<p>You are given an integer array <code>nums</code> and two integers <code>indexDiff</code> and <code>valueDiff</code>.</p>

<p>Find a pair of indices <code>(i, j)</code> such that:</p>

<ul>
	<li><code>i != j</code>,</li>
	<li><code>abs(i - j) &lt;= indexDiff</code>.</li>
	<li><code>abs(nums[i] - nums[j]) &lt;= valueDiff</code>, and</li>
</ul>

<p>Return <code>true</code><em> if such pair exists or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
<strong>Output:</strong> true
<strong>Explanation:</strong> We can choose (i, j) = (0, 3).
We satisfy the three conditions:
i != j --&gt; 0 != 3
abs(i - j) &lt;= indexDiff --&gt; abs(0 - 3) &lt;= 3
abs(nums[i] - nums[j]) &lt;= valueDiff --&gt; abs(1 - 1) &lt;= 0
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
<strong>Output:</strong> false
<strong>Explanation:</strong> After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>
	<li><code>0 &lt;= valueDiff &lt;= 10<sup>9</sup></code></li>
</ul>

from sortedcontainers import SortedList


class Solution:
    def containsNearbyAlmostDuplicate(self, nums: List[int], indexDiff: int, valueDiff: int) -> bool:
        """
            brute force : n * indexDiff

            balanced binary search tree : n * log (index diff)

            best solution:
            buckets: n
            
        """

        return bucket_search(nums, indexDiff, valueDiff)

        sorted_list = SortedList()
        for i, num in enumerate(nums):
            max_val = valueDiff + num
            min_val = -valueDiff + num
            # range from 0 to len() inclusive
            idx1 = sorted_list.bisect_left(min_val)
            if idx1 < len(sorted_list) and sorted_list[idx1] <= max_val:
                return True
            
            if i >= indexDiff:
                # remember remover
                sorted_list.remove(nums[i - indexDiff])
            
            sorted_list.add(num)
        
        return False




def bucket_search(nums, indexDiff, valueDiff):
    min_val, max_val = min(nums), max(nums)
    # case valueDiff 0
    bucket_size = valueDiff + 1 # [0, size)
    size = math.floor((max_val - min_val) // bucket_size) + 1
    buckets = [-math.inf] * size

    # [-inf, 4]
    for i, num in enumerate(nums):
        idx = (num - min_val) // bucket_size
        if buckets[idx] != -math.inf:
            return True
        prev, nex = idx - 1, idx + 1

        if prev >= 0 and buckets[prev] > -math.inf and num - buckets[prev] <= valueDiff or \
        nex < size and buckets[nex] > -math.inf and buckets[nex] - num <= valueDiff:
            return True
        
        if i >= indexDiff:
            removal = (nums[i - indexDiff] - min_val) // bucket_size
            buckets[removal] = -math.inf
        
        buckets[idx] = num
    
    return False

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