# 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 <= 105`
• `1 <= nums[i] <= 105`

### 解題

func maximumSubarraySum(nums []int, k int) int64 {
m := make(map[int]int)
sum := 0
max := 0
for i := 0; i < len(nums); i++ {
sum += nums[i]
m[nums[i]]++
if i >= k - 1{
if len(m) == k && sum > max {
max = sum
}
sum -= nums[i - k + 1]
m[nums[i - k + 1]]--
if m[nums[i - k + 1]] == 0 {
delete(m, nums[i - k + 1])
}
}
}
return int64(max)
}