Find X-Sum of All K-Long Subarrays II
<p>You are given an array <code>nums</code> of <code>n</code> integers and two integers <code>k</code> and <code>x</code>.</p>

<p>The <strong>x-sum</strong> of an array is calculated by the following procedure:</p>

<ul>
	<li>Count the occurrences of all elements in the array.</li>
	<li>Keep only the occurrences of the top <code>x</code> most frequent elements. If two elements have the same number of occurrences, the element with the <strong>bigger</strong> value is considered more frequent.</li>
	<li>Calculate the sum of the resulting array.</li>
</ul>

<p><strong>Note</strong> that if an array has less than <code>x</code> distinct elements, its <strong>x-sum</strong> is the sum of the array.</p>

<p>Return an integer array <code>answer</code> of length <code>n - k + 1</code> where <code>answer[i]</code> is the <strong>x-sum</strong> of the <span data-keyword="subarray-nonempty">subarray</span> <code>nums[i..i + k - 1]</code>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,1,2,2,3,4,2,3], k = 6, x = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[6,10,12]</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>For subarray <code>[1, 1, 2, 2, 3, 4]</code>, only elements 1 and 2 will be kept in the resulting array. Hence, <code>answer[0] = 1 + 1 + 2 + 2</code>.</li>
	<li>For subarray <code>[1, 2, 2, 3, 4, 2]</code>, only elements 2 and 4 will be kept in the resulting array. Hence, <code>answer[1] = 2 + 2 + 2 + 4</code>. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.</li>
	<li>For subarray <code>[2, 2, 3, 4, 2, 3]</code>, only elements 2 and 3 are kept in the resulting array. Hence, <code>answer[2] = 2 + 2 + 2 + 3 + 3</code>.</li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [3,8,7,8,7,5], k = 2, x = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[11,15,15,15,12]</span></p>

<p><strong>Explanation:</strong></p>

<p>Since <code>k == x</code>, <code>answer[i]</code> is equal to the sum of the subarray <code>nums[i..i + k - 1]</code>.</p>
</div>

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

<ul>
	<li><code>nums.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= x &lt;= k &lt;= nums.length</code></li>
</ul>
Array; Hash Table; Sliding Window; Heap (Priority Queue)
class Solution {
public:
    vector<long long> findXSum(vector<int>& nums, int k, int x) {
        using pii = pair<int, int>;
        set<pii> l, r;
        long long s = 0;
        unordered_map<int, int> cnt;
        auto add = [&](int v) {
            if (cnt[v] == 0) {
                return;
            }
            pii p = {cnt[v], v};
            if (!l.empty() && p > *l.begin()) {
                s += 1LL * p.first * p.second;
                l.insert(p);
            } else {
                r.insert(p);
            }
        };
        auto remove = [&](int v) {
            if (cnt[v] == 0) {
                return;
            }
            pii p = {cnt[v], v};
            auto it = l.find(p);
            if (it != l.end()) {
                s -= 1LL * p.first * p.second;
                l.erase(it);
            } else {
                r.erase(p);
            }
        };
        vector<long long> ans;
        for (int i = 0; i < nums.size(); ++i) {
            remove(nums[i]);
            ++cnt[nums[i]];
            add(nums[i]);

            int j = i - k + 1;
            if (j < 0) {
                continue;
            }

            while (!r.empty() && l.size() < x) {
                pii p = *r.rbegin();
                s += 1LL * p.first * p.second;
                r.erase(p);
                l.insert(p);
            }
            while (l.size() > x) {
                pii p = *l.begin();
                s -= 1LL * p.first * p.second;
                l.erase(p);
                r.insert(p);
            }
            ans.push_back(s);

            remove(nums[j]);
            --cnt[nums[j]];
            add(nums[j]);
        }
        return ans;
    }
};
