LeetCode in Kotlin
3411. Maximum Subarray With Equal Products
Easy
You are given an array of positive integers nums.
An array arr is called product equivalent if prod(arr) == lcm(arr) * gcd(arr), where:
prod(arr)is the product of all elements ofarr.gcd(arr)is the GCD of all elements ofarr.lcm(arr)is the LCM of all elements ofarr.
Return the length of the longest product equivalent subarray of nums.
A subarray is a contiguous non-empty sequence of elements within an array.
The term gcd(a, b) denotes the greatest common divisor of a and b.
The term lcm(a, b) denotes the least common multiple of a and b.
Example 1:
Input: nums = [1,2,1,2,1,1,1]
Output: 5
Explanation:
The longest product equivalent subarray is [1, 2, 1, 1, 1], where prod([1, 2, 1, 1, 1]) = 2, gcd([1, 2, 1, 1, 1]) = 1, and lcm([1, 2, 1, 1, 1]) = 2.
Example 2:
Input: nums = [2,3,4,5,6]
Output: 3
Explanation:
The longest product equivalent subarray is [3, 4, 5].
Example 3:
Input: nums = [1,2,3,1,4,5,1]
Output: 5
Constraints:
2 <= nums.length <= 1001 <= nums[i] <= 10
Solution
import kotlin.math.max
class Solution {
private fun gcd(a: Int, b: Int): Int {
return if (b == 0) a else gcd(b, a % b)
}
private fun lcm(a: Int, b: Int): Int {
return (a / gcd(a, b)) * b
}
fun maxLength(nums: IntArray): Int {
val n = nums.size
var maxL = 0
for (i in 0..<n) {
var currGCD = nums[i]
var currLCM = nums[i]
var currPro = nums[i]
for (j in i + 1..<n) {
currPro *= nums[j]
currGCD = gcd(currGCD, nums[j])
currLCM = lcm(currLCM, nums[j])
if (currPro == currLCM * currGCD) {
maxL = max(maxL, j - i + 1)
}
}
}
return maxL
}
}