Category: C++ Coding Test Tutorials

Problem Description

Jumong’s command is one of the algorithm problems based on the situations he encounters during military operations.
Jumong wants to give specific orders to the soldiers in his unit. However, the soldiers may not fully understand Jumong’s commands.
Considering this situation, you need to solve the problem as described below.

Problem Definition

        Jumong is trying to issue commands to N soldiers. Each soldier can receive a command exactly once, and this command is 
        in the following format. Each soldier is numbered from 1 to N.

        Command: `i j k`
        
        - Meaning: Soldiers from soldier number i to soldier number j must follow the command,
        - Soldier number k can ignore it.

        If the number of soldiers following the command is greater than that of soldier number k, print "YES", 
        otherwise print "NO".

         N, M
         i, j, k
    

Solution Process

To solve this problem, you can approach it in the following steps.

Step 1: Understanding and Analyzing the Problem

The numbers of the soldiers and the range for executing the command are given to solve the problem.
The given i and j represent the range of soldiers who must follow the command, while k is a specific soldier within this range who will not follow the command.
Therefore, by checking the number of soldiers from i to j and whether soldier k is included, you can print “YES” or “NO”.

Step 2: Designing the Algorithm

You can use the following algorithm to solve the problem.

1. Input N and M.
2. Input the command M times.
3. For each command, perform the following tasks:
• Calculate the number of soldiers in the range [i, j].
• Check if soldier k is included in the range.
• If the number of soldiers following the command is greater than that of soldier k, print “YES”, otherwise print “NO”.

Step 3: Implementing C++ Code

Now let’s implement the algorithm designed above into C++ code.

        #include 
        using namespace std;

        int main() {
            int N, M;
            cin >> N >> M; // Input number of soldiers and commands

            for (int q = 0; q < M; q++) { // Repeat for each command
                int i, j, k;
                cin >> i >> j >> k; // Input command
                int command_count = j - i + 1; // Calculate number of soldiers from i to j

                if (command_count > 1) { // More than one soldier must be present.
                    if (k >= i && k <= j) {
                        cout << "NO" << endl; // Soldier k does not follow the command
                    } else {
                        cout << "YES" << endl; // Other soldiers excluding soldier k follow the command.
                    }
                } else {
                    cout << "NO" << endl; // If there is 1 or fewer soldiers following the command
                }
            }

            return 0;
        }
    

Step 4: Explaining the Code

I will provide a brief explanation of each part of the code.

• #include <iostream>: The input-output stream library of C++.
• using namespace std;: Uses the std namespace to enhance code readability.
• int N, M;: Integer variables that store the number of soldiers and commands, respectively.
• cin >> N >> M;: Input the number of soldiers and commands.
• for (int q = 0; q < M; q++): Repeats M times to process each command.
  • cin >> i >> j >> k;: Inputs i, j, k values for each command.
  • int command_count = j - i + 1;: Calculates the number of soldiers following the command.
  • if (command_count > 1): The count of following soldiers must be greater than 1 for the command to be valid.
    • if (k >= i && k <= j): Print “NO” if soldier k is included in the range.
    • cout << "YES" << endl;: Print “YES” if soldier k is outside the range.
  • Each command output moves to the next line using endl.

Step 5: Test Cases

To validate the accuracy of the program, several test cases must be considered. Below is an example of input and output.

        Input:
        5 3
        1 3 2
        1 5 5
        2 4 3

        Output:
        NO
        YES
        YES
    

Conclusion

In this C++ coding test course, we learned the algorithmic problem-solving process and C++ programming techniques through Jumong’s command problem.
The key to the problem was to accurately understand the command given and efficiently implement an algorithm to process it.
I would like to emphasize again the importance of considering various cases when solving a given problem.
The process of solving such problems can often help us significantly in solving real-world issues.

In the next course, we will deal with more complex algorithm problems. Thank you!

Hello, everyone! In this session, we will talk about preparing for coding tests using C++. Specifically, we will solve a problem to deepen our understanding of the Combination algorithm. Combinations are a very useful algorithm for finding ways to select items that meet specific conditions.

Definition of Combination

Combination refers to the way of selecting a specific number of elements from a given set. For example, when selecting 2 elements from the set {A, B, C}, the possible selections are {A, B}, {A, C}, and {B, C}. The important point here is that the order of the selected elements does not matter. This is a significant distinction from permutations.

