C++ Coding Test Course, Try

1. Introduction

In today’s lecture, we will take an in-depth look at the Trie data structure using the C++ programming language,
and solve algorithm problems utilizing this structure. A Trie is an optimized data structure for storing and
searching strings, useful for solving various problems such as dictionary lookups, auto-completion, and prefix searches.

2. Introduction to Trie Data Structure

A Trie stores characters of strings at each node, and the paths between nodes represent the prefixes of specific strings.
This data structure allows for efficient string searching, with a basic time complexity of O(m),
where m is the length of the string to be searched.

2.1 Components of a Trie

The basic components of a Trie are as follows:

  • Node: Each node contains a character and child nodes.
  • Insert: Adds a new string to the Trie.
  • Search: Searches for a specific string in the Trie.
  • Delete: Deletes a specific string from the Trie.

2.2 Advantages and Disadvantages of a Trie

The advantage of a Trie is that it makes prefix searches easy. However, a disadvantage is that it can use a lot
of memory and can be complex to implement.

3. Problem Definition

The following is an algorithm problem.
Please write a program that outputs all words with a specific prefix from the given list of strings.

Problem Description

Input:
– Word list words: [“apple”, “app”, “application”, “bat”, “ban”, “band”]
– Specific prefix prefix: “ap”

Output:
– Strings with the prefix “ap”: [“apple”, “app”, “application”]

4. Problem Solving Process

4.1 Defining the Trie Class

First, we need to define a Trie node and implement the Trie class. Below is a basic implementation in C++.


#include 
#include 
#include 
#include 

using namespace std;

class TrieNode {
public:
    unordered_map children;
    bool isEndOfWord;

    TrieNode() : isEndOfWord(false) {}
};

class Trie {
private:
    TrieNode* root;

public:
    Trie() {
        root = new TrieNode();
    }

    void insert(const string& word) {
        TrieNode* node = root;
        for (char c : word) {
            if (!node->children[c]) {
                node->children[c] = new TrieNode();
            }
            node = node->children[c];
        }
        node->isEndOfWord = true;
    }

    void searchByPrefix(const string& prefix, vector& result, string current, TrieNode* node) {
        if (node->isEndOfWord) {
            result.push_back(current);
        }

        for (const auto& pair : node->children) {
            searchByPrefix(prefix, result, current + pair.first, pair.second);
        }
    }

    vector getWordsWithPrefix(const string& prefix) {
        TrieNode* node = root;
        vector result;

        for (char c : prefix) {
            if (!node->children[c]) {
                return result; // No words with the given prefix
            }
            node = node->children[c];
        }

        searchByPrefix(prefix, result, prefix, node);
        return result;
    }
};

    

4.2 Implementing the Main Function

Now, let’s implement the main function to insert words into the Trie and search for words with the prefix.


int main() {
    Trie trie;
    vector words = {"apple", "app", "application", "bat", "ban", "band"};

    // Insert words into the trie
    for (const string& word : words) {
        trie.insert(word);
    }

    // Prefix to search
    string prefix = "ap";
    vector result = trie.getWordsWithPrefix(prefix);

    // Print the result
    cout << "Words with prefix '" << prefix << "': ";
    for (const string& word : result) {
        cout << word << " ";
    }
    cout << endl;

    return 0;
}

    

5. Code Execution Results

When the above code is executed, the following results will be printed.
All words corresponding to the given prefix “ap” will be output as a list.


Words with prefix 'ap': apple app application

    

6. Conclusion

In this lecture, we implemented the Trie data structure using C++ and solved the problem of searching for
words with a specific prefix. Tries are very useful data structures for solving string-related algorithm problems,
and can be used in various application areas. In future lectures,
I hope to learn about various data structures and techniques that can serve as references for solving various
algorithm problems. Thank you.

C++ Coding Test Course, Two Pointers

What is Two Pointers?

The two pointers technique is one of the methods for efficient algorithm design. This method is mainly useful for finding pairs of data or subarrays that satisfy specific conditions in sorted data structures. Essentially, it uses two pointers to point to the start and end of an array or two different elements, which helps solve the problem.

The two pointers technique allows us to solve various data problems with a time complexity of O(n). It is commonly used in problems involving consecutive arrays, sum problems, and subarray problems. This method is designed to find answers that meet the problem’s conditions by adjusting the positions of the two pointers.

Problem Description: Finding a Subarray with a Specific Sum

Problem: Given an integer array nums, write an algorithm to find two numbers that add up to a given integer target. Each element can be used only once. Return the indices of the two numbers in the form of an array.

Example Input:

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

Example Output:

                [0, 1]
            

Explanation: nums[0] + nums[1] = 2 + 7 = 9.

