Contains Duplicate III Python Solution

lcpy gen -n 220   # by problem number
lcpy gen -s contains_duplicate_iii   # by problem name

Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output: true
Explanation: We can choose (i, j) = (0, 3).
i != j, abs(i - j) <= indexDiff, abs(nums[i] - nums[j]) <= valueDiff

Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
Output: false
Explanation: No pair of indices satisfies all three conditions.

class Solution:
    # Time: O(n)
    # Space: O(min(n, index_diff))
    def contains_nearby_almost_duplicate(
        self, nums: list[int], index_diff: int, value_diff: int
    ) -> bool:
        width = value_diff + 1
        buckets: dict[int, int] = {}
        for i, num in enumerate(nums):
            if i > index_diff:
                del buckets[nums[i - index_diff - 1] // width]
            bucket = num // width
            if bucket in buckets:
                return True
            if bucket - 1 in buckets and num - buckets[bucket - 1] <= value_diff:
                return True
            if bucket + 1 in buckets and buckets[bucket + 1] - num <= value_diff:
                return True
            buckets[bucket] = num
        return False