1711

1711. Count Good Meals (Medium)

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

You can pick any two different foods to make a good meal.

Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i​​​​​​th​​​​​​​​ item of food, return the number of different good meals you can make from this list modulo 109 + 7.

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

 

Example 1:

Input: deliciousness = [1,3,5,7,9]
Output: 4
Explanation: 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.

Example 2:

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

 

Constraints:

• 1 <= deliciousness.length <= 105
• 0 <= deliciousness[i] <= 220

Related Topics:
Array, Hash Table, Two Pointers

Similar Questions:

• Two Sum (Easy)
• Max Number of K-Sum Pairs (Medium)

Solution 1.

Use an unordered_map<int, int> m to store the frequency of each number in array A (i.e. m[a] is the frequency of a).

Since A[i] is in range [0, 2^20], the sum of two numbers is in range [0, 2^21].

For each pair a, cnt in m, we try to find if b = sum - a is in m as well, where sum is power of 2.

To avoid repetitive computation, we skip a b value if b < a.

If a == b, add cnt * (cnt - 1) / 2 to answer.

Otherwise, add cnt * m[b] to answer.

// OJ: https://leetcode.com/problems/count-good-meals/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int countPairs(vector<int>& A) {
        long mod = 1e9 + 7, ans = 0;
        unordered_map<int, int> m;
        for (int n : A) m[n]++;
        for (auto &[a, cnt] : m) {
            for (int i = 0; i <= 21; ++i) {
                long b = (1 << i) - a;
                if (b < a || m.count(b) == 0) continue;
                long c = m[b];
                if (a == b) ans = (ans + (c * (c - 1) / 2) % mod) % mod;
                else ans = (ans + cnt * c % mod) % mod; 
            }
        }
        return ans;
    }
};