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Ways to Split Array Into Good Subarrays

Medium

Asked by:

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Topics:

Arrays

You are given a binary array nums.

A subarray of an array is good if it contains exactly one element with the value 1.

Return an integer denoting the number of ways to split the array nums into good subarrays. As the number may be too large, return it modulo 10⁹ + 7.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [0,1,0,0,1]
Output: 3
Explanation: There are 3 ways to split nums into good subarrays:
- [0,1] [0,0,1]
- [0,1,0] [0,1]
- [0,1,0,0] [1]

Example 2:

Input: nums = [0,1,0]
Output: 1
Explanation: There is 1 way to split nums into good subarrays:
- [0,1,0]

Constraints:

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

Try coding it in LeetCode

Ways to Split Array Into Good Subarrays

Medium

Asked by:

28 views

Topics:

Arrays

You are given a binary array nums.

A subarray of an array is good if it contains exactly one element with the value 1.

Return an integer denoting the number of ways to split the array nums into good subarrays. As the number may be too large, return it modulo 10⁹ + 7.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [0,1,0,0,1]
Output: 3
Explanation: There are 3 ways to split nums into good subarrays:
- [0,1] [0,0,1]
- [0,1,0] [0,1]
- [0,1,0,0] [1]

Example 2:

Input: nums = [0,1,0]
Output: 1
Explanation: There is 1 way to split nums into good subarrays:
- [0,1,0]

Constraints:

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

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 the values within the array and the length of the array? Can the values be negative, zero, or non-integers?
What defines a 'good' subarray according to the problem statement? Could you provide a more precise definition or example?
If the array cannot be split into 'good' subarrays, what should the function return? Should I return an empty list, null, or throw an exception?
Are overlapping subarrays allowed? Or must the split result in contiguous, non-overlapping subarrays covering the entire original array?
If multiple valid splits exist, is any valid split acceptable as output, or is there a particular criterion for choosing the 'best' split?

Brute Force Solution

Approach

The brute force approach to splitting an array into good subarrays involves checking every possible way to divide the array. It's like trying every single combination of cuts to see which ones work. We generate all splits and then validate each one.

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

Consider all the possible starting points for the first subarray. It could start at the very beginning, or one position later, or even further along.
For each starting point, consider all the possible ending points for the first subarray. This defines the length of the first subarray.
Check if the first subarray meets the 'good' criteria. If it doesn't, move on to the next possible split.
If the first subarray *is* good, repeat the process for the remaining part of the array. Find all the possible 'good' subarrays that can follow the first one.
Keep doing this until the entire original array is split into 'good' subarrays.
Every time you find a way to split the *entire* array into consecutive 'good' subarrays, count that as one valid way to split the array.
In the end, the total number of valid ways you've counted is the answer.

Code Implementation

def ways_to_split_array_into_good_subarrays(numbers) -> int:
    array_length = len(numbers)
    number_of_ways = 0

    def is_subarray_good(subarray) -> bool:
        return sum(subarray) > 0

    def solve(current_index, current_split) -> None:
        nonlocal number_of_ways

        # If we've reached the end of the array, it's a valid split.
        if current_index == array_length:
            number_of_ways += 1
            return

        # Iterate through possible end points.
        for next_index in range(current_index + 1, array_length + 1):
            subarray = numbers[current_index:next_index]

            # Only proceed if the current subarray is good.
            if is_subarray_good(subarray):

                # Explore further splits.
                solve(next_index, current_split + [subarray])

    # Start the recursion.
    solve(0, [])

    return number_of_ways

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach examines all possible splits of the array. For an array of size n, there are 2^(n-1) possible ways to split it (each element has the potential to be the start of a new subarray). For each split, we potentially iterate through the subarrays created to validate each to determine if it is a 'good' subarray. Thus, the total time complexity is exponential in the worst case because we explore nearly all possible combinations. This considers ALL subsets.

Space Complexity

O(N) – The brute force approach, as described, implicitly uses recursion to explore all possible splits. In the worst-case scenario, where the first subarray is of size 1, then 2, then 3, and so on, the recursion might go as deep as N, where N is the length of the input array, because at each recursive call, the array is split and the remaining array is passed to the next function call. Each level of recursion requires storing a stack frame with local variables and the return address. Therefore, the maximum depth of the recursion stack can be N, resulting in O(N) space complexity due to the call stack.

Optimal Solution

Approach

The problem asks us to count the number of ways to divide an input into 'good' sections. The key insight involves realizing 'good' sections are determined by where zeros are, and how many ones are in between them. We can leverage this to calculate the number of ways to make the split efficiently.