Problem Solving Process

1. Understanding the Problem

The goal of this problem is to find the indices of two numbers in the given array. The key point of the problem is that each number can only be used once. Therefore, it is necessary to create combinations to find the specific sum.

2. Approach

We will use the two pointers technique to solve the problem. Even if we sort the array, we need to keep track of the original indices. Additionally, since the array is sorted, we can adjust the positions of the two pointers to find the optimal solution.

3. Algorithm Design

  1. Sort the array.
  2. Set up two pointers: one pointing to the first element of the array and the other to the last element.
  3. If the sum reaches target, return the corresponding indices.
  4. If the sum is less than target, move the left pointer to the right to increase the sum.
  5. If the sum is greater than target, move the right pointer to the left to decrease the sum.
  6. Repeat the above steps until the two pointers overlap.

4. C++ Code Implementation

                
                #include 
                #include 
                #include 

                using namespace std;

                vector twoSum(vector& nums, int target) {
                    unordered_map numMap;
                    vector result;

                    for (int i = 0; i < nums.size(); i++) {
                        int complement = target - nums[i];
                        if (numMap.find(complement) != numMap.end()) {
                            result.push_back(numMap[complement]);
                            result.push_back(i);
                            return result;
                        }
                        numMap[nums[i]] = i;
                    }
                    return result; // In case no result is found
                }

                int main() {
                    vector nums = {2, 7, 11, 15};
                    int target = 9;
                    vector indices = twoSum(nums, target);
                    
                    cout << "[" << indices[0] << ", " << indices[1] << "]" << endl;
                    return 0;
                }
                
            

5. Code Explanation

The above C++ code iterates through the given nums array to find the indices of the two numbers that can create the target value. It uses a hash map (unordered_map) to maintain a time complexity of O(n). Each time we select the first number, we check if its complement (target - nums[i]) exists in the hash map, and if it does, we return the two indices as the result.

The hash map allows for fast retrieval of each number, enabling an efficient way to find the answer by iterating through the array only once.

6. Complexity Analysis

The time complexity of the above algorithm is O(n) and the space complexity is O(n). The hash map records the complement numbers and allows for efficient searching. Since arrays are often unsorted, using a hash map for access is advantageous.

Conclusion

The two pointers technique is very useful in array problems. Finding two numbers that satisfy a specific sum in an unsorted array, like in this problem, is a frequently encountered issue. Using a hash map to optimize the algorithm is also a good strategy.

Practice solving various problems using the two pointers technique and try to extend and apply it to other issues.

C++ Coding Test Course, Preparing for Departure

Hello everyone! In this post, we will explore coding tests for those preparing for employment with C++.

The Importance of Coding Tests

Recently, many companies present algorithm problems for technical interviews. In this process, we must understand algorithms and be able to implement them in C++. A deep understanding of algorithms and data structures is key to achieving good results in interviews.

Problem Description

Problem: Sum of Two Numbers

Given an integer array and an integer target value, this problem requires returning the indices of all pairs that make up the target value by selecting two numbers from the array. The input consists of an integer array nums and an integer target value target.

Input Example

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

Output Example

    [0, 1]

Approaches to Solve the Problem

There are several approaches to solve this problem. Here, we will explain two approaches: Brute Force and Using a Hashmap.

1. Brute Force Approach

The brute force method involves using two nested loops to compare all possible pairs of numbers in the array. The time complexity of this method is O(n^2).

    int twoSum(vector& nums, int target) {
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                if (nums[i] + nums[j] == target) {
                    return {i, j};
                }
            }
        }
        return {};
    }

2. Using a Hashmap

A more efficient method can be achieved using a hashmap. Using a hashmap, you can quickly find the number needed for each number in the array to reach the target value. The time complexity of this method is O(n).

    vector twoSum(vector& nums, int target) {
        unordered_map num_map;
        for (int i = 0; i < nums.size(); i++) {
            int complement = target - nums[i];
            if (num_map.find(complement) != num_map.end()) {
                return {num_map[complement], i}; // Found the two numbers
            }
            num_map[nums[i]] = i; // Store the number with its index
        }
        return {};
    }

Problem Solving Process

Now, let’s solve the problem based on the above two approaches.

Brute Force Method

First, we will execute the brute force approach. We will simply check all combinations of the given array through double loops and compare whether the sum of two numbers equals the target value.

Using Hashmap

Next, using the hashmap method, we scan the array once, find the needed value, and return the index when found.