Problem Description

Problem: Sum of Combinations

Given an integer array nums and an integer target, find and output all combinations that can sum up to target. Each number in the combination can be used multiple times, and combinations with the same numbers should not be included multiple times.

Input Example

nums = [2, 3, 6, 7]
target = 7

Output Example

[[2, 2, 3], [7]]

Problem Solving Process

1. Understanding the Problem

To solve the problem, we will use the Backtracking algorithm. Backtracking explores possible choices during the problem-solving process and, if a choice is invalid or does not lead to an optimal solution, goes back to a previous step to try another choice.

2. Algorithm Design

The algorithm for this problem consists of the following steps.

1. Add the current combination to the array.
2. If the sum of the combination equals target, add that combination to the result array.
3. If the current combination exceeds target, terminate the recursive call.
4. Add possible numbers to the combination and make a recursive call to include the same number.
5. Finally, remove the last selected number to return to the previous state.

3. Implementing the Code

Now, let’s look at the code implemented in C++.

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

void backtrack(vector<int> &combination, vector<vector<int>> &result, vector<int> &nums, int target, int start) {
    if (target == 0) {
        result.push_back(combination);
        return;
    }
    
    for (int i = start; i < nums.size(); i++) {
        if (nums[i] > target) continue; // Skip if the current number exceeds target
        combination.push_back(nums[i]); // Add number to the combination
        backtrack(combination, result, nums, target - nums[i], i); // Recursive call
        combination.pop_back(); // Remove the last number to backtrack
    }
}

vector<vector<int>> combinationSum(vector<int> &nums, int target) {
    vector<vector<int>> result;
    vector<int> combination;
    backtrack(combination, result, nums, target, 0);
    return result;
}

int main() {
    vector<int> nums = {2, 3, 6, 7};
    int target = 7;
    vector<vector<int>> result = combinationSum(nums, target);
    
    for (const auto &comb : result) {
        cout << "[ ";
        for (int num : comb) {
            cout << num << " ";
        }
        cout << "]" << endl;
    }
    
    return 0;
}

4. Explanation of the Code

The above code is a function called combinationSum that takes a given array of numbers and a target value, returning an array of all possible combinations. Inside the function, a recursive function called backtrack is called to generate combinations.

• backtrack function: This function stores the current combination in an array, updates the target value, and recursively explores new combinations.
• Conditional Statement: When the target is 0, the current combination is added to the result array, and if the target is less than the current number, the exploration is terminated.
• Adding Combinations: After adding the current number to the combination, the target value is reduced, and a recursive call is made.
• Removing Combinations: To backtrack, the last number is removed from the combination to return to the previous state.

5. Time Complexity Analysis

The time complexity of this problem is proportional to the number of combinations in the worst case. Therefore, it can be quite inefficient depending on the input size and number of possibilities. However, such algorithms can be very useful in many situations, so it is important to understand and practice them well.

6. Conclusion

Through this lecture, we learned to understand and solve combination problems in C++. Since combination problems can be utilized in various situations, it is important to understand the given context well and choose the appropriate algorithm. I hope you can further enhance your coding skills through various practice problems.

Additional Examples and Exercises

Try solving the additional examples below to practice more diverse combination problems:

• Example 1: What are the possible combinations for nums = [1, 2, 3] and target = 4?
• Example 2: What are the possible combinations for nums = [3, 5, 7] and target = 8?

It is important to continuously practice to enhance your coding skills. I hope you gain experience by solving various problems!

References

If you want more detailed algorithms and in-depth content about C++, please refer to the materials below:

• C++ Reference
• LeetCode
• GeeksforGeeks

We wish you the best in your coding studies!

1. Problem Definition

This is a problem to find out which pebbles would be the best choice when trying to take out m pebbles from a given n pebbles.
This problem can be solved using a Greedy Algorithm. The input consists of the weight of each pebble,
and the total weight of the pebbles taken out must not exceed K.

2. Understanding the Problem

Input:
– First line: Number of pebbles n (1 ≤ n ≤ 100)
– Second line: Number of pebbles to take out m (1 ≤ m ≤ n)
– Third line: Weights of n pebbles (1 ≤ weight ≤ 1000)
Output:
– Calculate the total weight of the taken pebbles such that it is maximized to K.

