C++ Coding Test Course, Stack and Queue

Let’s explore stacks and queues, which are one of the data structures frequently encountered in coding tests. Both data structures differ in how they store and process elements and can be selectively used depending on specific problems. In this course, we will look into the basic concepts of stacks and queues, algorithm problems utilizing these structures, and how to implement these problems in C++.

Stack

A stack is a structure for storing data based on the principle of Last In First Out (LIFO). This means that the most recently added data will be the first to be removed. The stack supports two main operations:

  • push: Adds data to the top of the stack.
  • pop: Removes data from the top of the stack and returns that data.

Examples of Stack Utilization

Stacks are useful for tasks like parentheses checking, reversing sets, and implementing backtracking functionality.

Queue

A queue is a structure for storing data based on the principle of First In First Out (FIFO). This means that the data that enters first will leave first. The main operations for queues are similarly as follows:

  • enqueue: Adds data to the rear of the queue.
  • dequeue: Removes data from the front of the queue and returns that data.

Examples of Queue Utilization

Queues are used in print job management, process scheduling, and implementing BFS (Breadth-First Search) algorithms.

Problem: Valid Parentheses String Check

This problem involves using a stack to check if the parentheses in a given string are valid. The definition of a valid parentheses string is as follows:

  • All opened parentheses must have a corresponding closing parenthesis.
  • Opened parentheses must not appear after their closing parentheses.

Problem Description

Given a string, you need to check if it is a valid parentheses string. The following cases can be considered “valid”:

  • “()”
  • “(())”
  • “()()”
  • “(()(()))”

Conversely, the following cases can be considered “invalid”:

  • “)(“
  • “(()”
  • “())”
  • “(()))”

Solution Process

To solve this problem, we will use a stack. Using the basic principle of a stack, we will add opened parentheses to the stack, and when a closed parenthesis appears, we will pop the opened parenthesis from the stack.

  1. Create an empty stack.
  2. Read the input string one character at a time.
  3. If the read character is ‘(‘, push it onto the stack.
  4. If the read character is ‘)’, check if the stack is not empty, then perform a pop operation on the stack. If the stack is empty, the string is invalid.
  5. After reading the entire string, if the stack is empty, the string is valid. If it’s not empty, the string is invalid.

C++ Code Implementation

The C++ code that implements the above algorithm is as follows:


#include <iostream>
#include <stack>
#include <string>

bool isValidParentheses(const std::string& str) {
    std::stack s;
    
    for (char c : str) {
        if (c == '(') {
            s.push(c);
        } else if (c == ')') {
            // If the stack is empty, it's an invalid parenthesis
            if (s.empty()) {
                return false;
            }
            s.pop(); // In a valid case, remove the opened parenthesis
        }
    }
    
    // The stack must be empty at the end to be valid
    return s.empty();
}

int main() {
    std::string input;
    std::cout << "Enter the parentheses string: ";
    std::cin >> input;

    if (isValidParentheses(input)) {
        std::cout << "It is a valid parentheses string." << std::endl;
    } else {
        std::cout << "It is an invalid parentheses string." << std::endl;
    }
    
    return 0;
}

Result Testing

You can test various input strings with the above code. For example:

  • Input: () – Output: “It is a valid parentheses string.”
  • Input: (()) – Output: “It is a valid parentheses string.”
  • Input: ()) – Output: “It is an invalid parentheses string.”
  • Input: (())) – Output: “It is an invalid parentheses string.”

Summary and Tips

Stacks and queues play important roles in various algorithm problems. Stacks are primarily used for problems requiring LIFO structure, while queues are used for problems requiring FIFO structure. Through the problem-solving processes discussed in this course, I hope you have learned how to utilize stacks more effectively.

Tip: Make sure to clearly understand the concepts of stacks and queues and consider the time complexity of operations as you tackle various problems. It is necessary to try diverse and complex problems to find your own optimal solutions.

C++ Coding Test Course, Finding the Sum of Numbers

Hello! In this course, we will solve the problem of finding the sum of numbers using C++. This course is designed to build basic to intermediate algorithmic problem-solving skills in C++. We will cover problem definition, input and output, implementation methods, example solutions, optimization strategies, and finally, anticipated questions and answers. So, shall we begin?

