 5;
        int[] result1 = solution.FindTwoElementsWithTargetModulo(numbers1, target1);
        Console.WriteLine(string.Join(", ", result1)); // Expected output: 0, 4

        // Test case 2
        int[] numbers2 = { 1, 2, 3, 7 };
        int target2 = 5;
        int[] result2 = solution.FindTwoElementsWithTargetModulo(numbers2, target2);
        Console.WriteLine(string.Join(", ", result2)); // Expected output: 0, 2

        // Test case 3
        int[] numbers3 = { 3, 8, 12, 5 };
        int target3 = 10;
        int[] result3 = solution.FindTwoElementsWithTargetModulo(numbers3, target3);
        Console.WriteLine(string.Join(", ", result3)); // Expected output: -1
    }
}
        

6. Conclusion

In this lecture, we learned how to optimize space and time by using a hash set to solve the problem of finding the sum modulo.
This approach is very useful for solving various problems that may appear in coding tests.
I hope you gain confidence by solving more algorithm problems.