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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Bytedance SWE coding round with a tricky binary string splitting problem. The time pressure was real and from what I gathered most people don't even finish it, which is both reassuring and kind of stressful to know going in.

Questions Asked (1)

Q1

Given a binary string, split it into exactly 3 non-empty contiguous parts such that each part contains the same number of '1's. Return the number of valid ways to do this, modulo 10^9 + 7.

Algorithms & Data Structures
Author's notes

The zero case is what gets people.

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AI HintsAI Generated

Suggested Approach

First, count the total number of '1's; if it's not divisible by 3, return 0. Then, find the positions of the first and last '1' in each third, and the number of valid splits is the product of the number of zeros between the first and second parts and between the second and third parts. Use modular arithmetic to handle large numbers.

Pro tip: Clarify edge cases early, such as when the string has no '1's or fewer than three '1's, and mention that the solution runs in O(n) time and O(1) space.

1. Count total ones

Count the total number of '1's in the string. If the count is not divisible by 3, return 0 immediately.

2. Identify boundaries

If the count is 0, return (n-1)*(n-2)/2 mod MOD. Otherwise, find the indices of the first and last '1' in each of the three parts by scanning and counting ones.

3. Count valid splits

The number of ways to split is the product of the number of zeros between the first and second parts and between the second and third parts. Compute this product modulo 10^9+7.

4. Handle edge cases

Ensure the string has at least three '1's if total ones > 0, and handle the all-zeros case separately.

Key Points to Mention

  • Total number of '1's must be divisible by 3, otherwise no valid split exists.
  • For each part, the number of '1's is exactly total_ones/3.
  • The valid split points are determined by the zeros between the first and second parts and between the second and third parts.
  • The answer is the product of the number of zeros in these two gaps, modulo 10^9+7.
  • Time complexity O(n) and space complexity O(1).
  • Special case: if the string contains no '1's, the number of ways is C(n-1, 2) = (n-1)*(n-2)/2.

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