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
class Solution {
    public long maximumSubarraySum(int[] nums, int k) {
        int n = nums.length;
        Map<Integer, Integer> cnt = new HashMap<>(k);
        long s = 0;
        for (int i = 0; i < k; ++i) {
            cnt.merge(nums[i], 1, Integer::sum);
            s += nums[i];
        }
        long ans = cnt.size() == k ? s : 0;
        for (int i = k; i < n; ++i) {
            cnt.merge(nums[i], 1, Integer::sum);
            if (cnt.merge(nums[i - k], -1, Integer::sum) == 0) {
                cnt.remove(nums[i - k]);
            }
            s += nums[i] - nums[i - k];
            if (cnt.size() == k) {
                ans = Math.max(ans, s);
            }
        }
        return ans;
    }
}