Problem Definition

Write a program to calculate the sum of the integers from 1 to a given natural number N.

For example, if N is 5, the output should be 15, which is the result of 1 + 2 + 3 + 4 + 5.

Input and Output

Input

A single integer N (1 ≤ N ≤ 100,000) is given.

Output

The sum of the integers up to N is printed in one line.

Problem-Solving Process

Step 1: Understanding the Algorithm

To find the sum of numbers, you can generally use a loop, adding each number while iterating up to the given natural number N. However, using a mathematical formula can provide a more efficient solution.

Step 2: Mathematical Approach

The formula for the sum from 1 to N is as follows:

Sum = N * (N + 1) / 2

Using this formula, we can solve the problem with O(1) time complexity. The space complexity also remains O(1).

Step 3: Implementing C++ Code

Now let’s move on to the implementation step. Below is the C++ code to solve the problem.


#include <iostream>
using namespace std;

int main() {
    int N;
    cout << "Please enter the value of N: ";
    cin >> N;

    // Sum calculation using the formula
    long long sum = static_cast(N) * (N + 1) / 2;
    
    cout << "The sum of integers from 1 to " << N << " is: " << sum << "." << endl;

    return 0;
}

        

Step 4: Example Input and Output

After carefully reviewing the code, let’s check if the results are correct with a few examples.

Example 1:

Input: 5

Output: The sum of integers from 1 to 5 is: 15.

Example 2:

Input: 10

Output: The sum of integers from 1 to 10 is: 55.

Example 3:

Input: 100

Output: The sum of integers from 1 to 100 is: 5050.

Optimization Strategies

This problem has been solved using a mathematical formula with O(1), so no further optimization is necessary. However, if the range of input data is reduced, using loops to calculate the sum may become inefficient.

For example, if N is very large, using a loop can take a long time and may result in a timeout error. The mathematical formula is especially useful in resolving such issues.

Anticipated Questions and Answers

Q1: What if N is negative?

A1: The conditions for this problem restrict N to be a natural number greater than or equal to 1, so negative input will not be provided. It’s advisable to thoroughly check the input range.

Q2: Can this be implemented in languages other than C++?

A2: Yes, the algorithm for calculating the sum of numbers can be similarly applied in other programming languages. However, the code needs to be adjusted to fit the syntax of each language.

Q3: Are there no variations of this problem?

A3: This problem is a basic sum problem. If you want a variation, you could consider problems that sum only even or odd numbers, for example.

Conclusion

Through this course, we learned how to calculate the sum of numbers using C++. While it seems like a simple problem, I hope this has helped you understand the importance of selecting an efficient algorithm. In the next course, we will cover more complex problems, so please look forward to it!

C++ coding test course, finding the order of permutations

Hello, everyone! Today, we will take a deep dive into the Finding the Order of Permutations problem, which frequently appears in C++ coding tests. This problem will be very helpful in understanding the concept of permutations and developing combinatorial problem-solving skills. Permutations refer to all possible arrangements of a given set of numbers, and problems utilizing this concept are applied in various fields.

Problem Description

We will explore the concept of permutations and C++ implementation through the process of solving a problem as follows:

Problem: Given integers N and K, find the K-th permutation among all permutations that can be made using the numbers from 1 to N. Assume K starts from 1.

Input Format

  • The first line contains integers N (1 ≤ N ≤ 10) and K (1 ≤ K ≤ N!).

Output Format

Output the K-th permutation on a single line, separated by spaces.

Example

Input:
3 2

Output:
2 1 3

Approach to the Problem

To solve this problem, we can approach it with the following steps:

  1. Understand the basic concept of permutations.
  2. Design a method to find the K-th permutation when N numbers are given.
  3. Implement the method of generating permutations in code.
  4. Return the K-th permutation.

Method of Generating Permutations

There are various methods for generating permutations; here we will use the method utilizing factorials. The permutations of N numbers yield N! combinations, and the algorithm for generating these combinations helps to efficiently calculate the K-th permutation directly.

Factorial Calculation

The factorial N! is defined as N * (N-1) * (N-2) * … * 1. For example, 3! = 6. Through this, we can calculate the number of possible choices for selecting each number given N.

