Count the Number of Ideal Arrays

Hard

Google

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Topics:

Dynamic Programming

You are given two integers n and maxValue, which are used to describe an ideal array.

A 0-indexed integer array arr of length n is considered ideal if the following conditions hold:

Every arr[i] is a value from 1 to maxValue, for 0 <= i < n.
Every arr[i] is divisible by arr[i - 1], for 0 < i < n.

Return the number of distinct ideal arrays of length n. Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: n = 2, maxValue = 5
Output: 10
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (5 arrays): [1,1], [1,2], [1,3], [1,4], [1,5]
- Arrays starting with the value 2 (2 arrays): [2,2], [2,4]
- Arrays starting with the value 3 (1 array): [3,3]
- Arrays starting with the value 4 (1 array): [4,4]
- Arrays starting with the value 5 (1 array): [5,5]
There are a total of 5 + 2 + 1 + 1 + 1 = 10 distinct ideal arrays.

Example 2:

Input: n = 5, maxValue = 3
Output: 11
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (9 arrays): 
   - With no other distinct values (1 array): [1,1,1,1,1] 
   - With 2^nd distinct value 2 (4 arrays): [1,1,1,1,2], [1,1,1,2,2], [1,1,2,2,2], [1,2,2,2,2]
   - With 2^nd distinct value 3 (4 arrays): [1,1,1,1,3], [1,1,1,3,3], [1,1,3,3,3], [1,3,3,3,3]
- Arrays starting with the value 2 (1 array): [2,2,2,2,2]
- Arrays starting with the value 3 (1 array): [3,3,3,3,3]
There are a total of 9 + 1 + 1 = 11 distinct ideal arrays.

Constraints:

2 <= n <= 10⁴
1 <= maxValue <= 10⁴

Count the Number of Ideal Arrays

Hard

Google

2 views

Topics:

Dynamic Programming

You are given two integers n and maxValue, which are used to describe an ideal array.

A 0-indexed integer array arr of length n is considered ideal if the following conditions hold:

Every arr[i] is a value from 1 to maxValue, for 0 <= i < n.
Every arr[i] is divisible by arr[i - 1], for 0 < i < n.

Return the number of distinct ideal arrays of length n. Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: n = 2, maxValue = 5
Output: 10
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (5 arrays): [1,1], [1,2], [1,3], [1,4], [1,5]
- Arrays starting with the value 2 (2 arrays): [2,2], [2,4]
- Arrays starting with the value 3 (1 array): [3,3]
- Arrays starting with the value 4 (1 array): [4,4]
- Arrays starting with the value 5 (1 array): [5,5]
There are a total of 5 + 2 + 1 + 1 + 1 = 10 distinct ideal arrays.

Example 2:

Input: n = 5, maxValue = 3
Output: 11
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (9 arrays): 
   - With no other distinct values (1 array): [1,1,1,1,1] 
   - With 2^nd distinct value 2 (4 arrays): [1,1,1,1,2], [1,1,1,2,2], [1,1,2,2,2], [1,2,2,2,2]
   - With 2^nd distinct value 3 (4 arrays): [1,1,1,1,3], [1,1,1,3,3], [1,1,3,3,3], [1,3,3,3,3]
- Arrays starting with the value 2 (1 array): [2,2,2,2,2]
- Arrays starting with the value 3 (1 array): [3,3,3,3,3]
There are a total of 9 + 1 + 1 = 11 distinct ideal arrays.

Constraints:

2 <= n <= 10⁴
1 <= maxValue <= 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 is the maximum value of `n` and `maxValue`? This will help me understand the scale of the problem.
Are `n` and `maxValue` guaranteed to be positive integers?
What should I return if no ideal arrays exist for the given `n` and `maxValue`?
Is an ideal array strictly increasing, or is a non-decreasing sequence acceptable?
Can you provide a simple example input (e.g., small `n` and `maxValue`) and its expected output to ensure I understand the problem correctly?

Brute Force Solution

Approach

The brute force approach involves trying out every single possible array configuration. We generate each potential array and check if it meets our specific criteria. If it does, we count it; if not, we discard it and move on to the next possibility.

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

Start by trying the simplest array: one element long.
Then, try arrays that are two elements long, then three, and so on, up to the specified length.
For each array we create, check that each number is a whole multiple of the number before it.
Also make sure that each number in our array is not bigger than our specified maximum number.
If an array meets both of these requirements, we count it as 'ideal'.
Keep doing this for all possible array combinations within the length and maximum number constraints.
Finally, report the total number of 'ideal' arrays we found.

Code Implementation

def count_ideal_arrays_brute_force(max_value, array_length):
    total_ideal_arrays = 0

    def is_ideal(array):
        for index in range(1, len(array)):
            if array[index] % array[index - 1] != 0:
                return False

        return True

    def generate_arrays(current_array):
        nonlocal total_ideal_arrays
        
        if len(current_array) == array_length:
            # Check if the generated array meets our criteria
            if is_ideal(current_array):
                total_ideal_arrays += 1

            return

        if not current_array:
            start = 1
        else:
            start = current_array[-1]

        for next_value in range(start, max_value + 1):
            # Ensure we do not exceed the max value constraint

            generate_arrays(current_array + [next_value])

    # Start generating arrays from an empty list
    generate_arrays([])

    return total_ideal_arrays

Big(O) Analysis

Time Complexity

O(m^n * k) – The described brute force approach explores all possible arrays. We iterate through lengths from 1 to n (the target length). For each length, we potentially iterate up to m (the maximum value) for each element in the array. Therefore, the number of possible arrays we check is on the order of m^n. Additionally, for each potential array, we perform k checks to ensure the multiple requirement is met, where k is the length of the potential array being checked (k <= n). The runtime is then approximately m^n * k. Thus the time complexity is O(m^n * k), where m is the maximum number, n is the length, and k is the array length.

