Ways to Split Array Into Good Subarrays

Ways to Split Array Into Good Subarrays

What is this problem about?

This problem involves splitting an array into one or more subarrays. A "good" subarray is defined as one that contains exactly one '1'. You need to find the total number of ways to split the entire array such that every resulting subarray is a good subarray. If the array contains no 1s, the number of ways is 0.

Why is this asked in interviews?

Flipkart and other e-commerce tech teams often ask this to test basic Combinatorics and Array Traversal. It’s a logic-heavy problem disguised as an array manipulation task. It requires the candidate to realize that the split points only matter between consecutive 1s. This tests your ability to simplify a complex-sounding requirement into a simple mathematical multiplication.

Algorithmic pattern used

The pattern is Math and Counting. The split can happen anywhere between two consecutive 1s. If there are k zeros between two 1s, there are k+1 possible places to put a "divider." To find the total ways, you multiply the number of possible positions between every pair of consecutive 1s. This is an O(N) approach as you only need to find the indices of the 1s.

Example explanation

Array: [1, 0, 0, 1, 0, 1].

1. Locate the 1s at indices 0, 3, and 5.
2. Between the first 1 (idx 0) and second 1 (idx 3), there are 2 zeros. This gives us 3 possible split points:
   • [1], [0, 0, 1, ...]
   • [1, 0], [0, 1, ...]
   • [1, 0, 0], [1, ...]
3. Between the second 1 (idx 3) and third 1 (idx 5), there is 1 zero. This gives us 2 possible split points.
4. Total ways = 3 * 2 = 6.

Common mistakes candidates make

• Ignoring the "no 1s" case: Forgetting to return 0 if the array doesn't have any 1s.
• Off-by-one in counting zeros: Miscounting the gaps between the 1s.
• Adding instead of multiplying: Using addition to combine the possibilities between gaps instead of the product rule of combinatorics.

Interview preparation tip

Always look for the "bottleneck" or the "constraint" in split problems. Here, the constraint is that each subarray must have exactly one '1'. By focusing on the 1s, you transform a potentially difficult DP problem into a simple multiplication problem.