LeetCode in Kotlin
1425. Constrained Subsequence Sum
Hard
Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied.
A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.
Example 1:
Input: nums = [10,2,-10,5,20], k = 2
Output: 37
Explanation: The subsequence is [10, 2, 5, 20].
Example 2:
Input: nums = [-1,-2,-3], k = 1
Output: -1
Explanation: The subsequence must be non-empty, so we choose the largest number.
Example 3:
Input: nums = [10,-2,-10,-5,20], k = 2
Output: 23
Explanation: The subsequence is [10, -2, -5, 20].
Constraints:
1 <= k <= nums.length <= 105-104 <= nums[i] <= 104
Solution
class Solution {
fun constrainedSubsetSum(nums: IntArray, k: Int): Int {
val n = nums.size
var res = Int.MIN_VALUE
val mono = ArrayList<IntArray>()
for (i in 0 until n) {
var take = nums[i]
while (mono.isNotEmpty() && i - mono.first()[0] > k) {
mono.removeFirst()
}
if (mono.isNotEmpty()) {
val mx = Math.max(0, mono.first()[1])
take += mx
}
while (mono.isNotEmpty() && take > mono.last()[1]) {
mono.removeLast()
}
mono.add(intArrayOf(i, take))
res = Math.max(res, take)
}
return res
}
}