How do you solve Ways to Split Array Into Good Subarrays on LeetCode?

Ways to Split Array Into Good Subarrays (LeetCode #2750) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying key positions (indices of '1's) simplifies the problem.

What is the time complexity of Ways to Split Array Into Good Subarrays?

The optimal solution for Ways to Split Array Into Good Subarrays runs in O(n) time and uses O(n) space. This complexity is linear because we only traverse the array a couple of times: once to find indices of '1's and once to calculate the gaps.

What is the brute force solution for Ways to Split Array Into Good Subarrays?

The brute force approach for Ways to Split Array Into Good Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible way to split the array into subarrays and counts those that are good. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Ways to Split Array Into Good Subarrays?

Ways to Split Array Into Good Subarrays uses the following concepts: Array, Math, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Ways to Split Array Into Good Subarrays?

Always start with a brute force solution to understand the problem better. Look for patterns in the data, such as the positions of '1's. Practice optimizing your solutions to improve efficiency.

What are common mistakes when solving Ways to Split Array Into Good Subarrays?

Not considering cases where there are no '1's in the array. Overcomplicating the counting of good subarrays instead of focusing on the gaps.

#2750

Ways to Split Array Into Good Subarrays

Medium

ArrayMathDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach identifies the positions of '1's and calculates the number of ways to split the array based on the gaps between them. This reduces the problem to linear time complexity.

⚙️

Algorithm

4 steps

1Step 1: Identify the indices of all '1's in the array.
2Step 2: Calculate the gaps between consecutive '1's to determine how many zeros are between them.
3Step 3: For each gap, the number of ways to split is (gap + 1) because you can place a split before each zero.
4Step 4: Multiply the number of ways for all gaps together and return the result modulo 10^9 + 7.

solution.py12 lines

1# Full working Python code
2
3def count_good_splits(nums):
4    MOD = 10**9 + 7
5    indices = [i for i, x in enumerate(nums) if x == 1]
6    if len(indices) == 0:
7        return 0
8    ways = 1
9    for i in range(1, len(indices)):
10        gap = indices[i] - indices[i-1] - 1
11        ways = (ways * (gap + 1)) % MOD
12    return ways

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Complexity note: This complexity is linear because we only traverse the array a couple of times: once to find indices of '1's and once to calculate the gaps.

1Identifying key positions (indices of '1's) simplifies the problem.
2The number of ways to split is determined by the gaps between '1's.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.