Class Solution
-
- All Implemented Interfaces:
public final class Solution3321 - Find X-Sum of All K-Long Subarrays II.
Hard
You are given an array
numsofnintegers and two integerskandx.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
xdistinct elements, its x-sum is the sum of the array.Return an integer array
answerof lengthn - k + 1whereanswer[i]is the x-sum of the subarraynums[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 subarraynums[i..i + k - 1].Constraints:
nums.length == n<code>1 <= n <= 10<sup>5</sup></code>
<code>1 <= numsi<= 10<sup>9</sup></code>
1 <= x <= k <= nums.length
-
-
Constructor Summary
Constructors Constructor Description Solution()
-