The Number of Good Subsets

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Dynamic ProgrammingBit Manipulation

You are given an integer array nums. We call a subset of nums good if its product can be represented as a product of one or more distinct prime numbers.

For example, if nums = [1, 2, 3, 4]:
- [2, 3], [1, 2, 3], and [1, 3] are good subsets with products 6 = 2*3, 6 = 2*3, and 3 = 3 respectively.
- [1, 4] and [4] are not good subsets with products 4 = 2*2 and 4 = 2*2 respectively.

Return the number of different good subsets in nums modulo 10⁹ + 7.

A subset of nums is any array that can be obtained by deleting some (possibly none or all) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Example 1:

Input: nums = [1,2,3,4]
Output: 6
Explanation: The good subsets are:
- [1,2]: product is 2, which is the product of distinct prime 2.
- [1,2,3]: product is 6, which is the product of distinct primes 2 and 3.
- [1,3]: product is 3, which is the product of distinct prime 3.
- [2]: product is 2, which is the product of distinct prime 2.
- [2,3]: product is 6, which is the product of distinct primes 2 and 3.
- [3]: product is 3, which is the product of distinct prime 3.

Example 2:

Input: nums = [4,2,3,15]
Output: 5
Explanation: The good subsets are:
- [2]: product is 2, which is the product of distinct prime 2.
- [2,3]: product is 6, which is the product of distinct primes 2 and 3.
- [2,15]: product is 30, which is the product of distinct primes 2, 3, and 5.
- [3]: product is 3, which is the product of distinct prime 3.
- [15]: product is 15, which is the product of distinct primes 3 and 5.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 30

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers in the input array? Can they be negative or zero?
How large can the input array be? What is the expected time complexity?
What constitutes a 'good' subset, according to the problem definition (e.g., is it clearly defined if there are specific mathematical constraints)?
Are duplicate numbers allowed in the input array, and if so, how should they be handled when determining 'good' subsets?
If no 'good' subsets exist, what should the function return?

Brute Force Solution

Approach

The brute force approach involves checking every possible combination of numbers from the given list to form subsets. For each subset, we determine if it is a 'good' subset according to the problem's definition. We then count how many of these 'good' subsets we found.

Here's how the algorithm would work step-by-step:

Start by considering the possibility of including or excluding each number in the list.
Think about building subsets by systematically adding or not adding each number, one by one, from the original list.
For every subset you create, check if it meets the criteria to be considered a 'good' subset.
If a subset is 'good', increase the counter that tracks the number of 'good' subsets.
Repeat this process for every possible combination of numbers to build every possible subset.
Finally, report the total count of 'good' subsets that you found after checking all the possibilities.

Code Implementation

def the_number_of_good_subsets_brute_force(numbers):
    number_of_good_subsets = 0
    number_of_numbers = len(numbers)

    # Iterate through all possible subsets
    for i in range(1 << number_of_numbers):
        subset = []
        for j in range(number_of_numbers):
            # Check if the j-th bit is set in i
            if (i >> j) & 1:
                subset.append(numbers[j])

        # Handle empty subsets
        if not subset:
            continue

        is_good_subset = True
        product_of_subset = 1

        # Calculate product to identify good subset.
        for number in subset:
            product_of_subset *= number

        # Checking for perfect square.
        root = product_of_subset**0.5
        if root == int(root):
            is_good_subset = False

        # Increment if the subset is good
        if is_good_subset:
            number_of_good_subsets += 1

    return number_of_good_subsets

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach involves generating all possible subsets of the input array. For an array of size n, there are 2^n possible subsets (each element can either be present or absent in a subset). For each subset, we perform some operations to check if it is a 'good' subset which would take at least O(1). Therefore, since we need to iterate through all 2^n subsets, the time complexity is O(2^n).

Space Complexity

O(N) – The provided brute force approach recursively explores all possible subsets. Each number in the input list of size N is considered for inclusion or exclusion in a subset. This leads to a recursion tree with a maximum depth of N, where N is the size of the input list. Each level of recursion requires storing function call information on the call stack, resulting in O(N) space complexity due to the call stack. No other significant auxiliary data structures are used.

Optimal Solution

Approach

This problem asks us to count special groups of numbers. Instead of checking every possible group which would take forever, we identify the prime numbers that cause issues and use combinations to figure out the count efficiently.

Here's how the algorithm would work step-by-step:

First, recognize that the only numbers that matter are those formed by the product of unique prime numbers less than 31. Other numbers won't lead to good subsets.
Count how many times each valid number appears in the input. For example, count the number of 2s, the number of 3s, the number of 5s, and so on, up to the number 30 (2 * 3 * 5).
Handle the number 1 separately. Each time you consider adding a new valid number to a subset, you can either include or not include the 1s already counted. This multiplies your eventual answer by 2 to the power of the count of 1s.
Using a 'dynamic programming' strategy, build up the answer. Start by considering the smallest valid number. How many valid subsets can you make with just that number?
Move on to the next valid number. Now, you can either create new subsets using just this number, or you can add this number to the previously created subsets (which contained the smaller numbers). But be sure the product stays a 'good subset' and no number used so far shares any prime factors.
Repeat this process for all valid numbers, keeping track of the number of good subsets that can be made so far.
The final result is the total number of good subsets we've built up, taking into account the contribution of the '1' values earlier.