3. Example

Example Input:
5
3
1 2 3 4 5
Example Output:
12 (3 + 4 + 5)

4. Algorithm Design

To solve this problem, first, the weights of the given pebbles need to be sorted, and then the process of selecting the m heaviest pebbles should be performed.
At this point, the total weight of the selected pebbles should be calculated and displayed as the result. The steps of the algorithm are as follows.

• Sort the weights of the pebbles in ascending order.
• Select the last m pebbles from the sorted list.
• Sum the weights of the selected pebbles and display the result.

5. Code Implementation

Now, let’s implement the C++ code based on the above algorithm.

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

int main() {
    int n, m;
    std::cout << "Please enter the number of pebbles: ";
    std::cin >> n;
    std::cout << "Please enter the number of pebbles to take out: ";
    std::cin >> m;

    std::vector<int> weights(n);
    std::cout << "Please enter the weights of the pebbles (separated by spaces): ";
    for (int i = 0; i < n; ++i) {
        std::cin >> weights[i];
    }

    // Sort the weights of the pebbles
    std::sort(weights.begin(), weights.end());

    // Select the heaviest m pebbles
    int total_weight = 0;
    for (int i = n - m; i < n; ++i) {
        total_weight += weights[i];
    }

    std::cout << "Total weight of selected pebbles: " << total_weight << std::endl;

    return 0;
}

6. Code Explanation

The above C++ code takes input for the number of pebbles and the number of pebbles to be selected, and stores the weights of each pebble in an array.
Then, it uses the STL sort function to sort the weights. It selects the heaviest m pebbles from the sorted array to calculate and display the total weight.

7. Algorithm Complexity Analysis

The time complexity of this algorithm is O(n log n), where n is the number of pebbles. The sorting process takes the largest share of the computation time.
The space complexity is O(n), which is the memory needed to store the weights of the input pebbles.

8. Problem Solving Reflection

This problem is a good practice problem for learning the basics of greedy algorithms. Through the pebble selection problem,
one can learn methods for making optimized choices. Similar problems frequently appear in actual coding tests, so
repetitive practice is necessary. I hope you gain more experience by solving various problems in the future.

Hello, everyone! Today, we will delve into the topic of “Finding Non-Perfect Squares” through a coding test course using C++. This problem is very useful for understanding the basic concepts of algorithms and can frequently appear in actual coding interviews.

Problem Definition

There is a given integer array. The problem is to count how many perfect squares (like 2 squared, 3 squared, etc.) are in a specific range when extracting integers from that array. To be precise, it’s about finding non-perfect squares from the array.

Problem Description

Input:
- Integer n: Size of the array
- Integer array A[n]: An array consisting of n integers
- Integer m: Start of the range (inclusive)
- Integer p: End of the range (inclusive)

Output:
- Count of non-perfect squares

Example:
Input
n = 5
A = [1, 2, 3, 4, 5]
m = 1
p = 5

Output
3  // 2 and 3 are not perfect squares, while 1 and 4 are perfect squares.

Problem Solving Strategy

To solve the problem, we will follow these steps:

1. Create a function to check if a number is a perfect square by examining all numbers in the array.
2. Count the numbers that are non-perfect squares among the numbers in the given range [m, p].
3. Output the result.

Perfect Square Determination Function

To determine if a number is a perfect square, we can take the square root of each number, convert it to an integer, and then square it again to check if it equals the original number. In Python, you could use the following code:

bool isPerfectSquare(int x) {
    int s = sqrt(x);
    return (s * s == x);
}

In C++, you can use the cmath library to utilize the sqrt() function. To count non-perfect squares, you can use a for loop to check the numbers in the specified range.

C++ Code Implementation

Now, let’s implement the C++ code based on what we’ve discussed.

#include 
#include 
#include 
using namespace std;

bool isPerfectSquare(int x) {
    if (x < 0) return false; // Negative numbers are not perfect squares.
    int s = sqrt(x);
    return (s * s == x);
}

int countNonPerfectSquares(const vector& arr, int m, int p) {
    int count = 0;
    for (int num : arr) {
        if (num >= m && num <= p && !isPerfectSquare(num)) {
            count++;
        }
    }
    return count;
}

int main() {
    int n, m, p;
    cout << "Enter the size of the array (n): ";
    cin >> n;
    vector arr(n);
    
    cout << "Enter the elements of the array: ";
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }

    cout << "Enter the start of the range (m): ";
    cin >> m;
    cout << "Enter the end of the range (p): ";
    cin >> p;

    int result = countNonPerfectSquares(arr, m, p);
    cout << "Count of non-perfect squares: " << result << endl;

    return 0;
}

Code Explanation

The above C++ code works as follows:

1. It receives the size of the array and its elements from the user.
2. It accepts the start and end of the range.
3. It calls the countNonPerfectSquares() function to calculate the count of non-perfect squares in the given range.
4. It outputs the result.

