LeetCode in Kotlin
1793. Maximum Score of a Good Subarray
Hard
You are given an array of integers nums (0-indexed) and an integer k.
The score of a subarray (i, j) is defined as min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1). A good subarray is a subarray where i <= k <= j.
Return the maximum possible score of a good subarray.
Example 1:
Input: nums = [1,4,3,7,4,5], k = 3
Output: 15
Explanation: The optimal subarray is (1, 5) with a score of min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15.
Example 2:
Input: nums = [5,5,4,5,4,1,1,1], k = 0
Output: 20
Explanation: The optimal subarray is (0, 4) with a score of min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 2 * 1040 <= k < nums.length
Solution
class Solution {
fun maximumScore(nums: IntArray, k: Int): Int {
var i = k
var j = k
var res = nums[k]
var min = nums[k]
var goLeft: Boolean
while (i >= 1 || j < nums.size - 1) {
// sub array [i...j] is already traversed. Either goLeft or goRight to increase the
// sequence
goLeft = if (i == 0) {
false
} else if (j == nums.size - 1) {
true
} else {
nums[j + 1] <= nums[i - 1]
}
min = if (goLeft) Math.min(min, nums[i - 1]) else Math.min(min, nums[j + 1])
if (goLeft) {
while (i >= 1 && min <= nums[i - 1]) {
i--
}
} else {
while (j < nums.size - 1 && min <= nums[j + 1]) {
j++
}
}
res = Math.max(res, min * (j - i + 1))
}
return res
}
}