A Little More Specifically

When listing the numbers from 1 to N, the first number can be chosen based on K. For instance, if K is 2, we find the K-th permutation by considering the number of ways to choose the next number based on each chosen first number.

C++ Code Implementation

Now, let’s write the C++ code based on the above approach. This code will output the K-th permutation using the given N and K.


#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// Factorial calculation function
long long factorial(int n) {
    if (n == 0) return 1;
    long long result = 1;
    for (int i = 1; i <= n; i++) {
        result *= i;
    }
    return result;
}

// Function to find the K-th permutation
vector<int> getKthPermutation(int n, int k) {
    vector<int> numbers;
    for (int i = 1; i <= n; i++) {
        numbers.push_back(i);
    }
    
    k--; // 0-indexing
    vector<int> permutation;

    for (int i = 0; i < n; i++) {
        long long fact = factorial(n - 1 - i);
        int index = k / fact;
        permutation.push_back(numbers[index]);
        numbers.erase(numbers.begin() + index); // Remove the selected number from the list
        k %= fact;
    }

    return permutation;
}

int main() {
    int n, k;
    cin >> n >> k;

    vector<int> kthPermutation = getKthPermutation(n, k);
    
    for (int num : kthPermutation) {
        cout << num << " ";
    }
    
    return 0;
}
    

Code Explanation

  1. Factorial Function: Calculates and returns the factorial of the given N.
  2. K-th Permutation Function: Stores numbers from 1 to N in a vector and iterates to find the K-th permutation.
  3. Calculates the number of possible scenarios at each step to decide which number to select.
  4. The selected number is removed from the list, and K is updated for the next number selection.

Testing and Result Verification

In the above code, input N and K, and check if it outputs the K-th permutation correctly. For example, entering N=3 and K=2 should output 2 1 3.

Conclusion

Today, we learned about how to find the order of permutations using C++. This algorithm can enhance understanding of permutations and help solve various algorithmic problems. Since it is a commonly tested type in coding interviews, make sure to practice sufficiently.

Thank you!

C++ Coding Test Course, Making the Maximum Value by Grouping Numbers

Problem Description

You need to combine the given integers appropriately to create one maximum value while receiving multiple integers. The integers follow these rules:

  • Integers can include negative and positive values, and 0 also exists.
  • Multiplying a negative with another negative creates a positive, allowing for larger numbers to be made.
  • Positive numbers should only be multiplied with numbers greater than 1, otherwise, their sum will be larger. 1 is used for addition.
  • 0 does not affect the maximum value but can be multiplied by negative numbers.

For example, if the given integers are {1, 2, 0, -1, -2}, the maximum value would be
(2 * 1) + (0) + ((-1) * (-2)) = 2 + 0 + 2 = 4.

Input

The first line contains an integer N (1 ≤ N ≤ 100,000),
and the second line contains N integers arr[i] (-1,000 ≤ arr[i] ≤ 1,000).

Output

Output the maximum value of the integers in one line.

Problem Solving Process

Step 1: Reading Input

Read the input integers and convert them into a format that the computer can understand. In C++,
vector is used to store the integers.
All values are processed by storing them in one list.

Step 2: Separating Integers

First, separate the integers into positive, negative, and 0. This allows for various cases
to be handled efficiently. The goal is to group negative numbers to create positives,
and combine positives to form the maximum value.

Step 3: Handling Positives

Sort the list of positive numbers and start multiplying the larger values at the end first. Only
values greater than 1 should be combined, so it’s best to group with 1.

Step 4: Handling Negatives

Handle negatives by combining them in pairs to create the maximum value. For example,
multiplying two negatives yields a positive, so they should be combined.

Step 5: Calculating and Outputting the Result

Ultimately, store the maximum value obtained through all combinations and output it.
This process can be easily implemented using the C++ code below.