Test Cases

Now, let’s run a few test cases to verify that the code works correctly.

Example 1:
Input:
5
1 2 3 4 5
1
5

Output:
3 // [2, 3, 5] are not perfect squares.

Example 2:
Input:
6
-1 0 1 2 3 4
0
4

Output:
3 // [0, 2, 3] are not perfect squares.

Conclusion

Today, we discussed the topic of “Finding Non-Perfect Squares” in our C++ coding test course. This problem was a good opportunity to understand the concepts of perfect and non-perfect squares and to implement logic for evaluating specific ranges within an array. I hope you can learn how to solve algorithmic problems through this code and also prepare for actual coding interviews.

I hope this article has been helpful to you. I look forward to seeing you next time with more interesting and informative topics!

Hello, everyone. In this lecture, we will discuss the algorithm problem “Making an Integer Equal to 1,” which can be implemented in C++. This problem is one of the types frequently encountered in coding tests for job preparation and requires various ways of thinking. Therefore, while solving this problem, you can enhance your basic algorithm design and trimming skills, as well as improve your C++ programming abilities.

Problem Description

Given an integer x, we want to make this integer equal to 1 using the following two operations:

• If x is even, divide x by 2.
• If x is odd, subtract 1 from x.

Only one operation can be performed at a time, and the goal is to find the minimum number of times to make x equal to 1 by repeating the above operations. The range of x is from 1 to 10⁶.

Input

x = 10

Output

result = 5

Approach to the Problem

To solve this problem, we need to calculate the minimum number of operations while transforming x to 1, either through a recursive approach or a loop. The following steps can be considered to implement the algorithm.

1. Basic Algorithm Design

Initially, determine whether x is odd or even and apply the appropriate operation based on that condition. Continue this process and increment the count until x becomes 1. While recursion can be used, a for loop might be simpler.

2. Time Complexity Analysis

The time complexity of this algorithm depends on the value of x. Since x can be up to 10^6, the operations will be performed fewer than 20 times at most. This is a very efficient method.

C++ Code Implementation

Now let’s implement the above algorithm in C++. The code below calculates the minimum number of operations required to make the given integer equal to 1.

#include 
using namespace std;

int makeOne(int x) {
    int count = 0; // Variable to count the number of operations
    while (x > 1) { // Continue looping while x is greater than 1
        if (x % 2 == 0) { // If x is even
            x /= 2; // Divide x by 2
        } else { // If x is odd
            x -= 1; // Subtract 1 from x
        }
        count++; // Increase operation count
    }
    return count; // Return the final operation count
}

int main() {
    int x;
    cout << "Enter an integer: ";
    cin >> x; // Get an integer input from the user
    int result = makeOne(x); // Call the makeOne function
    cout << "Minimum number of operations to make the integer equal to 1: " << result << endl; // Output the result
    return 0;
}

Code Explanation

The code above defines a function makeOne that takes the given integer x as an argument and calculates the minimum number of operations needed to make that integer equal to 1. Using a while loop, the operations are continuously performed while x is greater than 1, and the operation count is recorded at each iteration. As a result, it ultimately returns the total number of operations required to make x equal to 1.

Testing and Validation

Now, let's verify if the code works correctly with several test cases.

Enter an integer: 10
Minimum number of operations to make the integer equal to 1: 5

Enter an integer: 15
Minimum number of operations to make the integer equal to 1: 8

Enter an integer: 1
Minimum number of operations to make the integer equal to 1: 0

Conclusion

In this lecture, we have solved the algorithm problem "Making an Integer Equal to 1." Through this problem, we covered the basics of C++ syntax, algorithm design, and code implementation methods. It is essential to constantly think about various situations encountered during the process of solving algorithm problems and the ways to overcome them. I encourage you to solve various problems to enhance your problem-solving skills. I will return with beneficial content in the next lecture. Thank you!