<p>Given an integer array nums, return <em>the number of <strong>AND triples</strong></em>.</p>

<p>An <strong>AND triple</strong> is a triple of indices <code>(i, j, k)</code> such that:</p>

<ul>
	<li><code>0 &lt;= i &lt; nums.length</code></li>
	<li><code>0 &lt;= j &lt; nums.length</code></li>
	<li><code>0 &lt;= k &lt; nums.length</code></li>
	<li><code>nums[i] &amp; nums[j] &amp; nums[k] == 0</code>, where <code>&amp;</code> represents the bitwise-AND operator.</li>
</ul>

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

<pre>
<strong>Input:</strong> nums = [2,1,3]
<strong>Output:</strong> 12
<strong>Explanation:</strong> We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 &amp; 2 &amp; 1
(i=0, j=1, k=0) : 2 &amp; 1 &amp; 2
(i=0, j=1, k=1) : 2 &amp; 1 &amp; 1
(i=0, j=1, k=2) : 2 &amp; 1 &amp; 3
(i=0, j=2, k=1) : 2 &amp; 3 &amp; 1
(i=1, j=0, k=0) : 1 &amp; 2 &amp; 2
(i=1, j=0, k=1) : 1 &amp; 2 &amp; 1
(i=1, j=0, k=2) : 1 &amp; 2 &amp; 3
(i=1, j=1, k=0) : 1 &amp; 1 &amp; 2
(i=1, j=2, k=0) : 1 &amp; 3 &amp; 2
(i=2, j=0, k=1) : 3 &amp; 2 &amp; 1
(i=2, j=1, k=0) : 3 &amp; 1 &amp; 2
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> 27
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt; 2<sup>16</sup></code></li>
</ul>

class Solution:
    def countTriplets(self, nums):
        # Initialize a defaultdict to memoize bitwise AND counts
        memo = defaultdict(int)
        # Find the maximum length of binary representation among all numbers
        maxVal = len(bin(max(nums))) - 1

        # Build the memo dictionary with counts of bitwise AND results
        for first in nums:
            for second in nums:
                memo[first & second] += 1

        # Initialize a Trie data structure
        head = Trie()

        # Function to reverse and pad a binary representation with zeros
        def getReverseBin(num):
            bin_val = bin(num)[2:][::-1]
            bin_val += (maxVal - len(bin_val)) * "0"
            return bin_val

        # Build Trie using the memoized bitwise AND counts
        for key, value in memo.items():
            node = head
            bin_val = getReverseBin(key)
            node.count += value

            # Traverse the Trie, updating counts for each node
            for string in bin_val:
                if string not in node.data:
                    node.data[string] = Trie()
                node = node.data[string]
                node.count += value
            node.end = True

        # Dynamic Programming function to count triplets based on Trie
        def getCount(nodes, s, remaining):
            if remaining == 0:
                count = 0
                for node in nodes:
                    count += node.count
                return count
            elif not s:
                return 0
            else:
                if s[0] == ".":
                    temp = []
                    for node in nodes:
                        temp.extend(node.data.values())
                    return getCount(temp, s[1:], remaining)
                else:
                    temp = []
                    for node in nodes:
                        if "0" not in node.data:
                            continue
                        temp.append(node.data["0"])
                    return getCount(temp, s[1:], remaining-1)

        ans = 0

        # Iterate over each number in the input array
        for num in nums:
            # Convert the binary representation, replacing "0" with "."
            bin_val = getReverseBin(num)
            bin_val = bin_val.replace("0", ".")
            # Count triplets using the Trie and update the result
            remaining = bin_val.count("1")
            ans += getCount([head], bin_val, remaining)

        return ans


class Trie:
    def __init__(self):
        # Initialize Trie node with data, count, and end indicator
        self.data = {}
        self.count = 0
        self.end = False
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