Code Implementation

def the_number_of_good_subsets(numbers):
    prime_numbers = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
    mask_for_number = [
        0, 1, 2, 4, 0, 8, 16, 32, 0, 64, 128,
        0, 256, 512, 0, 1024, 2048, 4096, 0, 8192, 16384,
        0, 32768, 65536, 0, 131072, 262144, 0, 524288, 1048576, 0
    ]
    modulo = 10**9 + 7

    number_counts = [0] * 31
    for number in numbers:
        number_counts[number] += 1

    number_of_ones = number_counts[1]
    dp_table = [0] * (1 << len(prime_numbers))
    dp_table[0] = 1

    # Iterate through possible numbers to form subsets
    for number in range(2, 31):
        if number_counts[number] == 0:
            continue

        mask = mask_for_number[number]
        if mask == 0:
            continue

        # Build new valid subsets
        for subset_mask in range(1 << len(prime_numbers)):
            if (subset_mask & mask) == 0:
                dp_table[subset_mask | mask] += dp_table[subset_mask] * number_counts[number]
                dp_table[subset_mask | mask] %= modulo

    answer = 0
    for subset_mask in range(1, 1 << len(prime_numbers)):
        answer += dp_table[subset_mask]
        answer %= modulo

    # Account for all possible combinations with '1'
    power_of_two = pow(2, number_of_ones, modulo)
    answer *= power_of_two
    answer %= modulo

    # Need to return the total number of subsets
    return answer

Big(O) Analysis

Time Complexity

O(m * 2^k) – The solution iterates through each of the m unique valid numbers (formed by products of primes < 31) to count their occurrences in the input array, which takes O(m) time. The core dynamic programming part involves considering all possible combinations of these valid numbers to form 'good subsets'. Since each valid number can either be included or not included in a subset, the number of possible subsets is 2^k, where k is the number of valid numbers. For each valid number, the algorithm checks against existing subsets, giving a O(m * 2^k) complexity where m is the number of valid numbers (at most 30) and k depends on valid numbers. While m is small, the 2^k factor dominates, making O(m * 2^k) the time complexity.

Space Complexity

O(2^10) – The dynamic programming approach described in the steps uses an array to store the number of good subsets possible for each mask of prime factors. Since we only consider numbers formed by the product of unique prime numbers less than 31, and there are 10 such primes (2, 3, 5, 7, 11, 13, 17, 19, 23, 29), the number of possible masks is 2^10. Therefore, the auxiliary space required to store the dynamic programming table is proportional to 2^10, which is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 1, as the empty subset is considered 'good'.
Array containing only the number 1	Return 2^(length of array) - 1, as any combination is valid.
Array with large prime numbers	Precompute prime factors for each number to improve performance, as repeated calculations will lead to timeouts.
Array with maximum possible input size and mostly prime numbers	Employ memoization/dynamic programming to avoid redundant calculations of valid subset count and bitmask combinations, to avoid time limit exceeding.
Array contains a large number of elements with the same prime factors	Handle potential integer overflow by using modular arithmetic during multiplication of counts.
Input array contains numbers whose product exceeds the maximum integer value	Use appropriate data types (e.g., long long) to store intermediate product values to prevent overflows and incorrect results.
No good subset exists	The algorithm should correctly return 0 if after processing the entire array, no good subset is found (other than the empty set, which has count of 1).
Array containing numbers whose prime factorization results in bitmask conflicts (same bitmask represents two distinct numbers)	Since numbers with the same bitmask are treated the same way, handle the duplicates by multiplying the count of possible subsets by 2^(duplicates) during the dp update stage.

Null or empty input array

How to Handle:

Return 1, as the empty subset is considered 'good'.

Array containing only the number 1

How to Handle:

Return 2^(length of array) - 1, as any combination is valid.

Array with large prime numbers

How to Handle:

Precompute prime factors for each number to improve performance, as repeated calculations will lead to timeouts.

Array with maximum possible input size and mostly prime numbers

How to Handle:

Employ memoization/dynamic programming to avoid redundant calculations of valid subset count and bitmask combinations, to avoid time limit exceeding.

Array contains a large number of elements with the same prime factors

How to Handle:

Handle potential integer overflow by using modular arithmetic during multiplication of counts.

Input array contains numbers whose product exceeds the maximum integer value

How to Handle:

Use appropriate data types (e.g., long long) to store intermediate product values to prevent overflows and incorrect results.

No good subset exists

How to Handle:

The algorithm should correctly return 0 if after processing the entire array, no good subset is found (other than the empty set, which has count of 1).

Array containing numbers whose prime factorization results in bitmask conflicts (same bitmask represents two distinct numbers)

How to Handle:

Since numbers with the same bitmask are treated the same way, handle the duplicates by multiplying the count of possible subsets by 2^(duplicates) during the dp update stage.