Kotlin coding test course, calculating the remainder sum

Problem Description

You are given an array A consisting of N integers and an integer M, and you need to find the sum of the remainders of all elements in array A when divided by M. This will enhance your understanding of the properties of remainders and how to handle arrays efficiently, as well as improve your ability to use loops and conditionals in Kotlin.

Input Format

• The first line contains the size of the array N (1 ≤ N ≤ 10^6) and the integer M (1 ≤ M ≤ 1000).
• The second line contains N integers A₁, A₂, …, A_N. (0 ≤ A_i ≤ 10⁹)

Output Format

Print the sum of the remainders of all elements in array A when divided by M.

Example

    Input:
    5 3
    1 2 3 4 5

    Output:
    1
    

Problem Solving Process

To solve this problem, we can approach it by dividing each element of array A by M and summing all the remainders. This approach has a time complexity of O(N), making it very efficient and feasible within the given constraints.

Step-by-Step Procedure

• Store the values of N and M that are received as input.
• Receive the elements of array A as input.
• Calculate the remainders of all elements divided by M and obtain their sum.
• Print the result.

Kotlin Code Implementation

The following code is a Kotlin program that implements all the above procedures:

    fun main() {
        // Input
        val (n, m) = readLine()!!.split(" ").map { it.toInt() }
        val a = readLine()!!.split(" ").map { it.toInt() }

        // Calculate remainder sum
        var remainderSum = 0
        for (i in 0 until n) {
            remainderSum += a[i] % m
        }

        // Print result
        println(remainderSum)
    }
    

Code Explanation

In the above code, readLine() is used to receive input from the user. The first line inputs the array size N and M, while the second line inputs the elements of array A. Then, a for loop is used to calculate the remainders of all elements in A when divided by M and accumulates them in the remainderSum variable. Finally, the sum of the calculated remainders is printed.

Complexity Analysis

The time complexity of this problem is O(N) since all elements of the array are visited once. Furthermore, the space complexity is O(1), as only one additional variable is used to store the result, making it very efficient.

Optimization

The structure of the given problem is approached in the most optimized manner, and there is no need for further optimization. This is because all elements are iterated through only once to obtain the result.

Conclusion

This course has taught us how to efficiently handle arrays through the problem of calculating the sum of remainders, as well as the basic input/output and loop usage in Kotlin. Practicing such foundational problems will greatly help in building the basics for coding tests. I hope you continue to solve various problems and enhance your skills.