<p>A <strong>good meal</strong> is a meal that contains <strong>exactly two different food items</strong> with a sum of deliciousness equal to a power of two.</p>

<p>You can pick <strong>any</strong> two different foods to make a good meal.</p>

<p>Given an array of integers <code>deliciousness</code> where <code>deliciousness[i]</code> is the deliciousness of the <code>i<sup>​​​​​​th</sup>​​​​</code>​​​​ item of food, return <em>the number of different <strong>good meals</strong> you can make from this list modulo</em> <code>10<sup>9</sup> + 7</code>.</p>

<p>Note that items with different indices are considered different even if they have the same deliciousness value.</p>

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

<pre>
<strong>Input:</strong> deliciousness = [1,3,5,7,9]
<strong>Output:</strong> 4
<strong>Explanation: </strong>The good meals are (1,3), (1,7), (3,5) and, (7,9).
Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.
</pre>

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

<pre>
<strong>Input:</strong> deliciousness = [1,1,1,3,3,3,7]
<strong>Output:</strong> 15
<strong>Explanation: </strong>The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.</pre>

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

<ul>
	<li><code>1 &lt;= deliciousness.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= deliciousness[i] &lt;= 2<sup>20</sup></code></li>
</ul>

class Solution {
public:
    int countPairs(vector<int>& nums) {
        int mod = 1e9 + 7;
        map<int, int> mp;
        vector<int> powers;

        int cnt = 0;

        // Generate all powers of 2 up to 2^21
        for (int i = 0; i <= 21; i++) {
            powers.push_back(1 << i); // 1 << i is equivalent to 2^i
        }

        // Populate the map with the frequency of each element in nums
        for (auto val : nums) mp[val]++;

        // Iterate through each element in nums
        for (int val : nums) {
            mp[val]--; // Decrease the count of the current value
            if(mp[val]==0)mp.erase(val);
            // Check against all powers of 2
            for (int pow : powers) {
                int target = pow - val; // The value needed to make the sum a power of 2

                // If the target exists in the map, add its count to the result
                if (mp.find(target) != mp.end()) {
                    cnt = (cnt + mp[target]) % mod;
                }
            }

            // If the count becomes 0, erase it from the map
            if (mp[val] == 0) mp.erase(val);
        }
        return cnt; // Return the final count modulo 10^9 + 7
    }
};

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