LeetCode in Kotlin
2873. Maximum Value of an Ordered Triplet I
Easy
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 <= 1001 <= nums[i] <= 106
Solution
import kotlin.math.max
class Solution {
fun maximumTripletValue(nums: IntArray): Long {
val n = nums.size
val iNumMaxs = IntArray(n)
var prev = 0
for (i in 0 until n) {
if (nums[i] <= prev) {
iNumMaxs[i] = prev
} else {
iNumMaxs[i] = nums[i]
prev = iNumMaxs[i]
}
}
var result: Long = 0
var kNumMax = nums[n - 1]
for (j in n - 2 downTo 1) {
result = max(
result.toDouble(),
((iNumMaxs[j - 1] - nums[j]).toLong() * kNumMax).toDouble()
).toLong()
if (nums[j] > kNumMax) {
kNumMax = nums[j]
}
}
return result
}
}