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
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:
1 <= n == nums.length <= 50
1 <= nums[i] <= 50
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 $\textit{nums}$.
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