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2653. Sliding Subarray Beauty

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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.

  • 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.length 
  • 1 <= n <= 105
  • 1 <= k <= n
  • 1 <= 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

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