Count Ways to Make Array With Product
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Topics:
ArraysDynamic ProgrammingRecursion
You are given a 2D integer array, queries. For each queries[i], where queries[i] = [ni, ki], find the number of different ways you can place positive integers into an array of size ni such that the product of the integers is ki. As the number of ways may be too large, the answer to the ith query is the number of ways modulo 109 + 7.
Return an integer array answer where answer.length == queries.length, and answer[i] is the answer to the ith query.
Example 1:
Input: queries = [[2,6],[5,1],[73,660]] Output: [4,1,50734910] Explanation: Each query is independent. [2,6]: There are 4 ways to fill an array of size 2 that multiply to 6: [1,6], [2,3], [3,2], [6,1]. [5,1]: There is 1 way to fill an array of size 5 that multiply to 1: [1,1,1,1,1]. [73,660]: There are 1050734917 ways to fill an array of size 73 that multiply to 660. 1050734917 modulo 109 + 7 = 50734910.
Example 2:
Input: queries = [[1,1],[2,2],[3,3],[4,4],[5,5]] Output: [1,2,3,10,5]
Constraints:
1 <= queries.length <= 1041 <= ni, ki <= 104
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 are the possible values for elements within the input array and the product value? Are they all positive integers, or can they be negative or zero?
- What is the maximum size of the input array?
- If there are no combinations of numbers in the input array that multiply to the target product, what should the function return?
- Is the order of factors in the product important (e.g., is [2, 3] the same as [3, 2])?
- Can I reuse numbers from the array multiple times to achieve the target product?
Brute Force Solution
Approach
The brute force method involves exploring every single potential combination to find the correct solution. In this case, we'll try every possible way to distribute some quantity among different categories. We check each arrangement to see if it meets our desired product, and if it does, we count it.
Here's how the algorithm would work step-by-step:
- Think of having a certain number of identical items to distribute into distinct containers.
- Start by putting all the items into the first container and none in the others, then check if the resulting product meets the target.
- Next, shift one item from the first container to the second, and then to the third, and so on, always calculating the product and checking against the target.
- Repeat the process by shifting two items, then three, and so on, exploring all possible distributions between the containers.
- Each time we find a distribution where the product matches our target, we increment a counter.
- After trying every single possible distribution of items into containers, the final count represents the total number of ways to achieve the desired product.
Code Implementation
def count_ways_to_make_array_with_product_brute_force(slots_available, target_product): number_of_ways = 0
# Iterate through all possible distributions of values into slots
def find_ways(current_slot_index, remaining_product, current_combination):
nonlocal number_of_ways
# Base case: all slots filled
if current_slot_index == slots_available:
if remaining_product == 1:
number_of_ways += 1
return
# Iterate through all possible values for the current slot
for current_value in range(1, target_product + 1):
# If the remaining product is divisible, proceed with the recursion.
if remaining_product % current_value == 0:
find_ways(current_slot_index + 1, remaining_product // current_value, current_combination + [current_value])
find_ways(0, target_product, [])
# Return the final count of valid ways.
return number_of_waysBig(O) Analysis
Time Complexity
O(target^(num_slots-1)) – The brute force approach explores every possible distribution of factors that multiply to the target value across num_slots. In the worst-case scenario, each slot can potentially take on any value from 1 up to the target value. The number of possible combinations grows exponentially with the target value and the number of slots. Specifically, the dominant factor is driven by the number of ways to partition 'target' into 'num_slots' parts. This exhaustive search effectively explores all combinations, leading to a time complexity of approximately O(target^(num_slots-1)), where 'target' is the product we aim to achieve, and 'num_slots' is the number of array elements.
Space Complexity
O(1) – The described brute force method iterates through different combinations of items in containers, adjusting counts and checking the product. The primary space used involves storing a counter for the number of valid arrangements and possibly a few integer variables to represent the number of items in each container. The number of these counters/variables does not depend on the input size, such as the target product or the number of containers. Therefore, the auxiliary space complexity is constant.
Optimal Solution
Approach
This problem asks us to find the number of ways to split a target product among different groups. The optimal solution cleverly combines prime factorization with a stars and bars technique from combinatorics to avoid brute-force calculation.
Here's how the algorithm would work step-by-step:
- First, break down the target product into its prime factors. This reveals the basic building blocks of any possible multiplication.
- For each group, decide how many of each prime factor to assign to it. This is like distributing identical items (prime factors) into distinct containers (groups).
- Use the stars and bars method from combinatorics to figure out how many ways you can distribute each prime factor among the groups.
- Multiply the number of ways to distribute each prime factor together. This gives you the total number of ways to achieve the target product across all the groups.
