The challenge
Given an array X of positive integers, its elements are to be transformed by running the following operation on them as many times as required:
if X[i] > X[j] then X[i] = X[i] - X[j]
When no more transformations are possible, return its sum (“smallest possible sum”).
For instance, the successive transformation of the elements of input X = [6, 9, 21] is detailed below:
X_1 = [6, 9, 12] # -> X_1[2] = X[2] - X[1] = 21 - 9
X_2 = [6, 9, 6] # -> X_2[2] = X_1[2] - X_1[0] = 12 - 6
X_3 = [6, 3, 6] # -> X_3[1] = X_2[1] - X_2[0] = 9 - 6
X_4 = [6, 3, 3] # -> X_4[2] = X_3[2] - X_3[1] = 6 - 3
X_5 = [3, 3, 3] # -> X_5[1] = X_4[0] - X_4[1] = 6 - 3
The returning output is the sum of the final transformation (here 9).
Example
solution([6, 9, 21]) #-> 9
Solution steps
[6, 9, 12] #-> X[2] = 21 - 9
[6, 9, 6] #-> X[2] = 12 - 6
[6, 3, 6] #-> X[1] = 9 - 6
[6, 3, 3] #-> X[2] = 6 - 3
[3, 3, 3] #-> X[1] = 6 - 3
Additional notes
There are performance tests consisted of very big numbers and arrays of size at least 30000. Please write an efficient algorithm to prevent timeout.
The solution in Kotlin
Option 1:
fun solution(n: LongArray): Long {
fun gcd(a: Long, b: Long): Long = if (b == 0L) a else gcd(b, a % b)
return n.toList().reduce { acc, l -> gcd(acc, l) } * n.count()
}
Option 2:
fun solution(numbers: LongArray): Long {
var gcdVal : Long = numbers[0]
for (i in numbers.indices) {
gcdVal = gcd(gcdVal, numbers[i])
}
return gcdVal * numbers.size
}
fun gcd(a: Long, b: Long): Long {
return if (a == 0L) b else gcd(b % a, a)
}
Option 3:
fun solution(numbers: LongArray): Long {
var array = numbers.toTypedArray()
while (!array.all { i -> i == array[0] } ) {
val min = array.min()!!
array = array.map { i ->
val mod = i % min
return@map if (mod != 0L) mod else min
}.toTypedArray()
}
return array.sum()
}
Test cases to validate our solution
import kotlin.test.assertEquals
import org.junit.Test
class TestExample {
@Test
fun `Basic tests`() {
assertEquals(9, solution(longArrayOf(6,9,21)))
assertEquals(3, solution(longArrayOf(1,21,55)))
assertEquals(5, solution(longArrayOf(3,13,23,7,83)))
assertEquals(923, solution(longArrayOf(71,71,71,71,71,71,71,71,71,71,71,71,71)))
assertEquals(22, solution(longArrayOf(11,22)))
assertEquals(2, solution(longArrayOf(5,17)))
assertEquals(12, solution(longArrayOf(4,16,24)))
assertEquals(9, solution(longArrayOf(9)))
}
}
