Find All Possible Stable Binary Arrays II

Hard

Asked by:

24 views

Topics:

Dynamic Programming

You are given 3 positive integers zero, one, and limit.

A binary array arr is called stable if:

The number of occurrences of 0 in arr is exactly zero.
The number of occurrences of 1 in arr is exactly one.
Each subarray of arr with a size greater than limit must contain both 0 and 1.

Return the total number of stable binary arrays.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: zero = 1, one = 1, limit = 2

Output: 2

Explanation:

The two possible stable binary arrays are [1,0] and [0,1].

Example 2:

Input: zero = 1, one = 2, limit = 1

Output: 1

Explanation:

The only possible stable binary array is [1,0,1].

Example 3:

Input: zero = 3, one = 3, limit = 2

Output: 14

Explanation:

All the possible stable binary arrays are [0,0,1,0,1,1], [0,0,1,1,0,1], [0,1,0,0,1,1], [0,1,0,1,0,1], [0,1,0,1,1,0], [0,1,1,0,0,1], [0,1,1,0,1,0], [1,0,0,1,0,1], [1,0,0,1,1,0], [1,0,1,0,0,1], [1,0,1,0,1,0], [1,0,1,1,0,0], [1,1,0,0,1,0], and [1,1,0,1,0,0].

Constraints:

1 <= zero, one, limit <= 1000

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 are the constraints on `n` and `k`? Specifically, what are the maximum possible values for `n` and `k`, and is `k` guaranteed to be less than or equal to `n`?
If no stable binary array is possible for the given `n` and `k`, what value should I return? Should I return 0, null, or throw an exception?
Is `k` guaranteed to be non-negative? Can `k` be 0?
Could you provide a few examples of stable binary arrays and unstable binary arrays for given `n` and `k` to ensure my understanding of the stability condition is correct?
Are there any memory constraints I should be aware of, given the potential scale of `n` and `k`?

Brute Force Solution

Approach

The brute force strategy involves checking every single possible arrangement of zeros and ones in the array. We generate each potential array and then determine if that specific arrangement meets the stability criteria. If it does, we count it; otherwise, we discard it and move on to the next possibility.

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

Imagine listing every single possible combination of zeros and ones given the total number of spaces you have.
For each of those combinations, check to see if it's 'stable'. This means checking if every 'one' has at least one adjacent 'zero'.
If a specific combination does have every 'one' next to a 'zero', then that combination is 'stable', so count it.
If a specific combination does not have every 'one' next to a 'zero', then that combination is not 'stable', so ignore it.
Continue until you have looked at absolutely every possible combination of zeros and ones.
The final count is the number of stable combinations.

Code Implementation

def find_all_stable_binary_arrays_brute_force(number_of_zeros, number_of_ones):
    stable_arrays = []
    current_array = []

    def is_stable(array):
        for index in range(1, len(array) - 1):
            if array[index] == 0 and array[index - 1] == 0 and array[index + 1] == 0:
                return False
            if array[index] == 1 and array[index - 1] == 1 and array[index + 1] == 1:
                return False
        return True

    def find_all_stable_arrays_recursive(zeros_left, ones_left):
        nonlocal stable_arrays, current_array

        if zeros_left == 0 and ones_left == 0:
            if is_stable(current_array):
                stable_arrays.append(current_array[:])
            return

        # Check if placing a zero still results in a possible stable array
        if zeros_left > 0:
            current_array.append(0)
            find_all_stable_arrays_recursive(zeros_left - 1, ones_left)
            current_array.pop() # Backtrack

        # Check if placing a one still results in a possible stable array
        if ones_left > 0:
            current_array.append(1)

            # Recursively explore the solution space by adding more values
            find_all_stable_arrays_recursive(zeros_left, ones_left - 1)
            current_array.pop() # Backtrack to previous state

    find_all_stable_arrays_recursive(number_of_zeros, number_of_ones)
    return stable_arrays

Big(O) Analysis

Time Complexity

O(2^n * n) – The algorithm generates all possible binary arrays of length n, which takes O(2^n) time, as there are 2^n possible arrays. For each generated array, it then checks if it's stable, which involves iterating through the array to see if each 'one' has an adjacent 'zero'. This stability check takes O(n) time in the worst case. Therefore, the overall time complexity is O(2^n * n).

