Category: C++ Coding Test Tutorials

Hello! In this lecture, we will cover the coding test problem “Sorting Numbers 1” using C++. This problem is a good foundation for sorting algorithms and allows us to compare the efficiency of various algorithms. We will examine the process of solving the problem step by step.

Problem Description

The problem is as follows:

Given N numbers, sort them in ascending order and print them.
(The integer N is between 1 and 1,000,000.)

Input example:

5
5
3
2
4
1

Output example:

1
2
3
4
5

Problem Analysis

This problem serves as a fundamental problem for sorting algorithms, where you simply need to sort and print the given numbers. It is ideal to use an efficient algorithm with a time complexity of O(N log N). The most commonly used sorting algorithms include Quick Sort, Merge Sort, and Heap Sort. In this problem, we can conveniently use the `std::sort` function from C++’s STL (Standard Template Library) to solve the problem easily.

Solve Process

1. Input Processing

First, we need to decide how to handle the input. In C++, we can use cin to read numbers from standard input. We will first input the number of values and then read the numbers into a vector.

2. Selecting a Sorting Algorithm

There are various sorting algorithms mentioned above, but it’s most convenient to use the std::sort function from C++ STL. This function internally uses the Quick Sort algorithm, providing an average performance of O(N log N).

3. Outputting the Sorted Result

To print the sorted results, we need to use a loop again. Since we need to line break with each printed number, we can utilize cout.

4. Code Implementation

Based on the above contents, let’s write the full code.

#include 
#include 
#include  // Include required for std::sort

using namespace std;

int main() {
    int N;
    cin >> N; // Input the number of values

    vector numbers(N); // Declare vector to store the numbers

    // Input the numbers
    for (int i = 0; i < N; i++) {
        cin >> numbers[i];
    }

    // Sort in ascending order
    sort(numbers.begin(), numbers.end());

    // Output the sorted numbers
    for (int i = 0; i < N; i++) {
        cout << numbers[i] << endl;
    }

    return 0;
}

Code Explanation

Let’s examine the code line by line:

• #include <iostream>: Library for input and output.
• #include <vector>: Library for using dynamic arrays.
• #include <algorithm>: Library for using STL algorithm functions.
• using namespace std;: Added to use the std namespace.
• int N;: Declares a variable to store the number of values.
• vector numbers(N);: Declares a vector to store N integers.
• cin: Used to receive input from the user for N numbers.
• sort(numbers.begin(), numbers.end());: Sorts the numbers in the vector in ascending order.
• cout: Outputs the sorted results.

Test Cases

Now that we have written the code, we will run it with various test cases. Below are some examples.

Example 1

Input:

5
5
3
2
4
1

Output:

1
2
3
4
5

Example 2

Input:

10
9
8
7
6
5
4
3
2
1
0

Output:

0
1
2
3
4
5
6
7
8
9

Example 3

Input:

3
1
1
1

Output:

1
1
1

Performance Analysis

Now let’s analyze the performance of the algorithm. std::sort has an average performance of O(N log N). Thus, even in cases where we input integers within the range of 1 to 1,000,000, it will operate quickly enough. While various sorting algorithms can also be used, std::sort is the most suitable for meeting the problem’s requirements.

Conclusion

In this lecture, we covered the “Sorting Numbers 1” problem using C++. We learned the basics of sorting as well as how to utilize STL. In the next lecture, we will explore more complex data structures and algorithms. Thank you!

Sorting numbers in coding tests is a very basic yet important problem. In this article, we will provide an in-depth explanation of how to sort numbers using C++, guiding you through the process of solving the algorithm problem.

Problem Description

The task is to sort the given numbers and output them. The specific input and output conditions are as follows.

• The first line contains the number of elements to sort, N. (1 ≤ N ≤ 1,000,000)
• The second line contains N integers A[1], A[2], …, A[N]. (−1,000,000,000 ≤ A[i] ≤ 1,000,000,000)

The output must print N integers sorted in ascending order, each on a new line.

Example Input

        5
        5
        4
        3
        2
        1
        

Example Output

        1
        2
        3
        4
        5
        

Solution Process

There are various methods to solve the problem, but here we will use the sorting algorithm utilizing C++’s STL (Standard Template Library). The sort() function of STL has a fast and efficient sorting algorithm, typically with a time complexity of O(N log N).

