{"id":34556,"date":"2024-11-01T09:29:21","date_gmt":"2024-11-01T09:29:21","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34556"},"modified":"2024-11-01T11:40:46","modified_gmt":"2024-11-01T11:40:46","slug":"javascript-coding-test-course-making-an-integer-1","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34556\/","title":{"rendered":"Javascript Coding Test Course, Making an Integer 1"},"content":{"rendered":"<\/p>\n
Hello! In this course, we will discuss one of the JavaScript coding test problems, “Making an Integer Equal to 1”. This problem can be solved using various algorithmic techniques, and we will explore the process of coding and approaches together. I will provide an in-depth explanation of the problem approach, algorithm implementation, test cases, and optimization methods.<\/p>\n
Problem Description<\/h2>\n
Given an integer x<\/code>, find the minimum number of operations required to make this integer equal to 1 by performing one of the following two operations:<\/p>\n
\n
If it\u2019s even: x \u2192 x \/ 2<\/code><\/li>\n
If it\u2019s odd: x \u2192 x - 1<\/code><\/li>\n<\/ul>\n
For example, if x = 8<\/code>, it can be made into 1 through the following process:<\/p>\n
\n8 \u2192 4 \u2192 2 \u2192 1\n <\/pre>\n
In this case, the minimum number of operations is 3.<\/p>\n
Approach<\/h2>\n
To solve this problem, a state-based approach can be utilized. We can use BFS (Breadth-First Search) or DFS (Depth-First Search) to find the optimal path to make the given integer x<\/code> equal to 1. However, BFS is more suitable for this problem.<\/p>\n
The reason for using BFS is that there often are multiple paths to reach each state. BFS explores all possible paths at each step, enabling us to efficiently find the optimal path.<\/p>\n
BFS Algorithm Implementation<\/h2>\n
Now, let’s write the JavaScript code to solve the problem using the BFS algorithm. The code is as follows:<\/p>\n
\nfunction minStepsToOne(x) {\n const queue = [];\n const visited = new Set();\n \n queue.push(x);\n visited.add(x);\n let steps = 0;\n \n while (queue.length > 0) {\n const size = queue.length; \/\/ Number of nodes at the current level\n \n for (let i = 0; i < size; i++) {\n const current = queue.shift();\n \/\/ If the current number is 1, return steps\n if (current === 1) {\n return steps;\n }\n \n \/\/ If it's even\n if (current % 2 === 0 && !visited.has(current \/ 2)) {\n queue.push(current \/ 2);\n visited.add(current \/ 2);\n }\n \/\/ If it's odd\n if (current > 1 && !visited.has(current - 1)) {\n queue.push(current - 1);\n visited.add(current - 1);\n }\n }\n steps++; \/\/ End of one level\n }\n \n return steps;\n}\n\n\/\/ Example call\nconsole.log(minStepsToOne(8)); \/\/ Output: 3\n <\/code><\/pre>\n
Code Explanation<\/h2>\n
Now, let me explain the main points of the algorithm used in the code:<\/p>\n
\n
Using a Queue:<\/strong> BFS is implemented using the queue data structure. Each state is added to the queue and processed one by one.<\/li>\n
Visited Tracking:<\/strong> To prevent duplicate visits, the visited states are recorded in a Set<\/code> data structure.<\/li>\n
Step Counting:<\/strong> The number of steps is incremented every time we proceed to a new BFS level to keep track of the total number of operations.<\/li>\n<\/ul>\n
Test Cases<\/h2>\n
It\u2019s important to consider various test cases during the problem-solving process. Here are a few test cases:<\/p>\n
Though we can achieve efficient results with the BFS algorithm mentioned above, additional optimizations are possible. For instance, memoization can be used when storing states. This approach saves previously computed values to reduce unnecessary redundant calculations.<\/p>\n
Conclusion<\/h2>\n
In this course, we’ve solved the problem of “Making an Integer Equal to 1” using JavaScript. We discussed the algorithm utilizing BFS, the implementation of the code, test cases, and optimization methods. I hope you can showcase these problem-solving skills in your next coding test.<\/p>\n