Description
Given an integer array nums containing n integers, find the beauty of each subarray of size k.
The beauty of a subarray is the xth smallest integer in the subarray if it is negative, or 0 if there are fewer than x negative integers.
Return an integer array containing n - k + 1 integers, which denote the beauty of the subarrays in order from the first index in the array.
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A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,-1,-3,-2,3], k = 3, x = 2 Output: [-1,-2,-2] Explanation: There are 3 subarrays with size k = 3. The first subarray is[1, -1, -3]and the 2nd smallest negative integer is -1. The second subarray is[-1, -3, -2]and the 2nd smallest negative integer is -2. The third subarray is[-3, -2, 3]and the 2nd smallest negative integer is -2.
Example 2:
Input: nums = [-1,-2,-3,-4,-5], k = 2, x = 2 Output: [-1,-2,-3,-4] Explanation: There are 4 subarrays with size k = 2. For[-1, -2], the 2nd smallest negative integer is -1. For[-2, -3], the 2nd smallest negative integer is -2. For[-3, -4], the 2nd smallest negative integer is -3. For[-4, -5], the 2nd smallest negative integer is -4.
Example 3:
Input: nums = [-3,1,2,-3,0,-3], k = 2, x = 1 Output: [-3,0,-3,-3,-3] Explanation: There are 5 subarrays with size k = 2. For[-3, 1], the 1st smallest negative integer is -3. For[1, 2], there is no negative integer so the beauty is 0. For[2, -3], the 1st smallest negative integer is -3. For[-3, 0], the 1st smallest negative integer is -3. For[0, -3], the 1st smallest negative integer is -3.
Constraints:
n == nums.length1 <= n <= 1051 <= k <= n1 <= x <= k-50 <= nums[i] <= 50
Solution
Python3
from sortedcontainers import SortedList
class Solution:
def getSubarrayBeauty(self, nums: List[int], k: int, x: int) -> List[int]:
N = len(nums)
sl = SortedList()
res = []
for i, v in enumerate(nums):
if v < 0:
sl.add((v, i))
if i + 1 >= k:
if len(sl) == 0 or len(sl) < x:
res.append(0)
else:
res.append(sl[x - 1][0])
prevIndex = i - k + 1
if nums[prevIndex] < 0:
sl.remove((nums[prevIndex], prevIndex))
return res