LeetCode 1711. Count Good Meals Solution in Java, C++, Python & Go | Explanation + Code

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(n²), 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 × 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=2²⁰.

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 ≠ a, then add m × cnt[s - a] to the answer; if s - a = a, then add m × (m - 1) to the answer.

Finally, divide the answer by 2, modulo 10⁹ + 7, and return.

The time complexity is the same as the method above.

PythonJavaC++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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

Solution 2

PythonJavaC++Go

class Solution: def countPairs(self, deliciousness: List[int]) -> int: mod = 10**9 + 7 cnt = Counter(deliciousness) ans = 0 for i in range(22): s = 1 << i for a, m in cnt.items(): if (b := s - a) in cnt: ans += m * (m - 1) if a == b else m * cnt[b] return (ans >> 1) % mod(code-box)

class Solution { private static final int MOD = (int) 1e9 + 7; public int countPairs(int[] deliciousness) { Map<Integer, Integer> cnt = new HashMap<>(); for (int d : deliciousness) { cnt.put(d, cnt.getOrDefault(d, 0) + 1); } long ans = 0; for (int i = 0; i < 22; ++i) { int s = 1 << i; for (var x : cnt.entrySet()) { int a = x.getKey(), m = x.getValue(); int b = s - a; if (!cnt.containsKey(b)) { continue; } ans += 1L * m * (a == b ? m - 1 : cnt.get(b)); } } ans >>= 1; return (int) (ans % MOD); } }(code-box)

class Solution { public: const int mod = 1e9 + 7; int countPairs(vector<int>& deliciousness) { unordered_map<int, int> cnt; for (int& d : deliciousness) ++cnt[d]; long long ans = 0; for (int i = 0; i < 22; ++i) { int s = 1 << i; for (auto& [a, m] : cnt) { int b = s - a; if (!cnt.count(b)) continue; ans += 1ll * m * (a == b ? (m - 1) : cnt[b]); } } ans >>= 1; return ans % mod; } };(code-box)

func countPairs(deliciousness []int) (ans int) { cnt := map[int]int{} for _, d := range deliciousness { cnt[d]++ } const mod int = 1e9 + 7 for i := 0; i < 22; i++ { s := 1 << i for a, m := range cnt { b := s - a if n, ok := cnt[b]; ok { if a == b { ans += m * (m - 1) } else { ans += m * n } } } } ans >>= 1 return ans % mod }(code-box)