Number of Ways to Split Array

Medium

Number of Ways to Split Array

Medium

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 size of the input array, and what is the range of values for the numbers within the array? Can the numbers be negative or zero?
What constitutes a 'split'? Is it splitting into exactly two subarrays, or can it be split into more?
What is the expected return value if there are no valid splits possible?
Are there any constraints on the size or contents of the subarrays formed after the split?
Are there any specific criteria for choosing between multiple valid splits (e.g., should I prioritize splits with smaller subarrays on the left)?

Brute Force Solution

Approach

The brute force strategy for finding the number of ways to split an array involves trying every single possible split. We look at each split, assess if it meets our criteria, and count the splits that do. This exhaustive approach guarantees we find the right answer, albeit inefficiently.

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

Imagine a divider that can be placed at different points throughout the group of numbers.
Start by placing the divider after the first number, creating two groups: one with only the first number, and the rest of the numbers in the other group.
Check if the sum of the first group is bigger or equal than the sum of the second group. If it is, increase the count of successful splits.
Move the divider to the right, after the second number. Now the first group contains the first two numbers, and the other group contains the remaining numbers.
Again, check if the sum of the first group is bigger or equal than the sum of the second group. If it is, increase the count of successful splits.
Repeat this process, moving the divider one number at a time to the right, checking the sums and updating the count each time.
Stop when the divider reaches the second-to-last number (as the last group has to have at least one number).
The final count represents the total number of valid ways to split the group of numbers according to the rule.

Code Implementation

def number_of_ways_to_split_array_brute_force(numbers):
    number_of_successful_splits = 0
    number_of_elements = len(numbers)

    # Iterate up to the second to last element.
    for split_index in range(1, number_of_elements):

        first_group_sum = 0
        for index in range(split_index):
            first_group_sum += numbers[index]

        second_group_sum = 0
        for index in range(split_index, number_of_elements):
            second_group_sum += numbers[index]

        # Check if the first group's sum is greater or equal.
        if first_group_sum >= second_group_sum:
            # Count the successful splits
            number_of_successful_splits += 1

    return number_of_successful_splits

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array, considering each position as a potential split point. For each split point, it calculates the sum of the left subarray and the sum of the right subarray. Calculating each sum requires iterating through the respective subarray. In the worst case, for each of the n-1 possible split points, we may iterate through almost the entire array again to calculate the sums. Therefore, the time complexity is proportional to (n-1) * (average length of iteration for sum calculations), which approximates n * n/2. This simplifies to O(n²).

Space Complexity

O(1) – The described brute force solution only uses a few variables to keep track of the current split and calculate sums. Specifically, it doesn't create any auxiliary data structures that scale with the input array's size N. Therefore, the extra memory used is constant, independent of the input array's length.

Optimal Solution

Approach

The goal is to find how many ways you can cut an ordered list of numbers into three parts so that the sum of the first part is less than or equal to the sum of the second part, and the sum of the second part is less than or equal to the sum of the third part. The key is to efficiently calculate the sums without recomputing them every time we consider a new cut.

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

First, calculate the total sum of all the numbers in the list.
Then, start from the beginning and calculate the sum of the first part as you move along the list.
For each possible sum of the first part, calculate the remaining sum (total sum minus sum of the first part).
Figure out the minimum possible sum the second part can have (same as the sum of the first part).
Figure out the maximum possible sum the second part can have. This value will ensure the sum of the third part will be greater or equal to the second part.
Check how many ways you can divide the remaining numbers to achieve those sums for the second part (between the minimum and maximum you calculated), this is the number of valid splits for that first part.
Add up the number of valid splits for each possible first part. This gives you the total number of ways to split the list according to the rules.

Code Implementation

def number_of_ways_to_split_array(numbers):
    total_sum = sum(numbers)
    count_of_ways = 0
    first_part_sum = 0

    for i in range(len(numbers) - 2):
        first_part_sum += numbers[i]
        remaining_sum = total_sum - first_part_sum

        # The min second part sum is the first, to satisfy condition 1.
        min_second_part_sum = first_part_sum

        # Condition 2 requires third part to be >= second part.
        max_second_part_sum = (remaining_sum) // 2

        valid_splits = 0
        second_part_sum = 0

        for j in range(i + 1, len(numbers) - 1):
            second_part_sum += numbers[j]

            if (second_part_sum >= min_second_part_sum and
                    second_part_sum <= max_second_part_sum):

                # We increment here, a valid second part sum found.
                valid_splits += 1

        count_of_ways += valid_splits

    return count_of_ways

Big(O) Analysis

Time Complexity

O(n²) – The algorithm begins by calculating the total sum of the array, which takes O(n) time. Then, it iterates through the array to consider each element as the end of the first part. For each of these n elements, it determines the possible range of sums for the second part based on the first part's sum and the remaining total. It appears that for each possible first part, it implicitly iterates through potential second part sizes to count valid splits, leading to nested iterations in the worst case. Thus, the dominant operation involves iterating through a variable portion of the array for each of the n possible first splits resulting in roughly n * n/2 operations. Therefore, the overall time complexity is O(n²).

Space Complexity

O(1) – The provided solution primarily calculates sums and counts without using any auxiliary data structures that scale with the input size N (the number of elements in the list). While it involves variables to store the total sum, partial sums, minimum/maximum second part sums, and the count of valid splits, these are all constant-sized variables. Therefore, the space used remains constant regardless of the input size, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no splits possible.
Array with only one element	Return 0 as there is no valid split.
Array with all identical elements	The algorithm should correctly calculate splits based on the sum requirement, even with identical elements.
Array with negative numbers	The sum calculation must handle negative numbers correctly, and the split condition must consider the potentially negative sums.
Array with large numbers potentially leading to integer overflow in sums	Use a data type that can accommodate large sums, such as long, to prevent overflow.
Array where no valid split exists (left sum never less than or equal to right sum)	The algorithm should return 0 if no such split is found after iterating through all possible split points.
Maximum sized input array to assess time complexity and potential memory issues	The solution should have a time complexity of O(n) to handle large inputs efficiently by precalculating total sum.
Array containing zero values	Zeros should not affect the basic functionality of the split as they sum to zero.