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3318. Find X-Sum of All K-Long Subarrays I
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
xmost 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:
1 <= n == nums.length <= 501 <= nums[i] <= 501 <= 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:
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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 int s; public int[] findXSum(int[] nums, int k, int x) { int n = nums.length; int[] ans = new int[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 += p[0] * p[1]; l.add(p); } while (l.size() > x) { var p = l.pollFirst(); s -= 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 -= 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 += p[0] * p[1]; } else { r.add(p); } } } -
class Solution { public: vector<int> findXSum(vector<int>& nums, int k, int x) { using pii = pair<int, int>; set<pii> l, r; int 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 += 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 -= p.first * p.second; l.erase(it); } else { r.erase(p); } }; vector<int> 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 += p.first * p.second; r.erase(p); l.insert(p); } while (l.size() > x) { pii p = *l.begin(); s -= 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; } }; -
from sortedcontainers import SortedList 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