1711. Count Good Meals

Description

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

Solutions

Solution 1: Hash Table + Enumeration of Powers of Two

According to the problem, we need to count the number of combinations in the array where the sum of two numbers is a power of $2$. Directly enumerating all combinations has a time complexity of $O(n^2)$, which will definitely time out.

We can traverse the array and use a hash table $cnt$ to maintain the number of occurrences of each element $d$ in the array.

For each element, we enumerate the powers of two $s$ as the sum of two numbers from small to large, and add the number of occurrences of $s - d$ in the hash table to the answer. Then increase the number of occurrences of the current element $d$ by one.

After the traversal ends, return the answer.

The time complexity is $O(n \times \log M)$, where $n$ is the length of the array deliciousness, and $M$ is the upper limit of the elements. For this problem, the upper limit $M=2^{20}$.

We can also use a hash table $cnt$ to count the number of occurrences of each element in the array first.

Then enumerate the powers of two $s$ as the sum of two numbers from small to large. For each $s$, traverse each key-value pair $(a, m)$ in the hash table. If $s - a$ is also in the hash table, and $s - a \neq a$, then add $m \times cnt[s - a]$ to the answer; if $s - a = a$, then add $m \times (m - 1)$ to the answer.

Finally, divide the answer by $2$, modulo $10^9 + 7$, and return.

The time complexity is the same as the method above.

Python3JavaC++Go

class Solution:
    def countPairs(self, deliciousness: List[int]) -> int:
        mod = 10**9 + 7
        mx = max(deliciousness) << 1
        cnt = Counter()
        ans = 0
        for d in deliciousness:
            s = 1
            while s <= mx:
                ans = (ans + cnt[s - d]) % mod
                s <<= 1
            cnt[d] += 1
        return ans

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countPairs(int[] deliciousness) {
        int mx = Arrays.stream(deliciousness).max().getAsInt() << 1;
        int ans = 0;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int d : deliciousness) {
            for (int s = 1; s <= mx; s <<= 1) {
                ans = (ans + cnt.getOrDefault(s - d, 0)) % MOD;
            }
            cnt.merge(d, 1, Integer::sum);
        }
        return ans;
    }
}

class Solution {
public:
    const int mod = 1e9 + 7;

    int countPairs(vector<int>& deliciousness) {
        int mx = *max_element(deliciousness.begin(), deliciousness.end()) << 1;
        unordered_map<int, int> cnt;
        int ans = 0;
        for (auto& d : deliciousness) {
            for (int s = 1; s <= mx; s <<= 1) {
                ans = (ans + cnt[s - d]) % mod;
            }
            ++cnt[d];
        }
        return ans;
    }
};

func countPairs(deliciousness []int) (ans int) {
    mx := slices.Max(deliciousness) << 1
    const mod int = 1e9 + 7
    cnt := map[int]int{}
    for _, d := range deliciousness {
        for s := 1; s <= mx; s <<= 1 {
            ans = (ans + cnt[s-d]) % mod
        }
        cnt[d]++
    }
    return
}

Solution 2