- Take the result modulo a big prime number (like 10^9 + 7) to prevent integer overflow, which can happen with large factorials.
Code Implementation
def count_ways_to_make_array_with_product(number_of_queries, product, modulo_value):
def prime_factorization(product_value):
prime_factors_counts = {}
divisor = 2
while divisor * divisor <= product_value:
while product_value % divisor == 0:
prime_factors_counts[divisor] = prime_factors_counts.get(divisor, 0) + 1
product_value //= divisor
divisor += 1
if product_value > 1:
prime_factors_counts[product_value] = prime_factors_counts.get(product_value, 0) + 1
return prime_factors_counts
def combinations(total_items, number_of_groups, modulo_value):
if number_of_groups == 0 and total_items == 0:
return 1
if number_of_groups == 0:
return 0
numerator = 1
denominator = 1
for i in range(number_of_groups - 1):
numerator = (numerator * (total_items + number_of_groups - 1 - i)) % modulo_value
denominator = (denominator * (i + 1)) % modulo_value
return (numerator * pow(denominator, modulo_value - 2, modulo_value)) % modulo_value
results = []
for query in number_of_queries:
number_of_arrays, _ = query
prime_factors = prime_factorization(product)
total_ways = 1
# Iterate over each prime factor to calculate ways.
for prime_factor in prime_factors:
prime_count = prime_factors[prime_factor]
# Stars and bars to distribute prime factors among arrays.
ways = combinations(prime_count, number_of_arrays, modulo_value)
total_ways = (total_ways * ways) % modulo_value
results.append(total_ways)
return resultsBig(O) Analysis
Time Complexity
O(target + n * log(n)) – The dominant part of the algorithm involves prime factorization of the target value, which, in the worst case, can take O(target) time where target is the input target value. Calculating combinations using precomputed factorials, inverse factorials, and modular exponentiation is done for each prime factor of the target. The number of prime factors is bounded. Furthermore, the computation of factorials and inverse factorials takes O(n * log(n)) time where n is the number of groups specified in the input array (arr.length). Thus, the overall time complexity is O(target + n * log(n)).
Space Complexity
O(sqrt(targetProduct) + log(targetProduct)) – The space complexity is driven by two main factors: the prime factorization and the intermediate calculations within the stars and bars method. Prime factorization might require storing primes up to the square root of the target product (sqrt(targetProduct)). The stars and bars method involves calculations of factorials and their modular inverses, which might involve creating a table of size up to the largest power present in the prime factorization, which can be logarithmic relative to the targetProduct. Thus, the overall auxiliary space complexity is O(sqrt(targetProduct) + log(targetProduct)).
Edge Cases
| Case | How to Handle |
|---|---|
| Empty queries array | Return an empty result list since there are no queries to process. |
| Empty factors array | If product is 1, return 1, otherwise return 0, as no factors means only an array of 1s could achieve a non-1 product. |
| Large n (array length) and small product | The number of combinations could be very large even for a small target and the solution must efficiently calculate and handle potentially large factor counts. |
| Product equals 0 | If 0 is not a factor then no solution, if 0 is a factor then the number of ways to form the array is based on combinations with repetition of only the 0 factor. |
| All queries are the same | The solution should not recalculate identical queries, using memoization can improve the runtime. |
| Very large product | The dynamic programming table or combination calculations must handle potentially very large numbers to avoid integer overflow, likely requiring modulo arithmetic. |
| Product is 1 | The problem reduces to how many ways to create a length n array where all factors are 1 if 1 is present; otherwise return 0. |
| Factors array with duplicates | Duplicate factors should be accounted for in the combinations with repetition formula correctly. |
Empty queries array
How to Handle:
Return an empty result list since there are no queries to process.
Empty factors array
How to Handle:
If product is 1, return 1, otherwise return 0, as no factors means only an array of 1s could achieve a non-1 product.
Large n (array length) and small product
How to Handle:
The number of combinations could be very large even for a small target and the solution must efficiently calculate and handle potentially large factor counts.
Product equals 0
How to Handle:
If 0 is not a factor then no solution, if 0 is a factor then the number of ways to form the array is based on combinations with repetition of only the 0 factor.
All queries are the same
How to Handle:
The solution should not recalculate identical queries, using memoization can improve the runtime.
Very large product
How to Handle:
The dynamic programming table or combination calculations must handle potentially very large numbers to avoid integer overflow, likely requiring modulo arithmetic.
Product is 1
How to Handle:
The problem reduces to how many ways to create a length n array where all factors are 1 if 1 is present; otherwise return 0.
Factors array with duplicates
How to Handle:
Duplicate factors should be accounted for in the combinations with repetition formula correctly.
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