Program to count good meals with exactly two items in Python

Suppose we have an array deli where deli[i] is the deliciousness of the ith food. We need to find the number of different good meals we can make from this list. A good meal contains exactly two different food items with a sum of deliciousness that is a power of two. If the answer is too large, return the result modulo 10^9 + 7.

So, if the input is like deli = [1,7,3,6,5], then the output will be 3 because we can make pairs (1,3), (1,7) and (3,5) whose sums are powers of 2 (4, 8, and 8 respectively).

Algorithm

To solve this, we will follow these steps:

• Create a frequency map of all deliciousness values
• For each deliciousness value, check all possible powers of 2 (up to 2^21)
• Find the complement that makes the sum a power of 2
• Count valid pairs, handling duplicates carefully
• Return the result divided by 2 (since each pair is counted twice) modulo 10^9 + 7

Example

Let us see the following implementation to get better understanding ?

from collections import Counter

def solve(deli):
    m = 10**9 + 7
    count = Counter(deli)
    ans = 0
    
    for i in count:
        for n in range(22):
            j = (1 << n) - i  # 2^n - i
            if j in count:
                if i == j:
                    # Same values: choose 2 from count[i]
                    ans += count[i] * (count[i] - 1)
                else:
                    # Different values
                    ans += count[i] * count[j]
    
    return (ans // 2) % m

# Test the function
deli = [1, 7, 3, 6, 5]
result = solve(deli)
print(f"Input: {deli}")
print(f"Output: {result}")

Input: [1, 7, 3, 6, 5]
Output: 3

How It Works

The algorithm works by:

• Frequency counting: Using Counter to count occurrences of each deliciousness value
• Power of 2 checking: For each value i, we check all powers of 2 from 2^0 to 2^21
• Complement finding: For each power of 2, we find the complement j = 2^n - i
• Pair counting: If the complement exists, we count the valid pairs
• Duplicate handling: We divide by 2 at the end since each pair is counted twice

Edge Cases

Let's test with another example ?

from collections import Counter

def solve(deli):
    m = 10**9 + 7
    count = Counter(deli)
    ans = 0
    
    for i in count:
        for n in range(22):
            j = (1 << n) - i
            if j in count:
                if i == j:
                    ans += count[i] * (count[i] - 1)
                else:
                    ans += count[i] * count[j]
    
    return (ans // 2) % m

# Test with duplicate values
deli2 = [1, 1, 1, 3, 3]
result2 = solve(deli2)
print(f"Input: {deli2}")
print(f"Output: {result2}")
print("Valid pairs: (1,1), (1,1), (1,1), (1,3), (1,3), (1,3), (1,3), (1,3), (1,3)")

Input: [1, 1, 1, 3, 3]
Output: 12
Valid pairs: (1,1), (1,1), (1,1), (1,3), (1,3), (1,3), (1,3), (1,3), (1,3)

Conclusion

This solution efficiently counts good meals by using frequency mapping and checking all possible powers of 2. The time complexity is O(n + k × 22) where n is the array length and k is the number of unique values.