## [2461\. Maximum Sum of Distinct Subarrays With Length K](https://leetcode.com/problems/maximum-sum-of-distinct-subarrays-with-length-k/)

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`

#define ll long long
class Solution {
public:
    long long maximumSubarraySum(vector<int>& nums, int k) {
        int n = nums.size();
        ll res = 0, sum = 0;
        unordered_map<int, int> freq;
        for(int i = 0; i < k; i++) {
            ++freq[nums[i]];
            sum += nums[i];
        }

        // base case
        if(freq.size() == k)
            res = sum;

        int i = k;
        for(int i = k; i < n; i++) {
            ++freq[nums[i]];
            --freq[nums[i - k]];
            
            if(freq[nums[i - k]] == 0)
                freq.erase(nums[i - k]);
            
            sum += nums[i];
            sum -= nums[i - k];
            // 維持 freq 的長度要等於 3
            // 也就是 distinct element = 3
            if(freq.size() == k)
                res = max(res, sum);
        }
        return res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(1)$
:::