<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <code>true</code> <em>if there are two <strong>distinct indices</strong> </em><code>i</code><em> and </em><code>j</code><em> in the array such that </em><code>nums[i] == nums[j]</code><em> and </em><code>abs(i - j) &lt;= k</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,1], k = 3
<strong>Output:</strong> true
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,0,1,1], k = 1
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,1,2,3], k = 2
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &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>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        seen = {}
        # Iterate in reverse
        for i in range(len(nums) - 1, -1, -1):
            # Update window size if needed
            if (len(nums) - 1 - i > k):
                numToRemove = nums.pop()
                seen.pop(numToRemove)

            # Check if we've seen the number before
            if nums[i] not in seen:
                seen[nums[i]] = 1 # Value doesn't matter
            else:
                return True

        
24891