Category: C++ Coding Test Tutorials

One of the most important things when preparing for coding tests is understanding and optimizing problem-solving skills along with time complexity. This article will detail the process of solving problems considering time complexity using C++.

Problem: Longest Increasing Subsequence

This problem involves finding the length of the longest increasing subsequence in a given integer array.

Problem Description

Given an integer array nums, return the length of the longest increasing subsequence. A subsequence is formed by selected elements from the array that maintain their original order but do not need to be contiguous.

Example

    Input: nums = [10, 9, 2, 5, 3, 7, 101, 18]
    Output: 4
    Explanation: The longest increasing subsequence is [2, 3, 7, 101], with a length of 4.
    

Approach

This problem can be solved using a combination of Dynamic Programming and Binary Search. Considering time complexity, the following approaches can be utilized:

1. Approach using Dynamic Programming

First, we record the length of the increasing subsequence that ends with each element. Then, we can iterate through each element and compare it with all previous elements to update the length of the subsequence if possible.

2. Combined with Binary Search

Ultimately, we use an array to record the end elements of subsequences and apply binary search on this array to find the position to insert the current element. This method allows us to achieve more optimized performance.

Implementation

Below is the code to solve this problem using C++:

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

using namespace std;

int lengthOfLIS(vector<int>& nums) {
    if (nums.empty()) return 0;

    vector<int> dp;
    for (int num : nums) {
        auto it = lower_bound(dp.begin(), dp.end(), num);
        if (it == dp.end()) {
            dp.push_back(num);
        } else {
            *it = num;
        }
    }
    return dp.size();
}

int main() {
    vector<int> nums = {10, 9, 2, 5, 3, 7, 101, 18};
    cout << "Length of Longest Increasing Subsequence: " << lengthOfLIS(nums) << endl;
    return 0;
}
    

Time Complexity Analysis

The time complexity of this problem is O(n log n). A binary search is performed for each element to find the appropriate position, and the time complexity of binary search is log n. Therefore, the overall time complexity is O(n log n), and the space complexity is O(n).

Conclusion

For problems like finding the longest increasing subsequence in an array, it is important to combine Dynamic Programming and Binary Search to reduce time complexity. Utilizing useful functions and data structures provided by C++’s STL can help write more efficient code.

This article explained time complexity analysis and algorithm problem-solving methods using C++. I hope it will be helpful in your future coding test preparation!

Introduction

Coding tests play an important role in modern software engineering. Many companies conduct coding tests to assess job candidates’ algorithmic problem-solving skills, and in this process, the C++ language has gained popularity due to its efficiency and various features. This post provides a guide on what algorithms to use through algorithm problems using C++.

Problem Description

Here is the algorithm problem we will solve. Let’s take the Two Sum problem as an example.

Problem: Two Sum

Given an integer array nums and an integer target, return the indices of the two numbers such that they add up to target. You may not use the same element twice, and you can assume that each input would have exactly one solution.

Example:
    Input: nums = [2, 7, 11, 15], target = 9
    Output: [0, 1]
    Explanation: nums[0] + nums[1] = 2 + 7 = 9, so return [0, 1].
    

Problem Solving Strategy

To solve this problem, we can use several approaches:

1. Brute Force

The simplest way is to try all possible combinations. This method uses a double loop to check all possible pairs of numbers. The time complexity is O(n^2).

2. Using a Hash Map

By using a hash map to store each number as a key and its index as a value, we can execute it more efficiently. We traverse the array only once, calculating target - nums[i] for each element and check whether this value exists in the hash map. The time complexity is O(n).

3. Sorting and Binary Search

After sorting the numbers, we can use binary search, but the sorting process takes additional time, increasing the time complexity to O(n log n). Since this method is not the most efficient, we will choose the hash map approach.

Implementation

Now, let’s implement the optimal solution using a hash map in C++.

