How do you solve Count Good Meals on LeetCode?

Count Good Meals (LeetCode #1711) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Powers of two can be efficiently checked using bit manipulation.

What is the brute force solution for Count Good Meals?

The brute force approach for Count Good Meals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of food items to see if their deliciousness sums to a power of two. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Good Meals?

Count Good Meals uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Good Meals?

Always think about the constraints and how they affect your approach. Discuss your thought process and potential optimizations with the interviewer. Practice writing clean and efficient code, focusing on readability.

What are common mistakes when solving Count Good Meals?

Not considering pairs of the same deliciousness value correctly. Failing to account for the modulo operation when counting pairs.

#1711

Count Good Meals

Medium

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages a hash map to count occurrences of each deliciousness value and checks for pairs that sum to powers of two. This reduces the need for nested loops and allows us to find pairs in linear time.

⚙️

Algorithm

5 steps

1Step 1: Create a hash map to count occurrences of each deliciousness value.
2Step 2: Iterate through the hash map keys and for each key, check all powers of two up to 2^21.
3Step 3: For each power of two, calculate the complement needed to form that power with the current key.
4Step 4: If the complement exists in the hash map, calculate the number of good meals based on their counts.
5Step 5: Return the total count modulo 10^9 + 7.

solution.py18 lines

1# Full working Python code
2
3def count_good_meals(deliciousness):
4    MOD = 10**9 + 7
5    count = 0
6    freq = {}
7    for d in deliciousness:
8        freq[d] = freq.get(d, 0) + 1
9    powers_of_two = [1 << i for i in range(22)]
10    for key in freq:
11        for power in powers_of_two:
12            complement = power - key
13            if complement in freq:
14                if complement == key:
15                    count += (freq[key] * (freq[key] - 1)) // 2
16                else:
17                    count += freq[key] * freq[complement]
18    return count % MOD

ℹ

Complexity note: The time complexity is O(n) because we iterate through the array once to build the frequency map and then through the map to find pairs, which is efficient. The space complexity is O(n) due to the hash map storing counts of deliciousness values.

1Powers of two can be efficiently checked using bit manipulation.
2Using a hash map allows for quick lookups of counts, reducing time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.