Code Example

            
                #include <iostream>
                #include <vector>
                #include <algorithm>

                using namespace std;

                int main() {
                    int N;
                    cin >> N;

                    vector arr(N);
                    vector positive;
                    vector negative;

                    for (int i = 0; i < N; i++) {
                        cin >> arr[i];
                        if (arr[i] > 0) positive.push_back(arr[i]);
                        else if (arr[i] < 0) negative.push_back(arr[i]);
                    }

                    sort(positive.rbegin(), positive.rend());
                    sort(negative.begin(), negative.end());

                    long long result = 0;

                    for (int i = 0; i < positive.size(); i += 2) {
                        if (i + 1 < positive.size()) {
                            result += positive[i] * positive[i + 1];
                        } else {
                            result += positive[i];
                        }
                    }

                    for (int i = 0; i < negative.size(); i += 2) {
                        if (i + 1 < negative.size()) {
                            result += negative[i] * negative[i + 1];
                        }
                    }

                    cout << result << endl;

                    return 0;
                }
            
        

Conclusion

This problem is an important algorithmic challenge of creating maximum value from the given integers.
The approach above helps to effectively solve the problem.
The key is classifying integers and deriving results through appropriate mathematical operations. Additionally, learning to write code using C++’s STL can be applied to other problems as well.

C++ Coding Test Course, Sorting Numbers 2

Problem Description

The “Sorting Numbers 2” problem is about sorting the given sequence of numbers and outputting it. This problem aims to understand and implement common sorting algorithms.
Given N numbers, the task is to sort this sequence in ascending order and output the result. The numbers are integers that are greater than or equal to -1000 and less than or equal to 1000.

Input Format

The first line contains the number of elements N (1 ≤ N ≤ 100,000).
Starting from the second line, N lines will contain the numbers.

Output Format

Print the sorted numbers, one per line.

Example Input

5
5
4
3
2
1
    

Example Output

1
2
3
4
5
    

Problem Solving Process

1. Understanding the Problem

This problem is a basic number sorting problem, and various sorting algorithms can be applied.
In C++, the standard library’s sorting function can be used, providing an efficient solution.
However, implementing sorting algorithms manually is also a good practice, so I will introduce two methods: selection sort and insertion sort.

2. Basic Data Collection and Storage

To receive input data, an array or vector can be used. In C++, using a vector is common.
Here is how to store data using a vector.


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

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

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

3. Sorting Algorithms

Let’s implement two methods for sorting the vector that contains the input numbers, namely selection sort and insertion sort.

3-1. Selection Sort

Selection sort selects the smallest (or largest) element from the given array and moves it to the front. This is repeated until sorting is complete.
The time complexity of selection sort is O(N^2).


void selectionSort(vector<int> &arr) {
    int n = arr.size();
    for (int i = 0; i < n - 1; i++) {
        int minIndex = i;
        for (int j = i + 1; j < n; j++) {
            if (arr[j] < arr[minIndex]) {
                minIndex = j;
            }
        }
        swap(arr[i], arr[minIndex]); // Swap elements
    }
}
    

3-2. Insertion Sort

Insertion sort inserts each element of the array into its appropriate position to sort the array.
The time complexity of insertion sort is O(N^2), which is efficient when the data is nearly sorted.


void insertionSort(vector<int> &arr) {
    int n = arr.size();
    for (int i = 1; i < n; i++) {
        int key = arr[i];
        int j = i - 1;
        while (j >= 0 && arr[j] > key) {
            arr[j + 1] = arr[j]; // Move elements
            j--;
        }
        arr[j + 1] = key; // Insert
    }
}
    

4. Final Code

Let’s complete the final code by choosing one of the sorting algorithms above. Below is a method using the vector and the standard library’s sort function.


#include <iostream>
#include <vector>
#include <algorithm> // Includes sort() function
using namespace std;

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

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

    sort(numbers.begin(), numbers.end()); // Perform sorting

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

    return 0;
}
    

5. Time Complexity

By using the standard library’s sort function as shown above, sorting can be done with an average time complexity of O(N log N).
This is much more efficient than selection sort and insertion sort.

6. Conclusion

Through the “Sorting Numbers 2” problem, we practiced the basic concepts of sorting algorithms and how to use C++ vectors.
Understanding and implementing various sorting algorithms is a very useful process for improving programming skills.
It will enhance understanding of sorting algorithms commonly used in practice and lay the groundwork for solving more complex algorithms.