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 <= 1050 <= nums[i] <= 1When 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:
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:
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_waysThe 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:
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| 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. |