Maximum Sum of Distinct Subarrays With Length K
<p>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:</p>

<ul>
	<li>The length of the subarray is <code>k</code>, and</li>
	<li>All the elements of the subarray are <strong>distinct</strong>.</li>
</ul>

<p>Return <em>the maximum subarray sum of all the subarrays that meet the conditions</em><em>.</em> If no subarray meets the conditions, return <code>0</code>.</p>

<p><em>A <strong>subarray</strong> is a contiguous non-empty sequence of elements within an array.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,4,2,9,9,9], k = 3
<strong>Output:</strong> 15
<strong>Explanation:</strong> 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
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,4], k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Hash Table; Sliding Window
func maximumSubarraySum(nums []int, k int) (ans int64) {
	n := len(nums)
	cnt := map[int]int64{}
	var s int64
	for _, x := range nums[:k] {
		cnt[x]++
		s += int64(x)
	}
	if len(cnt) == k {
		ans = s
	}
	for i := k; i < n; i++ {
		cnt[nums[i]]++
		cnt[nums[i-k]]--
		if cnt[nums[i-k]] == 0 {
			delete(cnt, nums[i-k])
		}
		s += int64(nums[i] - nums[i-k])
		if len(cnt) == k && ans < s {
			ans = s
		}
	}
	return
}