<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 sorted sliding window
        # Keep queue of nums to pop off the sliding window, these are the nums exiting the sliding window
        # On every iteration, use the numbers next to the number just inserted to get the new diffs 
        # If new diff is less than valueDiff then return True
        # Size of the sliding window is indexDiff
        # O(log(N)) to insert into sliding window, which can happen N times so O(Nlog(N))

        window = []
        queue = Deque()

        def find_index(num):
            l, r = 0, len(window)
            while l < r:
                m = (l + r) // 2
                if window[m] < num:
                    l = m + 1
                else:
                    r = m
            
            return l

        def remove_from_window(num):
            window.pop(find_index(num))

        def add_to_window(num):
            ind = find_index(num)
            window.insert(ind, num)
            return ind

        for num in nums:
            if len(window) == indexDiff + 1:
                remove_from_window(queue.popleft())
            
            ind = add_to_window(num)
            if ind - 1 >= 0 and num - window[ind-1] <= valueDiff:
                return True
            
            if ind + 1 < len(window) and window[ind+1] - num <= valueDiff:
                return True
            
            queue.append(num)
            # print(queue, window)

        return False
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