{"id":34174,"date":"2024-11-01T09:25:07","date_gmt":"2024-11-01T09:25:07","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34174"},"modified":"2024-11-01T10:58:19","modified_gmt":"2024-11-01T10:58:19","slug":"c-coding-test-course-exploring-debugging-use-cases-2","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34174\/","title":{"rendered":"C++ Coding Test Course, Exploring Debugging Use Cases"},"content":{"rendered":"

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Author: [Your Name]<\/p>\n

Publication Date: [Publication Date]<\/p>\n<\/header>\n

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1. Introduction to the Algorithm Problem<\/h2>\n

In this post, we will examine an algorithm problem that can be solved using C++. Through this problem, we will explore the basic syntax of C++ and the approach to algorithms, as well as experience the process of solving problems using debugging techniques.<\/p>\n

Problem: Two Sum<\/h3>\n

The problem is to return the indices of two numbers from a given integer array that add up to a target integer. If such a pair does not exist, -1 should be returned.<\/p>\n

Example Input:<\/strong><\/p>\n

\n            nums = [2, 7, 11, 15]\n            target = 9\n            <\/pre>\n
Example Output:<\/strong><\/p>\n
\n            [0, 1]\n            <\/pre>\n
Since the sum of the two numbers 2 and 7 is 9, we return indices 0 and 1.<\/p>\n<\/section>\n
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2. Problem Approach<\/h2>\n
This problem is somewhat intuitive. You can check all pairs using nested loops, but this method has a time complexity of O(n^2). A better approach is to use a hashmap. By iterating through each number and storing it in the hashmap while checking the difference between the target and the current number, this method can be solved with a time complexity of O(n).<\/p>\n<\/section>\n
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3. C++ Code Implementation<\/h2>\n
Now, let’s implement this problem in C++.<\/p>\n
\n            #include \n            #include \n            #include \n            using namespace std;\n\n            vector twoSum(vector& nums, int target) {\n                unordered_map map; \/\/ Hashmap to store numbers and their indices\n                vector result;\n\n                for (int i = 0; i < nums.size(); i++) {\n                    int complement = target - nums[i]; \/\/ The value of target minus the current number\n                    if (map.find(complement) != map.end()) { \/\/ Search\n                        result.push_back(map[complement]);\n                        result.push_back(i);\n                        return result; \/\/ Return the result\n                    }\n                    map[nums[i]] = i; \/\/ Store the current number and its index in the hashmap\n                }\n\n                return {-1}; \/\/ In case there are no pairs\n            }\n\n            int main() {\n                vector nums = {2, 7, 11, 15};\n                int target = 9;\n                vector result = twoSum(nums, target);\n\n                cout << \"[\" << result[0] << \", \" << result[1] << \"]\" << endl;\n                return 0;\n            }\n            <\/pre>\n<\/section>\n
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4. Code Explanation<\/h2>\n
The code above first takes the integer array nums<\/code> and the target value target<\/code> as input. It creates a hashmap map<\/code> to store each number and its index using unordered_map<\/code>. Then, while iterating through the array:<\/p>\n