# LC 2461. Maximum Sum of Distinct Subarrays With Length K

### [Problem link](https://leetcode.com/problems/maximum-sum-of-distinct-subarrays-with-length-k/)

###### tags: `leedcode` `medium` `c++` `Sliding Window`

You are given an integer array <code>nums</code> and an integer <code>k</code>. Find the maximum subarray sum of all the subarrays of <code>nums</code> that meet the following conditions:

- The length of the subarray is <code>k</code>, 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 <code>0</code>.

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:**

- <code>1 <= k <= nums.length <= 10⁵</code>
- <code>1 <= nums[i] <= 10⁵</code>

## Solution 1 - Sliding Window
#### C++

class Solution {
public:
    long long maximumSubarraySum(vector<int>& nums, int k) {
        long ans = 0;
        long sum = 0;
        unordered_map<int, int> freq;

        for (int i = 0; i < nums.size(); i++) {
            sum += nums[i];
            freq[nums[i]]++;
            if (i < k - 1) {
                continue;
            }
            if (freq.size() == k) {
                ans = max(ans, sum);
            }
            sum -= nums[i - k + 1];
            if (--freq[nums[i - k + 1]] == 0) {
                freq.erase(nums[i - k + 1]);
            }
        }
        return ans;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
x