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

First, check if the entire input is all zeros. If so, there's only one way to split it (doing nothing), unless there is an error or something is not 'good' about this zero-only input.
Next, find all the locations where a zero is followed by a one. These are the critical points that dictate where splits *can* happen without violating rules about 'good' sections.
Multiply the number of options together. The final result will be the product of all of these counts. This works because choices at different points are independent. Multiplying them allows you to obtain the total number of combinations. This avoids needing to enumerate all possible splits.

Code Implementation

def ways_to_split_array(input_array):
    number_of_elements = len(input_array)
    number_of_ways = 1
    MODULO = 10**9 + 7

    # Check if the array is all zeros
    if all(element == 0 for element in input_array):
        return 1

    split_points = []
    for index in range(number_of_elements - 1):
        if input_array[index] == 0 and input_array[index + 1] == 1:
            # Found a valid split point
            split_points.append(index)

    # If there are no split points, there is only one way
    if not split_points:
        return 1

    previous_split_point = -1
    for split_point in split_points:
        ones_count = 0
        for index in range(previous_split_point + 1, split_point + 1):
            if input_array[index] == 1:
                ones_count += 1

        # The number of ways to split at this point
        number_of_ways = (number_of_ways * (ones_count + 1)) % MODULO
        previous_split_point = split_point

    # Multiply the possibilities at each split point.
    return number_of_ways

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n a maximum of two times. The first pass checks for the all-zeros case, which takes O(n) time. The second pass finds the locations where a zero is followed by a one, also taking O(n) time. Finally, multiplying the number of options together takes time proportional to the number of such locations, which is bounded by n. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm identifies the locations where a zero is followed by a one and stores them. This requires storing these indices in a data structure, likely a list or array. In the worst case, every element could be such a location, thus requiring space proportional to the input size N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 1, as there's one way to split an empty array (no splits).
Array with a single element	If the single element is odd, return 1; otherwise, return 0 (cannot split into good subarrays).
Array with all even numbers	Return 1, because no good subarrays can be formed, and thus there is one way (no splits).
Array with all odd numbers	Return the length of the array plus one, since every split yields good subarrrays.
Array containing leading even numbers followed by odd number(s)	The leading even numbers cannot be part of a good subarray, so the number of ways to split equals the number of ways to split the remaining array starting at the first odd number.
Very large array exceeding memory limits	Implement a dynamic programming approach that only stores the necessary previous states to avoid memory overflow.
Integer overflow when calculating the number of ways to split	Use modulo arithmetic with a prime number to prevent integer overflow during multiplication.
Array contains a large number of consecutive even numbers interspersed within the array	Optimize by efficiently skipping over the consecutive even number sequences and directly focusing on the odd number positions when calculating the splittings.

Empty input array

How to Handle:

Return 1, as there's one way to split an empty array (no splits).

Array with a single element

How to Handle:

If the single element is odd, return 1; otherwise, return 0 (cannot split into good subarrays).

Array with all even numbers

How to Handle:

Return 1, because no good subarrays can be formed, and thus there is one way (no splits).

Array with all odd numbers

How to Handle:

Return the length of the array plus one, since every split yields good subarrrays.

Array containing leading even numbers followed by odd number(s)

How to Handle:

The leading even numbers cannot be part of a good subarray, so the number of ways to split equals the number of ways to split the remaining array starting at the first odd number.

Very large array exceeding memory limits

How to Handle:

Implement a dynamic programming approach that only stores the necessary previous states to avoid memory overflow.

Integer overflow when calculating the number of ways to split

How to Handle:

Use modulo arithmetic with a prime number to prevent integer overflow during multiplication.

Array contains a large number of consecutive even numbers interspersed within the array

How to Handle:

Optimize by efficiently skipping over the consecutive even number sequences and directly focusing on the odd number positions when calculating the splittings.

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 the values within the array and the length of the array? Can the values be negative, zero, or non-integers?
What defines a 'good' subarray according to the problem statement? Could you provide a more precise definition or example?
If the array cannot be split into 'good' subarrays, what should the function return? Should I return an empty list, null, or throw an exception?
Are overlapping subarrays allowed? Or must the split result in contiguous, non-overlapping subarrays covering the entire original array?
If multiple valid splits exist, is any valid split acceptable as output, or is there a particular criterion for choosing the 'best' split?