Summary and Conclusion

In this post, we explored how to solve algorithm problems in C++ through a simple sum of two numbers problem. The brute force method is simple but inefficient, while the method using a hashmap is significantly more efficient. When encountering similar problems during coding tests, it’s essential to consider various approaches and select the optimal method while taking time complexity into account.

Preparing for coding tests while getting ready to leave a job can be challenging, but consistent practice and solving various problems can help improve your skills. Thank you!

Author: [Your Name]

Blog: [Your Blog Link]

C++ Coding Test Course, Time Machine to Get There Fast

Recently, many companies are evaluating applicants’ algorithmic thinking and problem-solving abilities through coding tests. In this course, we will explain in detail the process of understanding and solving an interesting algorithm problem called “Time Machine to Fast Travel”. In particular, we will seek efficient and optimized solutions using the C++ language.

Problem Description

You have a time machine that can transcend time. However, this time machine operates according to specific rules. The rules for operating the time machine are as follows:

  • You are currently at time t, and this time is a positive integer.
  • You can move k seconds ahead, and k is limited to a maximum of C.
  • After each move, you can determine k again based on the new time t + k.
  • You need to reach the target time N by making several moves.

You need to write a program to find the minimum number of moves required to reach the target time N, given the current time t, the target time N, and the maximum moving time C.

Input Format

The first line contains the current time t (1 ≤ t ≤ 105), the target time N (t ≤ N ≤ 105), and the maximum moving time C (1 ≤ C ≤ 1000).

Output Format

Print the minimum number of moves required to reach the target time N.

Example

    Input
    5 10 3

    Output
    2
    

Explanation: The customer starts at time 5, moves 3 seconds ahead (time 8), then moves 2 seconds ahead (time 10), making a total of 2 moves to reach the target time 10.

Problem Solving Strategy

This problem can be solved using BFS (Breadth-First Search). BFS is a very useful algorithm for exploring the shortest path in graphs or trees. The graph in this problem can be thought of as nodes composed of the current time and the target time, where each move acts as an edge of the graph.

Steps of the BFS Algorithm

  1. Start the BFS using a queue. The initial state is set to the current time t.
  2. Remove the current time from the queue and check if you have reached the target time N.
  3. From the current time, try all possible moves (from 1 second to C seconds) to calculate the new time, and check if this new time can reach the target time N.
  4. If the new time equals the target time N, print the number of moves. Otherwise, add the new time to the queue and continue searching.

C++ Code Implementation

    #include 
    #include 
    #include 
    using namespace std;

    int minMoves(int t, int N, int C) {
        vector visited(N + 1, false);
        queue> q; // {current_time, moves}
        q.push({t, 0});
        visited[t] = true;

        while (!q.empty()) {
            auto [current_time, moves] = q.front();
            q.pop();

            if (current_time == N) {
                return moves;
            }

            for (int jump = 1; jump <= C; ++jump) {
                int next_time = current_time + jump;
                if (next_time > N) {
                    continue; // Ignore if it exceeds the target time
                }

                if (!visited[next_time]) {
                    visited[next_time] = true;
                    q.push({next_time, moves + 1});
                }
            }
        }

        return -1; // In case of unreachable
    }

    int main() {
        int t, N, C;
        cin >> t >> N >> C;
        cout << minMoves(t, N, C) << endl;
        return 0;
    }
    

Code Explanation

The above C++ code implements the BFS algorithm to solve the given problem.

  1. The minMoves function takes the current time, target time, and maximum moving time as arguments. This function returns the minimum number of moves required to reach the target time N.
  2. First, initialize an array visited to store the visited status and prepare the queue.
  3. Remove the current time from the queue, and if you have reached the target time, return the number of moves.
  4. Try all possible moves up to a maximum of C seconds from the current time, adding new times to the queue.
  5. This process repeats, and once you reach the target time, the corresponding move count will be printed.

Result Verification

The program needs to be benchmarked and verified by running various inputs to ensure the results are as expected. For example:

    Input
    1 5 2

    Output
    3
    

By providing this input, we can examine each move path and confirm that the optimal solution is achieved.

Conclusion

In this course, we explored the process of solving the "Time Machine to Fast Travel" problem using C++. We learned how to find the shortest path through the BFS algorithm and gained experience solving problems through actual code. This knowledge will be greatly helpful for real coding tests or algorithm problem-solving. We look forward to tackling more interesting problems in the next course!

C++ Coding Test Course, Quick Sort

Hello! In this session, we will take a closer look at the Quick Sort algorithm using C++. Quick Sort is a sorting algorithm that frequently appears in various programming interviews and algorithm exams. In this article, we will master Quick Sort through its concept, implementation, time complexity, and example problems.

