LeetCode in Kotlin
2874. Maximum Value of an Ordered Triplet II
Medium
You are given a 0-indexed integer array nums.
Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.
The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].
Example 1:
Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77. It can be shown that there are no ordered triplets of indices with a value greater than 77.
Example 2:
Input: nums = [1,10,3,4,19]
Output: 133
Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133. It can be shown that there are no ordered triplets of indices with a value greater than 133.
Example 3:
Input: nums = [1,2,3]
Output: 0
Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
Constraints:
3 <= nums.length <= 1051 <= nums[i] <= 106
Solution
import kotlin.math.max
class Solution {
fun maximumTripletValue(nums: IntArray): Long {
val diff = IntArray(nums.size)
var tempMax = nums[0]
for (i in 1 until diff.size - 1) {
diff[i] = tempMax - nums[i]
tempMax = max(tempMax.toDouble(), nums[i].toDouble()).toInt()
}
var max = Long.MIN_VALUE
tempMax = nums[nums.size - 1]
for (i in nums.size - 2 downTo 1) {
max = max(max.toDouble(), (tempMax.toLong() * diff[i]).toDouble()).toLong()
tempMax = max(tempMax.toDouble(), nums[i].toDouble()).toInt()
}
return max(max.toDouble(), 0.0).toLong()
}
}