<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>

class Solution:
    def containsNearbyAlmostDuplicate(self, nums: List[int], indexDiff: int, valueDiff: int) -> bool:
        # keep track of index ordering i.e., queue/ordered dict => can remove value out of bounds
        # keep values ordered so can check if pair exists quickly -> O(log n)
        # use a binary search + list queue
        sorted_ns = []
        queue = []
        for n in nums:
            # print(f"before {n}: sorted_ns {sorted_ns} {queue}")
            # check if pair exists for n
            i = bisect.bisect(sorted_ns, n)
            if i - 1 >= 0 and n - sorted_ns[i-1] <= valueDiff:
                return True
            if i < len(sorted_ns) and sorted_ns[i] - n <= valueDiff:
                return True
            sorted_ns.insert(i, n)
            queue.append(n)
            # print(f"after: sorted_ns {sorted_ns} {queue}")
            if len(sorted_ns) > indexDiff:
                # remove first added item
                remove_n = queue[0]
                # find remove_n in bst
                remove_i = bisect.bisect_left(sorted_ns, remove_n)
                del sorted_ns[remove_i]
                queue = queue[1:]
        return False
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