Brute Force Solution

Approach

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

Consider all the possible starting points for the first subarray. It could start at the very beginning, or one position later, or even further along.
For each starting point, consider all the possible ending points for the first subarray. This defines the length of the first subarray.
Check if the first subarray meets the 'good' criteria. If it doesn't, move on to the next possible split.
If the first subarray *is* good, repeat the process for the remaining part of the array. Find all the possible 'good' subarrays that can follow the first one.
Keep doing this until the entire original array is split into 'good' subarrays.
Every time you find a way to split the *entire* array into consecutive 'good' subarrays, count that as one valid way to split the array.
In the end, the total number of valid ways you've counted is the answer.

Code Implementation

def ways_to_split_array_into_good_subarrays(numbers) -> int:
    array_length = len(numbers)
    number_of_ways = 0

    def is_subarray_good(subarray) -> bool:
        return sum(subarray) > 0

    def solve(current_index, current_split) -> None:
        nonlocal number_of_ways

        # If we've reached the end of the array, it's a valid split.
        if current_index == array_length:
            number_of_ways += 1
            return

        # Iterate through possible end points.
        for next_index in range(current_index + 1, array_length + 1):
            subarray = numbers[current_index:next_index]

            # Only proceed if the current subarray is good.
            if is_subarray_good(subarray):

                # Explore further splits.
                solve(next_index, current_split + [subarray])

    # Start the recursion.
    solve(0, [])

    return number_of_ways

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, check if the entire input is all zeros. If so, there's only one way to split it (doing nothing), unless there is an error or something is not 'good' about this zero-only input.
Next, find all the locations where a zero is followed by a one. These are the critical points that dictate where splits *can* happen without violating rules about 'good' sections.
Multiply the number of options together. The final result will be the product of all of these counts. This works because choices at different points are independent. Multiplying them allows you to obtain the total number of combinations. This avoids needing to enumerate all possible splits.

Code Implementation

def ways_to_split_array(input_array):
    number_of_elements = len(input_array)
    number_of_ways = 1
    MODULO = 10**9 + 7

    # Check if the array is all zeros
    if all(element == 0 for element in input_array):
        return 1

    split_points = []
    for index in range(number_of_elements - 1):
        if input_array[index] == 0 and input_array[index + 1] == 1:
            # Found a valid split point
            split_points.append(index)

    # If there are no split points, there is only one way
    if not split_points:
        return 1

    previous_split_point = -1
    for split_point in split_points:
        ones_count = 0
        for index in range(previous_split_point + 1, split_point + 1):
            if input_array[index] == 1:
                ones_count += 1

        # The number of ways to split at this point
        number_of_ways = (number_of_ways * (ones_count + 1)) % MODULO
        previous_split_point = split_point

    # Multiply the possibilities at each split point.
    return number_of_ways

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 1, as there's one way to split an empty array (no splits).
Array with a single element	If the single element is odd, return 1; otherwise, return 0 (cannot split into good subarrays).
Array with all even numbers	Return 1, because no good subarrays can be formed, and thus there is one way (no splits).
Array with all odd numbers	Return the length of the array plus one, since every split yields good subarrrays.
Array containing leading even numbers followed by odd number(s)	The leading even numbers cannot be part of a good subarray, so the number of ways to split equals the number of ways to split the remaining array starting at the first odd number.
Very large array exceeding memory limits	Implement a dynamic programming approach that only stores the necessary previous states to avoid memory overflow.
Integer overflow when calculating the number of ways to split	Use modulo arithmetic with a prime number to prevent integer overflow during multiplication.
Array contains a large number of consecutive even numbers interspersed within the array	Optimize by efficiently skipping over the consecutive even number sequences and directly focusing on the odd number positions when calculating the splittings.

Empty input array

How to Handle:

Return 1, as there's one way to split an empty array (no splits).

Array with a single element

How to Handle:

If the single element is odd, return 1; otherwise, return 0 (cannot split into good subarrays).

Array with all even numbers

How to Handle:

Return 1, because no good subarrays can be formed, and thus there is one way (no splits).

Array with all odd numbers

How to Handle:

Return the length of the array plus one, since every split yields good subarrrays.

Array containing leading even numbers followed by odd number(s)

How to Handle:

The leading even numbers cannot be part of a good subarray, so the number of ways to split equals the number of ways to split the remaining array starting at the first odd number.

Very large array exceeding memory limits

How to Handle:

Implement a dynamic programming approach that only stores the necessary previous states to avoid memory overflow.

Integer overflow when calculating the number of ways to split

How to Handle:

Use modulo arithmetic with a prime number to prevent integer overflow during multiplication.

Array contains a large number of consecutive even numbers interspersed within the array

How to Handle:

Optimize by efficiently skipping over the consecutive even number sequences and directly focusing on the odd number positions when calculating the splittings.