README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3321.Find%20X-Sum%20of%20All%20K-Long%20Subarrays%20II/README_EN.md	2598	Weekly Contest 419 Q4	Array Hash Table Sliding Window Heap (Priority Queue)

3321. Find X-Sum of All K-Long Subarrays II

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Description

You are given an array nums of n integers and two integers k and x.

The x-sum of an array is calculated by the following procedure:

• Count the occurrences of all elements in the array.
• Keep only the occurrences of the top x most frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent.
• Calculate the sum of the resulting array.

Note that if an array has less than x distinct elements, its x-sum is the sum of the array.

Return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1].

 

Example 1:

Input: nums = [1,1,2,2,3,4,2,3], k = 6, x = 2

Output: [6,10,12]

Explanation:

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

Example 2:

Input: nums = [3,8,7,8,7,5], k = 2, x = 2

Output: [11,15,15,15,12]

Explanation:

Since k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1].

 

Constraints:

• nums.length == n
• 1 <= n <= 105
• 1 <= nums[i] <= 109
• 1 <= x <= k <= nums.length

Solutions

Solution 1: Hash Table + Ordered Set

We use a hash table $\textit{cnt}$ to count the occurrences of each element in the window, an ordered set $\textit{l}$ to store the $x$ elements with the highest occurrences in the window, and another ordered set $\textit{r}$ to store the remaining elements.

We maintain a variable $\textit{s}$ to represent the sum of the elements in $\textit{l}$. Initially, we add the first $k$ elements to the window, update the ordered sets $\textit{l}$ and $\textit{r}$, and calculate the value of $\textit{s}$. If the size of $\textit{l}$ is less than $x$ and $\textit{r}$ is not empty, we repeatedly move the largest element from $\textit{r}$ to $\textit{l}$ until the size of $\textit{l}$ equals $x$, updating the value of $\textit{s}$ in the process. If the size of $\textit{l}$ is greater than $x$, we repeatedly move the smallest element from $\textit{l}$ to $\textit{r}$ until the size of $\textit{l}$ equals $x$, updating the value of $\textit{s}$ in the process. At this point, we can calculate the current window's $\textit{x-sum}$ and add it to the answer array. Then we remove the left boundary element of the window, update $\textit{cnt}$, and update the ordered sets $\textit{l}$ and $\textit{r}$, as well as the value of $\textit{s}$. Continue traversing the array until the traversal is complete.

The time complexity is $O(n \times \log k)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

Similar problems:

• 3013. Divide an Array Into Subarrays With Minimum Cost II

Python3

class Solution:
    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
        def add(v: int):
            if cnt[v] == 0:
                return
            p = (cnt[v], v)
            if l and p > l[0]:
                nonlocal s
                s += p[0] * p[1]
                l.add(p)
            else:
                r.add(p)

        def remove(v: int):
            if cnt[v] == 0:
                return
            p = (cnt[v], v)
            if p in l:
                nonlocal s
                s -= p[0] * p[1]
                l.remove(p)
            else:
                r.remove(p)

        l = SortedList()
        r = SortedList()
        cnt = Counter()
        s = 0
        n = len(nums)
        ans = [0] * (n - k + 1)
        for i, v in enumerate(nums):
            remove(v)
            cnt[v] += 1
            add(v)
            j = i - k + 1
            if j < 0:
                continue
            while r and len(l) < x:
                p = r.pop()
                l.add(p)
                s += p[0] * p[1]
            while len(l) > x:
                p = l.pop(0)
                s -= p[0] * p[1]
                r.add(p)
            ans[j] = s

            remove(nums[j])
            cnt[nums[j]] -= 1
            add(nums[j])
        return ans

Java

class Solution {
    private TreeSet<int[]> l = new TreeSet<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
    private TreeSet<int[]> r = new TreeSet<>(l.comparator());
    private Map<Integer, Integer> cnt = new HashMap<>();
    private long s;

    public long[] findXSum(int[] nums, int k, int x) {
        int n = nums.length;
        long[] ans = new long[n - k + 1];
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            remove(v);
            cnt.merge(v, 1, Integer::sum);
            add(v);
            int j = i - k + 1;
            if (j < 0) {
                continue;
            }
            while (!r.isEmpty() && l.size() < x) {
                var p = r.pollLast();
                s += 1L * p[0] * p[1];
                l.add(p);
            }
            while (l.size() > x) {
                var p = l.pollFirst();
                s -= 1L * p[0] * p[1];
                r.add(p);
            }
            ans[j] = s;

            remove(nums[j]);
            cnt.merge(nums[j], -1, Integer::sum);
            add(nums[j]);
        }
        return ans;
    }

    private void remove(int v) {
        if (!cnt.containsKey(v)) {
            return;
        }
        var p = new int[] {cnt.get(v), v};
        if (l.contains(p)) {
            l.remove(p);
            s -= 1L * p[0] * p[1];
        } else {
            r.remove(p);
        }
    }

    private void add(int v) {
        if (!cnt.containsKey(v)) {
            return;
        }
        var p = new int[] {cnt.get(v), v};
        if (!l.isEmpty() && l.comparator().compare(l.first(), p) < 0) {
            l.add(p);
            s += 1L * p[0] * p[1];
        } else {
            r.add(p);
        }
    }
}

C++

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;
    }
};

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