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

vector twoSum(vector& nums, int target) {
    unordered_map num_map; // Hash map to store numbers and their indices.
    vector result; // Vector to store results.

    for (int i = 0; i < nums.size(); i++) {
        int complement = target - nums[i]; // Calculate the required complement.
        if (num_map.find(complement) != num_map.end()) {
            result.push_back(num_map[complement]); // Add the index.
            result.push_back(i); // Add the current index.
            return result; // Return the result.
        }
        num_map[nums[i]] = i; // Add the number to the hash map.
    }
    return result; // Return an empty vector if no answer is found.
}

int main() {
    vector nums = {2, 7, 11, 15};
    int target = 9;
    vector result = twoSum(nums, target);
    
    cout << "[";
    for (int i = 0; i < result.size(); i++) {
        cout << result[i];
        if (i < result.size() - 1) cout << ", ";
    }
    cout << "]" << endl;
    return 0;
}
    

Code Explanation

The provided code has the following structure:

1. Including Header Files: Necessary header files are included. iostream is for input-output streams, vector is for dynamic arrays, and unordered_map is needed for using the hash map.
2. Function Definition: The twoSum function takes an integer array and a target value as arguments. It defines a hash map and traverses all elements of the array.
3. Checking using Hash Map: In each iteration, it calculates the required value and updates the result if it exists in the hash map.
4. Main Function: The main function defines the input values, calls the twoSum function, and outputs the result.

Time Complexity Analysis

The worst-case time complexity of this solution is O(n), where n is the size of the array. This is because adding and searching each element in the hash map takes an average of O(1) time. The additional space used is O(n) for the hash map.

Conclusion

Choosing the appropriate algorithm is very important when solving coding test problems in C++. In particular, analyzing the characteristics of the problem and the size of the data is crucial. In this post, we learned the optimal solution using a hash map through the Two Sum problem. This approach helps to solve many coding test problems that are straightforward yet efficient.

Additional Learning Resources

If you want to learn more advanced topics for solving algorithm problems, here are some resources to recommend:

• GeeksforGeeks – Covers various algorithms and problem-solving techniques.
• LeetCode – Offers a variety of coding test problems and topics.
• HackerRank – A platform to learn by solving various types of algorithms and problems.

Final Thoughts

Learning to solve algorithm problems requires continuous effort and practice. Trying to solve various problems and finding optimal solutions will greatly help improve your skills. Based on a deep understanding of C++ and algorithms, I wish you success in your coding tests!

1. Introduction: The Importance of Algorithms

In modern software development, algorithms are an essential component. Especially in coding tests, they serve as a crucial criterion for evaluating candidates’ problem-solving abilities. Therefore, enhancing algorithmic knowledge and problem-solving skills is very important. In this article, we will practice C++ coding through actual coding test problems and learn about time complexity notation in detail.

2. Problem: Sum of Two Numbers

The problem is as follows: Given an integer array nums and an integer target, implement a function twoSum that returns the indices of the two numbers such that they add up to target. Note that there is exactly one solution for each input, and you may not use the same element twice.

Function Signature:
vector twoSum(vector& nums, int target);

Example

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

Approach to Solving the Problem

This problem involves finding two numbers in an array that add up to a specific sum, and there are various approaches to tackle it. The simplest method is to use nested loops, while a more efficient approach is to use a hash map. We will compare the time complexities of each method and look for the optimal solution.

2.1. Brute Force Approach

The simplest approach is to compare all possible pairs of two numbers in the array nums to check if their sum equals target.
The time complexity of this method is O(n^2) because it requires n*(n-1)/2 operations to explore all pairs through two nested loops, where n is the length of the array.

vector 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 {}; // No solution
    }

2.2. Approach Using Hash Map

To solve this problem more efficiently, we can use a hash map. By maintaining a record of how many times each number appears, we can quickly find the needed value for the current number while iterating.
The time complexity of this method is O(n) because it requires a single pass through the array while updating the hash map, and finding the needed value takes constant time O(1).

vector twoSum(vector& nums, int target) {
        unordered_map numMap;
        for (int i = 0; i < nums.size(); i++) {
            int complement = target - nums[i];
            if (numMap.find(complement) != numMap.end()) {
                return {numMap[complement], i};
            }
            numMap[nums[i]] = i;
        }
        return {}; // No solution
    }

3. Understanding Time Complexity

Time complexity is a crucial factor when evaluating the efficiency of algorithms. There are several methods for analyzing time complexity, but commonly used notations include O(n), O(log n), and O(n^2). These indicate how many operations an algorithm performs based on the input size n.