Space Complexity

O(M^L) – The brute force approach generates all possible arrays of length up to L, where each element in the array can be a number up to M. The space required to store a single candidate array is O(L). However, the number of possible arrays grows exponentially with M and L. Since the algorithm potentially needs to store or generate all possible array configurations, the space complexity is approximately O(M^L) representing all possible array combinations that may be created and checked. Thus, auxiliary space used scales exponentially with both the maximum number (M) and the length of array (L).

Optimal Solution

Approach

The core idea is to realize that we don't need to explore every possible array. The count is heavily influenced by how many distinct prime factors a number has, and how those factors can be combined to construct an increasing array. This lets us use combinatorics to arrive at the answer efficiently.

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

First, identify the prime factors of each possible number within the given range.
Next, determine how many times each prime factor can appear in our array, since we know array elements must be multiples of their predecessors.
Recognize that building the array is like distributing prime factors across the array positions.
Now use combinations to figure out the number of ways to distribute each prime factor across the positions in the array.
Multiply together the number of combinations for each prime factor to get the total number of ideal arrays for that maximum number.
Finally sum these counts for each possible maximum number in the array to find the grand total.

Code Implementation

def count_the_number_of_ideal_arrays(maximum_number, array_length):
    modulus = 10**9 + 7

    def prime_factorization(number):
        factors = {}
        divisor = 2
        while divisor * divisor <= number:
            if number % divisor == 0:
                count = 0
                while number % divisor == 0:
                    number //= divisor
                    count += 1
                factors[divisor] = count
            divisor += 1
        if number > 1:
            factors[number] = 1
        return factors

    def combinations(number_of_items, number_to_choose, modulus):
        if number_to_choose < 0 or number_to_choose > number_of_items:
            return 0
        if number_to_choose == 0 or number_to_choose == number_of_items:
            return 1
        if number_to_choose > number_of_items // 2:
            number_to_choose = number_of_items - number_to_choose

        result = 1
        for i in range(number_to_choose):
            result = (result * (number_of_items - i)) % modulus
            power = pow(i + 1, modulus - 2, modulus)
            result = (result * power) % modulus
        return result

    total_ideal_arrays = 0
    for maximum_value in range(1, maximum_number + 1):
        prime_factors = prime_factorization(maximum_value)
        current_ideal_arrays = 1
        
        # Need to distribute prime factors, not actual number.
        for prime in prime_factors:
            exponent = prime_factors[prime]
            
            # Distribute exponents across array positions.
            current_ideal_arrays = (current_ideal_arrays * combinations(array_length + exponent - 1, exponent, modulus)) % modulus

        total_ideal_arrays = (total_ideal_arrays + current_ideal_arrays) % modulus

    return total_ideal_arrays

Big(O) Analysis

Time Complexity

O(n * log(n) + n * log(k)) – The algorithm iterates through each number from 1 to n to find its prime factors, which contributes a factor of log(n) due to prime factorization. This gives us an initial complexity of O(n * log(n)). Next, for each number from 1 to n, the code calculates combinations based on prime factor counts, where 'k' represents the length of the ideal array. Combination calculation will take log(k) since we are precomputing factorials. Summing this cost over each number from 1 to n adds O(n * log(k)) to the total runtime. Therefore, the combined time complexity is O(n * log(n) + n * log(k)).

Space Complexity

O(N) – The algorithm identifies prime factors for each number up to N, potentially storing them in a data structure like a list or dictionary. The size of this data structure is related to the maximum number N, as we need to store primes or their counts for each number up to N. Additionally, combinations calculations might involve memoization or temporary storage dependent on N. Therefore, the space complexity scales linearly with N.

Edge Cases

Case	How to Handle
n = 1, maxValue = any	Return 1, as an array of length 1 always has one ideal array
n = any, maxValue = 1	Return 1, as all elements must be 1 to maintain the ideal array property
n is very large, maxValue is small	The dynamic programming table might be large but the maxValue bound keeps the computation tractable
n is small, maxValue is very large	The dynamic programming table will be small related to n, but the maxValue range requires careful handling of integer overflow during computation of combinations
n = 0 or maxValue = 0	If n is 0, return 0; If maxValue is 0, return 0 if n > 0, and 1 if n=0 since empty array is considered valid
Potential integer overflow in combinations calculation	Use modulo arithmetic during calculations of combinations and intermediate products to prevent integer overflow
maxValue close to the maximum integer value	Ensure that calculations involving maxValue do not lead to integer overflow, even if we modulo later
Edge case with many factors	Memoization will ensure efficiency in computing the number of valid arrays for each specific maxValue