{"id":34348,"date":"2024-11-01T09:27:07","date_gmt":"2024-11-01T09:27:07","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34348"},"modified":"2024-11-01T10:57:36","modified_gmt":"2024-11-01T10:57:36","slug":"c-coding-test-course-preparing-for-departure","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34348\/","title":{"rendered":"C++ Coding Test Course, Preparing for Departure"},"content":{"rendered":"

<\/p>\n

Hello everyone! In this post, we will explore coding tests for those preparing for employment with C++.<\/p>\n

The Importance of Coding Tests<\/h2>\n

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.<\/p>\n

Problem Description<\/h2>\n

Problem: Sum of Two Numbers<\/h3>\n

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<\/code> and an integer target value target<\/code>.<\/p>\n

Input Example<\/h3>\n
\n    nums = [2, 7, 11, 15]\n    target = 9<\/pre>\n
Output Example<\/h3>\n
\n    [0, 1]<\/pre>\n
Approaches to Solve the Problem<\/h2>\n
There are several approaches to solve this problem. Here, we will explain two approaches: Brute Force<\/strong> and Using a Hashmap<\/strong>.<\/p>\n
1. Brute Force Approach<\/h3>\n
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).<\/p>\n
\n    int twoSum(vector& nums, int target) {\n        for (int i = 0; i < nums.size(); i++) {\n            for (int j = i + 1; j < nums.size(); j++) {\n                if (nums[i] + nums[j] == target) {\n                    return {i, j};\n                }\n            }\n        }\n        return {};\n    }<\/int><\/pre>\n
2. Using a Hashmap<\/h3>\n
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).<\/p>\n
\n    vector twoSum(vector& nums, int target) {\n        unordered_map num_map;\n        for (int i = 0; i < nums.size(); i++) {\n            int complement = target - nums[i];\n            if (num_map.find(complement) != num_map.end()) {\n                return {num_map[complement], i}; \/\/ Found the two numbers\n            }\n            num_map[nums[i]] = i; \/\/ Store the number with its index\n        }\n        return {};\n    }<\/int,><\/int><\/int><\/pre>\n
Problem Solving Process<\/h2>\n
Now, let’s solve the problem based on the above two approaches.<\/p>\n
Brute Force Method<\/h3>\n
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.<\/p>\n