動機

sliding window但是用heap去找window的最大最小

Problem

Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit.

 

Example 1:

Input: nums = [8,2,4,7], limit = 4Output: 2 Explanation: All subarrays are: [8] with maximum absolute diff |8-8| = 0 <= 4.[8,2] with maximum absolute diff |8-2| = 6 > 4. [8,2,4] with maximum absolute diff |8-2| = 6 > 4.[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.[2] with maximum absolute diff |2-2| = 0 <= 4.[2,4] with maximum absolute diff |2-4| = 2 <= 4.[2,4,7] with maximum absolute diff |2-7| = 5 > 4.[4] with maximum absolute diff |4-4| = 0 <= 4.[4,7] with maximum absolute diff |4-7| = 3 <= 4.[7] with maximum absolute diff |7-7| = 0 <= 4. Therefore, the size of the longest subarray is 2.

Example 2:

Input: nums = [10,1,2,4,7,2], limit = 5Output: 4 Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.

Example 3:

Input: nums = [4,2,2,2,4,4,2,2], limit = 0Output: 3

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 109
  • 0 <= limit <= 109

Sol

from sortedcontainers import SortedList
class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        win = SortedList()
        
        i, ret = 0, 0
        for j,n in enumerate(nums):
            win.add(n)
            
            while i<=j and win[-1]-win[0] > limit:
                win.remove(nums[i])
                i += 1
            ret = max(ret, j-i+1)
        return ret