• O(1): Constant time complexity – An algorithm that takes a fixed time regardless of input size.
• O(n): Linear time complexity – The running time of the function increases proportionally with input size.
• O(log n): Logarithmic time complexity – An efficient algorithm that increases relatively slowly as the input size increases.
• O(n^2): Quadratic time complexity – The time taken grows quadratically as the input size increases.

4. Analysis Results

For this problem, we chose an efficient algorithm using a hash map. Thanks to this approach, we managed to reduce the time complexity to O(n), allowing for more efficient processing as the amount of data increases. When selecting algorithms, it is essential to consider time complexity to choose the best one.

5. Conclusion

Solving algorithmic problems is a crucial aspect of coding tests, and understanding various approaches to tackling these problems is important. We can leverage data structures like hash maps to solve problems more efficiently.
Furthermore, understanding time complexity is essential when comparing the performance of algorithms. This knowledge will help you take the first step toward becoming a better developer.

6. Additional Resources

C++ is a language well-suited for implementing various data structures and algorithms. You can gain a deeper understanding through the following resources:

• C++ Tutorial
• GeeksforGeeks
• LeetCode

Hello! Today, we will conduct a coding test problem-solving course using C++. In this course, we will explore a problem of arranging given numbers in ascending order using a data structure known as a stack. The stack follows a LIFO (Last In First Out) structure, and we will investigate how to create a sequence that meets specific requirements using this property.

Problem Description

You need to sort the numbers of a given sequence in ascending order using a stack. Every time a specific number comes in, you must examine the current state of the stack and, if possible, create an ascending sequence. If it is impossible, an appropriate message should be displayed.

Input

• The first line contains an integer N (1 ≤ N ≤ 100,000). (N is the number of integers in the sequence)
• The second line contains N integers, i.e., the sequence. (1 ≤ each integer in the sequence ≤ 1,000,000)

Output

• If the sequence can be made in ascending order, print ‘YES’.
• If the sequence cannot be made in ascending order, print ‘NO’.

Example Input

        5
        4 3 6 8 5
        

Example Output

        NO
        

Solution Process

To solve this problem, we need to effectively utilize the main properties of a stack. We will read the numbers from the left in order and push them into the stack, checking whether the top number of the stack matches the number we need to sort currently. Below is a step-by-step explanation of this process.

Step 1: Input Processing

First, we read the number of input numbers and receive the sequence accordingly. We plan to push this sequence into the stack in order.

Step 2: Sequence Manipulation Using Stack

Every time a number is input, we need to perform two tasks.

• Push the current number onto the stack.
• If the top number of the stack is the next number in our target ascending sequence, pop it from the stack and output it as a result.

Step 3: Handling Impossible Cases

If we processed all the input sequence but were unable to output the entire target sequence, print ‘NO’ and terminate the process.

Step 4: Code Implementation

Now, let’s implement this process in C++ code.

#include <iostream>
#include <stack>
#include <vector>

using namespace std;

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

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

    stack s;
    vector result;

    int next = 1; // The number to be sorted in ascending order
    for (int i = 0; i < N; i++) {
        s.push(sequence[i]); // Push the current number onto the stack.

        // Check if the top of the stack matches next.
        while (!s.empty() && s.top() == next) {
            result.push_back(s.top()); // Add the top element of the stack to the result.
            s.pop(); // Pop the top from the stack.
            next++; // Proceed to the next number.
        }
    }

    // Check if all numbers have been sorted.
    if (result.size() == N) {
        cout << "YES" << endl; // Successfully created an ascending sequence.
    } else {
        cout << "NO" << endl; // Failed.
    }

    return 0;
}
        

Conclusion

In this course, we learned how to sort given numbers in ascending order using a stack. By leveraging the characteristics of the stack, we could manipulate numbers in real-time and check if the conditions were satisfied. Understanding data structures is crucial in coding test problems, so I encourage you to build your skills through various problems in the future.

Further Progress

There are various problems that utilize stacks. Next time, we will tackle problems using queues or other data structures. Thank you for your interest!