5 Best Ways to Count Good Meals in Python

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πŸ’‘ Problem Formulation: In the context of a computational problem, counting “good meals” typically refers to finding pairs of dishes that, when combined, satisfy a particular culinary criterion, such as having a cumulative quality score or meeting a specific nutrient profile. Given an array of dish quality scores, we want to identify and count all pairs whose sum is a perfect power of 2. For example, if we have dishes with scores [1,3,5,7,9], we desire an output that would tell us there are two good meal combinations: (1, 7) and (3, 5).

Method 1: Brute Force

This method involves checking every possible pair of dishes to see if their sum is a power of two. Although straightforward, it can be highly inefficient for large lists of dishes due to its O(n^2) time complexity. The function implementation would iterate through the array twice, using nested loops, to test all possible combinations.

Here’s an example:

def is_power_of_two(n):
    return n and (not(n & (n - 1)))

def count_good_meals(dishes):
    count = 0
    for i in range(len(dishes)):
        for j in range(i+1, len(dishes)):
            if is_power_of_two(dishes[i] + dishes[j]):
                count += 1
    return count

meals = [1,3,5,7,9]
print(count_good_meals(meals))

Output: 2

This code snippet defines a utility function is_power_of_two() to check if a number is a power of two. The main function count_good_meals() then iterates through each unique pair of dishes, checking if their sum is a power of two, and counts the number of such pairs.

Method 2: Using Sets

Using sets can significantly optimize the process by reducing the need for a nested loop. This method first calculates all the possible powers of two within a certain range. Then it traverses the list of dishes once, and for each dish, it checks if the difference between a power of two and the dish’s score is in the set of dish scores, keeping track of the pairs found this way. The time complexity of this approach is O(n*log(m)), where n is the number of dishes and m is the maximum value of a dish score.

Here’s an example:

def count_good_meals(dishes):
    powers_of_two = {2**i for i in range(22)}  # Assuming dishes score <= 10^6
    dishes_counter = Counter(dishes)
    count = 0
    for dish in dishes_counter:
        for power in powers_of_two:
            complement = power - dish
            if complement in dishes_counter:
                if complement == dish:
                    count += dishes_counter[dish] * (dishes_counter[dish] - 1) // 2
                else:
                    count += dishes_counter[dish] * dishes_counter[complement]
    return count // 2  # To adjust for double counting

from collections import Counter
meals = [1,3,5,7,9]
print(count_good_meals(meals))

Output: 2

This code uses a set to store the powers of two and a Counter from the collections module to track the occurrence of each dish score. It then iterates through the dishes and checks for complements that would sum to a power of two, adjusting for double counting at the end.

Method 3: Binary Search

A binary search approach involves first sorting the list of dish scores and then, for each dish, using binary search to find a complement that would sum to any power of two. This method can be more efficient than brute force and works well when there are many possible powers of two to check against, with a time complexity of O(n*log(n)) due to sorting.

Here’s an example:

from bisect import bisect_left

def count_good_meals(dishes):
    dishes.sort()
    count = 0
    for i, dish in enumerate(dishes):
        for power in [2**i for i in range(22)]:
            complement = power - dish
            idx = bisect_left(dishes, complement, i+1)
            if idx < len(dishes) and dishes[idx] == complement:
                count += 1
    return count

meals = [1,3,5,7,9]
print(count_good_meals(meals))

Output: 2

The code snippet sorts the dish array and iterates through each dish. For each dish, it looks for the complement that would make the sum a power of two using binary search via bisect_left() from the bisect module. It then counts the pairs found.

Method 4: Hash Map

Hash maps are an excellent way to improve lookup times. By using a hash map, you can keep a running total of the dishes you’ve seen and compare each new dish’s complement (power of two minus the dish score) against the hash map. This optimizes the time complexity to nearly O(n), assuming good hash function performance.

Here’s an example:

def count_good_meals(dishes):
    dishes_counter = Counter(dishes)
    count = 0
    seen = set()
    for dish in dishes:
        multiples = [2**i for i in range(22)]
        for m in multiples:
            if (m - dish) in seen:
                count += dishes_counter[m - dish]
        seen.add(dish)
    return count

from collections import Counter
meals = [1,3,5,7,9]
print(count_good_meals(meals))

Output: 2

The code uses a Counter to count occurrences of dishes and a set to track seen dishes. For every dish, it calculates possible complements by subtracting the dish score from each power of two and checking if the complement has been seen.

Bonus One-Liner Method 5: List Comprehension with Filtering

For those who love one-liners and are not working with a large data set, Python’s list comprehensions can be used to filter all the combinations at once. However, this method isn’t recommended for performance due to its O(n^2) time complexity.

Here’s an example:

meals = [1,3,5,7,9]
good_meal_count = sum(1 for i in range(len(meals)) 
                      for j in range(i+1, len(meals)) 
                      if bin(meals[i]+meals[j]).count('1') == 1)
print(good_meal_count)

Output: 2

This one-liner uses list comprehension to iterate all unique pairs, checking the sum’s binary representation for a single ‘1’, which indicates a power of two. While clever, it’s inefficient for larger lists due to the nested loops.

Summary/Discussion

  • Method 1: Brute Force. Simple to implement. Inefficient for large lists due to O(n^2) complexity.
  • Method 2: Using Sets. Utilizes sets for faster access. Much faster than brute force for larger inputs. Still has somewhat high time complexity O(n*log(m)) due to iteration over powers of two.
  • Method 3: Binary Search. Efficient for lists with many potential powers of two. Requires sorting the list first, which incurs O(n*log(n)) complexity.
  • Method 4: Hash Map. Fast and efficient. Nearly O(n) complexity under ideal conditions but may consume more memory to maintain the hash map.
  • Method 5: List Comprehension with Filtering. Elegant one-liner. Not practical for performance-critical applications due to O(n^2) complexity.