{"id":34918,"date":"2024-11-01T09:33:37","date_gmt":"2024-11-01T09:33:37","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34918"},"modified":"2024-11-01T11:45:44","modified_gmt":"2024-11-01T11:45:44","slug":"kotlin-coding-test-course-022-sorting-numbers-3","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34918\/","title":{"rendered":"kotlin coding test course, 022 sorting numbers 3"},"content":{"rendered":"

<\/p>\n

Problem Description<\/h2>\n

\n This problem is an algorithmic problem of sorting a given sequence. Specifically,
\n you need to write a program that receives N<\/strong> integers as input, sorts
\n them in ascending order, and then prints each integer on a separate line.\n <\/p>\n

Input Format<\/h2>\n

\n The first line contains an integer N<\/strong> (1 \u2264 N \u2264 1,000,000).
\n The second line contains N integers, each of which is
\n between -1,000,000,000<\/strong> and
\n 1,000,000,000<\/strong>.\n <\/p>\n

Output Format<\/h2>\n

\n Sort the input integers in ascending order and print each integer on
\n a separate line.\n <\/p>\n

Approach to the Problem<\/h2>\n

\n To solve this problem, you can use an algorithm to sort the array.
\n Kotlin provides a powerful library function for array sorting by default.
\n However, due to the large size of numbers, attention must be paid to time complexity.
\n A special sorting algorithm should be applied for optimal performance.\n <\/p>\n

Algorithm Selection<\/h2>\n

\n Given that the input size of the problem can go up to 1,000,000,
\n it is advisable to use efficient sorting algorithms like quicksort or
\n mergesort, which have a time complexity of O(N log N).
\n Additionally, in Kotlin, you can easily sort by utilizing the
\n sort()<\/code> method from the standard library.\n <\/p>\n

Code Implementation<\/h2>\n

Below is the solution code using Kotlin:<\/p>\n

\nfun main() {\n    val n = readLine()!!.toInt()  \/\/ Input N from the first line\n    val numbers = IntArray(n)      \/\/ Create an integer array\n\n    for (i in 0 until n) {\n        numbers[i] = readLine()!!.toInt()  \/\/ Input the next N integers\n    }\n\n    numbers.sort()  \/\/ Perform sorting\n\n    numbers.forEach { println(it) }  \/\/ Output the sorted numbers\n}\n    <\/code><\/pre>\n
Code Explanation<\/h2>\n
\n        1. Use readLine()!!.toInt()<\/code> to read integers from standard input.
\n        Read the value of N from the first line and store all subsequent N integers in the
\n        numbers<\/code> array.\n    <\/p>\n
\n        2. Use the numbers.sort()<\/code> method to sort
\n        the array in ascending order. This method uses
\n        an efficient sorting algorithm internally.\n    <\/p>\n
\n        3. Finally, print the sorted numbers line by line using
\n        forEach<\/code>.\n    <\/p>\n
Example of Results<\/h2>\n
\n        Below is an example input and output after running this program:\n    <\/p>\n
Input:<\/strong><\/p>\n
\n5\n5\n2\n3\n1\n4\n    <\/code><\/pre>\n
Output:<\/strong><\/p>\n
\n1\n2\n3\n4\n5\n    <\/code><\/pre>\n
Memory and Time Complexity<\/h2>\n
\n        This algorithm has a time complexity of O(N log N). This is because
\n        it sorts by dividing the input array size logarithmically.
\n        Memory usage requires O(N) to store the array.\n    <\/p>\n
Summary of Problem-Solving Process<\/h2>\n