LeetCode in Kotlin
2461. Maximum Sum of Distinct Subarrays With Length K
Medium
You are given an integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions:
- The length of the subarray is
k, and - All the elements of the subarray are distinct.
Return the maximum subarray sum of all the subarrays that meet the conditions__. If no subarray meets the conditions, return 0.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,5,4,2,9,9,9], k = 3
Output: 15
Explanation: The subarrays of nums with length 3 are:
-
[1,5,4] which meets the requirements and has a sum of 10.
-
[5,4,2] which meets the requirements and has a sum of 11.
-
[4,2,9] which meets the requirements and has a sum of 15.
-
[2,9,9] which does not meet the requirements because the element 9 is repeated.
-
[9,9,9] which does not meet the requirements because the element 9 is repeated.
We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions
Example 2:
Input: nums = [4,4,4], k = 3
Output: 0
Explanation: The subarrays of nums with length 3 are:
- [4,4,4] which does not meet the requirements because the element 4 is repeated. We return 0 because no subarrays meet the conditions.
Constraints:
1 <= k <= nums.length <= 1051 <= nums[i] <= 105
Solution
class Solution {
fun maximumSubarraySum(nums: IntArray, k: Int): Long {
val seen: MutableSet<Int> = HashSet()
var sum: Long = 0
var current: Long = 0
var i = 0
var j = 0
while (j < nums.size) {
while (seen.contains(nums[j])) {
val `val` = nums[i++]
seen.remove(`val`)
current -= `val`.toLong()
}
seen.add(nums[j])
current += nums[j].toLong()
j++
if (seen.size == k) {
sum = Math.max(sum, current)
current -= nums[i].toLong()
seen.remove(nums[i])
i++
}
}
return sum
}
}