<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 {
public:
    vector<pair<int,int>> v;
    int helper(int key,int i, vector<int>&nums)
    {
        int startIndex = v[key].second + 1;
        int count = v[key].first;
        for(int j=startIndex;j<=i;j++) count += ((nums[j-1]&key)==0);
        v[key] = {count,i};
        return v[key].first;
    }
    int countTriplets(vector<int>& nums) {
        int n = nums.size();
        int maxi = 0;
        for(int i=0;i<n;i++) maxi = max(maxi, nums[i]);
        v.resize(maxi+1,{0,0});
        int count = (nums[0]==0);
        for(int i=1;i<n;i++)
        {
            count += (nums[i]==0);
            count += 6*(helper(nums[i],i,nums));
            for(int j=i+1;j<n;j++) 
            {
                int key = nums[i]&nums[j];
                count += 6*(helper(key,i,nums));
            }
        }
        return count;
    }
};
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