Space Complexity

O(N) – The brute force approach generates every possible arrangement of zeros and ones of length N, where N is the size of the array. To check each arrangement for stability, the arrangement itself must be stored, requiring an array or list of size N. Although the individual arrangements are checked and discarded, the brute force methodology requires exploration of the space by creating the array of size N in each permutation step. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

To find all the stable binary arrays, we use a counting technique that avoids generating every possibility. We use a formula based on combinations to directly calculate how many stable arrays exist, making the process very efficient. This means we only have to do math rather than constructing arrays.

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

Understand that a stable binary array means no two zeros can be next to each other.
Consider how many ones and zeros we have available.
Recognize that each zero must be surrounded by ones (except possibly at the very beginning or end).
Figure out how many 'slots' the zeros can fit into between the ones.
Use a combinations formula (like n choose k) to figure out how many different ways we can place the zeros into those slots.
The result of this combination is the number of stable binary arrays that can be created with those numbers of ones and zeros.
Consider the edge cases where there are zero ones or zero zeros.

Code Implementation

def find_all_possible_stable_binary_arrays(zeros_count, ones_count):
    # Calculate the number of available slots for ones.
    number_of_slots = zeros_count + 1

    # If there are not enough slots, return 0.
    if ones_count > number_of_slots:
        return 0

    # Calculate combinations using dynamic programming
    combinations = [[0] * (ones_count + 1) for _ in range(number_of_slots + 1)]

    for number_of_items in range(number_of_slots + 1):
        combinations[number_of_items][0] = 1
        for number_to_choose in range(1, min(number_of_items, ones_count) + 1):
            combinations[number_of_items][number_to_choose] = combinations[number_of_items - 1][number_to_choose - 1] + combinations[number_of_items - 1][number_to_choose]

    # Return the number of possible combinations.
    return combinations[number_of_slots][ones_count]

Big(O) Analysis

Time Complexity

O(1) – The described approach uses a combination formula (n choose k) to directly calculate the number of stable arrays rather than generating them. The calculation of the combination can be considered a constant-time operation as the input sizes (number of ones and zeros) are parameters to the combination formula and do not significantly scale with the overall size of the array that *could* be built. Therefore, the overall time complexity is dominated by the mathematical calculation, which remains constant regardless of a theoretical input array size 'n'. Thus, the time complexity is O(1).

Space Complexity

O(1) – The provided explanation focuses on a counting technique that uses a combinations formula. It does not mention the creation of any auxiliary data structures like lists, arrays, or hash maps. The calculation likely involves storing a few integer variables to hold intermediate results during the combinations calculation, such as the number of ones, zeros, slots, and the result of the combination. Since the number of these variables is constant regardless of the input (number of ones and zeros), the space complexity is O(1).

Edge Cases

Case	How to Handle
n = 0, k = 0	Should return 1, as an empty array technically satisfies the stability condition with no ones.
n < k	Return 0, as it's impossible to have more ones than the length of the array.
k = 0	Return 1, an array of all zeros is always stable.
k = 1	Return n, as there are n possible positions for the single '1'.
k = n	If k = n, the array is all ones, which is stable, return 1.
k = n - 1	This case will never be stable, return 0.
n is large, k is small (e.g., n = 100, k = 2)	Ensure the solution uses dynamic programming or other efficient algorithms to avoid exponential time complexity; combinations (n-k+1 choose k) must be computed efficiently to prevent overflow.
Integer overflow in intermediate calculations (e.g., binomial coefficients)	Use a data type capable of holding large numbers (e.g., long long in C++, or a dedicated arbitrary-precision arithmetic library if necessary).

n = 0, k = 0

How to Handle:

Should return 1, as an empty array technically satisfies the stability condition with no ones.

n < k

How to Handle:

Return 0, as it's impossible to have more ones than the length of the array.

k = 0

How to Handle:

Return 1, an array of all zeros is always stable.

k = 1

How to Handle:

Return n, as there are n possible positions for the single '1'.

k = n

How to Handle:

If k = n, the array is all ones, which is stable, return 1.

k = n - 1

How to Handle:

This case will never be stable, return 0.

n is large, k is small (e.g., n = 100, k = 2)

How to Handle:

Ensure the solution uses dynamic programming or other efficient algorithms to avoid exponential time complexity; combinations (n-k+1 choose k) must be computed efficiently to prevent overflow.

Integer overflow in intermediate calculations (e.g., binomial coefficients)

How to Handle:

Use a data type capable of holding large numbers (e.g., long long in C++, or a dedicated arbitrary-precision arithmetic library if necessary).