1. Including Necessary Header Files

We include the necessary header files to use the STL sort function and vector.

        #include <iostream>
        #include <vector>
        #include <algorithm>
        using namespace std;
        

2. Writing the Main Function

We will write the main function that processes the input and outputs the results after sorting.

        int main() {
            int N;
            cin >> N;  // Input the number of elements
            vector<int> A(N);  // Declare a vector A of size N

            // Input N integers
            for(int i = 0; i < N; i++) {
                cin >> A[i];
            }

            // Sort vector A
            sort(A.begin(), A.end());

            // Output the sorted result
            for(const int &num : A) {
                cout << num << endl;
            }

            return 0;
        }
        

3. Explanation of the Code

The above code operates in the following steps:

1. It takes an integer N as input to determine the number of numbers to sort.
2. It creates a vector A of size N and inputs N integers.
3. It uses the STL’s sort() function to sort vector A.
4. It outputs the contents of the sorted vector.

Time Complexity Analysis

The time complexity of the algorithm to solve this problem is O(N log N). This increases in proportion to the number of elements N being input, but since it is based on the most commonly used sorting algorithm, quicksort, it achieves very fast performance in practice.

Space Complexity Analysis

The space complexity is O(N). The vector A allocates space for N numbers to store all inputted values. Using STL allows efficient space management, and by using a vector instead of an array, memory management becomes easier.

Conclusion

In this post, we experienced a basic number sorting problem using C++. Through the number sorting problem, which is frequently asked in coding tests, we learned how to use sorting algorithms with STL and analyzed time and space complexity. Understanding such basic problems will greatly help in solving more complex problems in the future.

Additional Practice Problems

For more practice, try the following variations:

• Write a program that removes duplicate numbers from the input and outputs the sorted result.
• Output the sorted numbers in descending order.
• Write a program that outputs the frequency of each number. (Count how many times each number appears)

Through such additional practice, you can gain a deeper experience.

Algorithms are one of the key concepts in computer science and are particularly important in solving programming problems.
In this course, we will discuss the problem of ‘finding prime numbers’ and explain the algorithm and C++ code to solve it in detail.
A prime number is a natural number that can only be divided by 1 and itself, and it is one of the fundamental concepts in mathematics.
However, there are various approaches to solving this problem, each with different time complexities.

Problem Description

Given an integer N, write a program to find all prime numbers from 1 to N.
A prime number is a number greater than 1 that has no divisors other than 1 and itself.

Input

The first line contains a natural number N (1 ≤ N ≤ 10^6).

Output

Print all prime numbers from 1 to N, one per line.

Example

Input:
    10

    Output:
    2
    3
    5
    7
    

Algorithm Approach

There are several methods to find prime numbers, but here we will use the classical method known as the ‘Sieve of Eratosthenes’.
This method can efficiently and relatively simply find prime numbers.
The Sieve of Eratosthenes works through the following procedure.

1. List all numbers from 1 to N.
2. Starting from 2, remove all multiples of that number.
3. Repeat the above process for the next remaining number.
4. Consider the remaining numbers as primes and output them.

Time Complexity

The Sieve of Eratosthenes algorithm has a time complexity of about O(N log log N).
It becomes more efficient as N increases, ensuring fast performance even for N = 10^6.

C++ Code

Below is the C++ code that implements the above algorithm.
Comments will be added to each step to aid understanding.

#include 
#include 
using namespace std;

int main() {
    int N;
    cin >> N; // Accept input from the user.

    // Create a vector of size N+1 and initialize all values to true.
    vector isPrime(N + 1, true);
    isPrime[0] = isPrime[1] = false; // 0 and 1 are not prime.

    // Iterate from 2 to the square root of N.
    for (int i = 2; i * i <= N; i++) {
        if (isPrime[i]) { // If i is prime,
            for (int j = i * i; j <= N; j += i) {
                isPrime[j] = false; // Multiples of i are not prime.
            }
        }
    }

    // Output the results.
    for (int i = 2; i <= N; i++) {
        if (isPrime[i]) {
            cout << i << "\n"; // Print if it is prime.
        }
    }

    return 0;
}
    

Code Explanation

- #include and #include :
Include C++ input, output, and vector libraries.
- int N;: Declare an integer variable to hold user input.
- vector isPrime(N + 1, true);:
Declare a vector to store the primality of numbers up to N,
initializing all values to true.
(0 and 1 are not primes, so set separately)
- for (int i = 2; i * i <= N; i++):
Start from 2 and iterate up to the square root of N to find primes.
- The inner for (int j = i * i; j <= N; j += i) loop removes multiples of i.
- Finally, output all indices where isPrime[i] is true.