1. Overview of Quick Sort

Quick Sort is a type of divide-and-conquer algorithm that efficiently sorts data. The algorithm works as follows:

  1. Select a pivot element from the given array.
  2. Divide the array into two sub-arrays based on the pivot.
    (Elements less than the pivot go to the left sub-array, while elements greater than the pivot go to the right sub-array.)
  3. Recursively perform Quick Sort on both the left and right sub-arrays.
  4. Once the sub-arrays are sorted, place the pivot in the middle to form the final sorted array.

1.1 Selecting a Pivot

There are several methods for selecting a pivot. Common methods include:

  • Selecting the first element as the pivot
  • Selecting the last element as the pivot
  • Selecting a random element as the pivot
  • Selecting the median as the pivot

It is important to choose an appropriate method as the pivot selection significantly impacts sorting performance.

1.2 Time Complexity of Quick Sort

The average time complexity of Quick Sort is O(n log n). However, in the worst-case scenario (e.g., when the array is already sorted), performance can degrade to O(n²). Good pivot selection methods are important to prevent this.

2. Problem Description

Now, let’s solve a problem using Quick Sort. Solve the following problem:

Problem: Given an array of integers, sort the array in ascending order using the Quick Sort algorithm.
Input: [10, 7, 8, 9, 1, 5]
Output: [1, 5, 7, 8, 9, 10]

3. Problem Solving Process

The steps to solve the problem are as follows:

3.1 Algorithm Implementation

First, let’s implement the Quick Sort algorithm in C++.

 
#include <iostream>
#include <vector>

using namespace std;

// Partition function
int partition(vector<int> &arr, int low, int high) {
    int pivot = arr[high]; // Select the last element as pivot
    int i = (low - 1); // Index for smaller elements

    for (int j = low; j < high; j++) {
        // If current element is smaller than the pivot
        if (arr[j] < pivot) {
            i++; // Increment index of smaller elements
            swap(arr[i], arr[j]); // Swap elements
        }
    }
    swap(arr[i + 1], arr[high]); // Swap the pivot to the correct location
    return (i + 1); // Return the new index of the pivot
}

// Quick Sort function
void quickSort(vector<int> &arr, int low, int high) {
    if (low < high) {
        // Perform partition and get the pivot index
        int pi = partition(arr, low, high);

        quickSort(arr, low, pi - 1); // Sort the left sub-array
        quickSort(arr, pi + 1, high); // Sort the right sub-array
    }
}

// Main function
int main() {
    vector<int> arr = {10, 7, 8, 9, 1, 5};
    int n = arr.size();

    quickSort(arr, 0, n - 1);

    cout << "Sorted array: ";
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
    return 0;
}
    

The code above implements the Quick Sort algorithm. The partition function divides the given array based on the pivot, and the quickSort function performs sorting recursively.

3.2 Code Explanation

Let’s take a step-by-step look at the code:

  • partition function:
    • Selects the last element as the pivot.
    • Divides the given array based on the pivot.
    • Moves the pivot to its sorted position.
    • Returns the new pivot index.
  • quickSort function:
    • Checks the base condition and makes recursive calls for sub-arrays.
    • Performs the process of “finding the pivot and sorting” for each sub-array.

3.3 Code Execution Result

When the code is executed, it produces the following output:

Sorted array: 1 5 7 8 9 10

We can confirm that the sorted array is printed correctly. This indicates that Quick Sort has been performed successfully.

4. Additional Considerations

There are several additional considerations when using Quick Sort:

4.1 Stability

Quick Sort is an unstable sorting algorithm. That is, the relative order of elements with the same value is not guaranteed after sorting. If stability is required, other sorting algorithms (e.g., Merge Sort) should be considered.

4.2 Memory Usage

Quick Sort operates recursively, using memory on the call stack. In the worst case, it can have O(n) space complexity, but on average, it is O(log n).

5. Quick Sort vs Other Sorting Algorithms

Compared to other sorting algorithms, Quick Sort has the following advantages and disadvantages:

5.1 Advantages

  • Typically offers fast performance with an average of O(n log n).
  • Requires less additional memory space.
  • Supports recursive and intuitive implementations.

5.2 Disadvantages

  • Can have O(n²) performance in the worst case.
  • Is an unstable sort and the relative order of identical elements may change.

6. Conclusion

In this article, we implemented the Quick Sort algorithm using C++ and explored various aspects. Quick Sort is an efficient and widely used sorting algorithm, making it a common topic in algorithm exams or coding interviews. Understanding and implementing Quick Sort will greatly aid in coding tests.

Finally, try solving more problems using Quick Sort. Considering various cases will help deepen your understanding of this algorithm.

Author: [Your Name] | Date: [Date of Writing]