Kotlin Coding Test Course, Understanding Time Complexity Notation

In coding testing, solving problems goes beyond simply presenting the correct answer; it is important to understand how efficiently that answer can be computed. Time complexity is a key factor in analyzing the performance of an algorithm. In this post, we will solve a practical algorithm problem using Kotlin and analyze the time complexity of this problem.

Problem: Find the Second Largest Number in Two Integer Arrays

The following is the problem given to us:

Problem Description: Given two integer arrays, return the second largest number among the elements of these two arrays. Both arrays have the same size and all elements are distinct.

Input

• First array: [3, 1, 4, 5, 9]
• Second array: [2, 6, 5, 3, 7]

Output

Second largest number from the two arrays: 7

Approach

To solve this problem, we need to combine all the elements of the two arrays and sort them, then find the second largest number. However, simply sorting can lead to a time complexity of O(n log n). Therefore, we need to think of a method that can solve it with a time complexity of O(n).

Algorithm Design

1. Combine the two arrays into a single list.

2. Find the maximum value in the list and determine the second largest value that is not equal to the maximum.

This approach finds the maximum value while iterating through the list once, and then finds the second maximum value, resulting in a time complexity of O(n).

Code Implementation

fun findSecondLargest(array1: IntArray, array2: IntArray): Int? {
    val combinedArray = array1 + array2
    var max = Int.MIN_VALUE
    var secondMax = Int.MIN_VALUE

    for (number in combinedArray) {
        if (number > max) {
            secondMax = max
            max = number
        } else if (number > secondMax && number != max) {
            secondMax = number
        }
    }
    return if (secondMax != Int.MIN_VALUE) secondMax else null
}

// Example usage
fun main() {
    val array1 = intArrayOf(3, 1, 4, 5, 9)
    val array2 = intArrayOf(2, 6, 5, 3, 7)

    val result = findSecondLargest(array1, array2)
    println("Second largest number: $result")
}

Time Complexity Analysis

The above code creates a new array by combining the two arrays and finds the maximum and second maximum values by traversing it once. Therefore, the total time complexity is calculated as follows:

• Array merging: O(n)
• Finding the maximum value: O(n)

Thus, the overall time complexity is O(n).

Conclusion

In this post, we learned how to solve algorithm problems using Kotlin and how to analyze time complexity efficiently. When designing algorithms, it is crucial to consider optimal performance, and mastering this approach will help in solving more complex problems.

I hope to continue exploring various Kotlin coding test problems and deepen my understanding of algorithms.