Conclusion

In this course, we introduced an algorithm to solve the problem of finding prime numbers and implemented it in C++. The Sieve of Eratosthenes is an efficient method for finding primes, and it is a common question in coding tests.
Familiarize yourself with the algorithm and code to confidently tackle coding tests!

Today, we will take a deep dive into one of the frequently asked algorithm problems for job seekers: ‘Finding the minimum value among prime and palindrome numbers’. Through this problem, we will review the concepts of primes and palindromes, and have a session to implement the algorithm using C++.

Problem Description

Given an integer N, the problem is to find the minimum integer that is both a prime and a palindrome within the range of N or greater.

Examples

• If N = 10, the prime and palindrome number is 11.
• If N = 30, the prime and palindrome number is 31.
• If N = 100, the prime and palindrome number is 101.

Problem Approach

A prime number is a natural number greater than 1 that has no divisors other than 1 and itself. A palindrome is a number that reads the same forwards and backwards. For example, 121, 12321, etc.

To solve the problem, we will follow these steps:

1. Start from N and examine all numbers.
2. Check if each number is prime.
3. If it’s prime, check if it’s a palindrome.
4. Find and return the minimum that satisfies all conditions.

C++ Implementation

Now, let’s write the code to solve this problem using the C++ language.

Code Explanation

#include 
#include 
#include 

using namespace std;

// Prime determination function
bool isPrime(int num) {
    if (num < 2) return false;
    for (int i = 2; i <= sqrt(num); ++i) {
        if (num % i == 0) return false;
    }
    return true;
}

// Palindrome determination function
bool isPalindrome(int num) {
    string str = to_string(num);
    int len = str.length();
    for (int i = 0; i < len / 2; ++i) {
        if (str[i] != str[len - i - 1]) return false;
    }
    return true;
}

// Function to find the minimum prime palindrome that is greater than or equal to N
int findMinPrimePalindrome(int N) {
    for (int i = N; ; ++i) {
        if (isPrime(i) && isPalindrome(i)) {
            return i;
        }
    }
}

int main() {
    int N;
    cout << "Enter an integer N: ";
    cin >> N;
    
    int result = findMinPrimePalindrome(N);
    cout << "Minimum value: " << result << endl;
    
    return 0;
}

    

Code Explanation

Through the above code, we can achieve the desired result through the following processes:

1. isPrime function: Determines whether the given integer is prime. It only checks numbers greater than or equal to 2 and checks for divisibility from 2 up to sqrt(num).
2. isPalindrome function: Determines whether the given integer is a palindrome. It converts the number to a string and compares characters from the front and back for matches.
3. findMinPrimePalindrome function: Starts from the given N and checks each number for prime and palindrome status. It returns immediately upon finding a number that is both.
4. main function: Takes an integer N from the user and finds and prints the minimum value based on that.

Example for Understanding

For example, if N=10, the function starts from 10 and checks 10, 11. Since 11 is both prime and a palindrome, it returns 11 as the minimum value.

Time Complexity

The time complexity of the algorithm we wrote varies depending on the input value N. Checking for primality takes time proportional to sqrt(N), and when considering the entire process of starting from N to find the minimum, the worst-case time complexity is O(N√N).

Points to Note While Solving

Finding prime and palindrome numbers requires an understanding of basic mathematics and string conversion. Additionally, a grasp of basic C++ syntax and data types is essential. To solve this problem, it's important to organize information well and practice a step-by-step approach.

Conclusion

Through this lecture, we had the opportunity to understand algorithms related to primes and palindromes and implement them ourselves. This problem is one of the useful algorithms that can be utilized in the field and can be beneficial while preparing for job interviews. I hope you will continue to tackle various algorithm problems as you prepare for coding tests.

