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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 ith 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 <= 1050 <= 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.
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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; } }; -
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 -
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 }