The Traveling Salesman Problem (TSP) is one of the classic algorithm problems, also known as the “Salesman’s Dilemma”.
The problem involves finding the shortest possible route that visits a set of cities and returns to the origin city.
It falls under the category of combinatorial optimization problems and is classified as NP-hard.
Therefore, while small datasets can be solved using brute force algorithms, larger datasets require more efficient approaches.

Problem Description

Given N cities and a distance matrix between each pair of cities, the goal is to calculate the sum of the distances of the shortest route that visits each city exactly once before returning to the starting point.
Below is the format of the problem.

        
        Input:
        N (number of cities)
        Distance matrix (nxn matrix)
        
        Output:
        Length of the shortest route
        
    

Example

Input Example:

        4
        0 10 15 20
        10 0 35 25
        15 35 0 30
        20 25 30 0
    

Output Example:
80

Problem Solving Process

There are several approaches to solving this problem, and here we will use two methods: brute force and dynamic programming (DP) to solve the problem.

1. Brute Force Method

The brute force method searches through all possible paths to find the shortest route.
It generates all permutations of cities and calculates the total distance for each path, then finds the minimum distance among them.

Algorithm Description

1. Generate all permutations of the cities.
2. Calculate the total distance for each permutation.
3. Find the minimum distance among the calculated distances.

C++ Code Example

        
        #include 
        #include 
        #include 
        #include 

        using namespace std;

        int calculateDistance(const vector>& dist, const vector& path) {
            int totalDistance = 0;
            for (int i = 0; i < path.size() - 1; i++) {
                totalDistance += dist[path[i]][path[i + 1]];
            }
            totalDistance += dist[path.back()][path[0]]; // Return to start
            return totalDistance;
        }

        int main() {
            int N;
            cin >> N;
            vector> dist(N, vector(N, 0));

            for (int i = 0; i < N; i++)
                for (int j = 0; j < N; j++)
                    cin >> dist[i][j];

            vector cities(N);
            for (int i = 0; i < N; i++) cities[i] = i;

            int minDistance = INT_MAX;

            do {
                int currentDistance = calculateDistance(dist, cities);
                minDistance = min(minDistance, currentDistance);
            } while (next_permutation(cities.begin() + 1, cities.end())); // Fix first city

            cout << minDistance << endl;

            return 0;
        }
        
    

2. Dynamic Programming (DP) Method

The dynamic programming method efficiently uses memory.
It represents cities using a bitmask to indicate whether each city has been visited, and utilizes recursion and memoization to compute the shortest route.
In this case, the complexity is reduced to O(n^2 * 2^n). Below is an explanation of the dynamic programming solution.

Algorithm Description

1. Use a bitmask to represent whether each city has been visited.
2. Initialize a minimum cost table and set it to 0.
3. Explore possible paths through a recursive function, updating the minimum distance value.

C++ Code Example

        
        #include 
        #include 
        #include 

        using namespace std;

        int tsp(int pos, int visited, const vector>& dist, vector>& memo) {
            if (visited == ((1 << dist.size()) - 1)) {
                return dist[pos][0]; // Return to the start city
            }

            if (memo[pos][visited] != -1) {
                return memo[pos][visited];
            }

            int minCost = INT_MAX;

            for (int city = 0; city < dist.size(); city++) {
                if ((visited & (1 << city)) == 0) {
                    int newCost = dist[pos][city] + tsp(city, visited | (1 << city), dist, memo);
                    minCost = min(minCost, newCost);
                }
            }

            return memo[pos][visited] = minCost;
        }

        int main() {
            int N;
            cin >> N;
            vector> dist(N, vector(N, 0));

            for (int i = 0; i < N; i++)
                for (int j = 0; j < N; j++)
                    cin >> dist[i][j];

            vector> memo(N, vector((1 << N), -1));

            cout << tsp(0, 1, dist, memo) << endl;

            return 0;
        }
        
    

Conclusion

In this tutorial, we explored the Traveling Salesman Problem implemented in C++, covering the problem definition, examples,
and detailed approaches using brute force and dynamic programming methods.
It is important to note that the brute force method should only be used when the number of cities is small, as it becomes less efficient with increasing numbers of cities.
Such algorithmic problems frequently appear in coding tests and interviews, so sufficient practice is required.