Google·Software Engineer·Onsite - Coding / Algorithms·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Google onsite coding round, AI/ML loop. One problem, the whole session, and it looked deceptively simple until I had to actually justify why two cuts are always enough.

Questions Asked (1)

Given a circular string of 'D' and 'R' characters with equal counts of each, find at most 2 cut positions that split the necklace into two pieces where each piece has the same number of D's and R's. You also need to justify why 2 cuts always suffice.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The part that tripped me up wasn't the sliding window, it was the verbal justification.

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AI HintsAI Generated

Suggested Approach

Model the circular string as a sequence of +1 (for 'D') and -1 (for 'R') values, and use prefix sums to identify cut positions where the cumulative sum returns to zero. The problem reduces to finding two cut points that partition the circle into two segments each with sum zero, which is always possible due to the balanced nature of the string.

Pro tip: Start by clarifying that the two pieces are contiguous arcs of the circle, and mention that the proof of sufficiency relies on the intermediate value theorem applied to the prefix sum function around the circle.

1. Reframe the problem

Convert the string into a numeric array where 'D' = +1 and 'R' = -1, and note that the total sum is zero because counts are equal.

2. Define prefix sums

Compute the prefix sum around the circle, treating it as a periodic function with period n, and identify points where the sum equals zero.

3. Find first cut

Choose any cut point where the prefix sum is zero; this ensures one piece (the arc between cuts) has balanced D's and R's.

4. Find second cut

From the first cut, traverse the circle and find another point where the prefix sum returns to zero, which must exist because the total sum is zero and the function is continuous in discrete steps.

5. Justify sufficiency

Explain that two cuts always suffice because the prefix sum function must cross zero at least twice (including the start/end) due to its periodic nature and zero total sum.

Key Points to Mention

Prefix sum technique for balanced parentheses or equal counts.
The circular string can be linearized by considering any starting point.
The intermediate value theorem (discrete version) guarantees a zero crossing.
Two cuts are sufficient because the problem reduces to partitioning a circular array into two zero-sum segments.
Time complexity: O(n) to compute prefix sums and find cuts.
Edge cases: all D's and R's alternating, or already balanced halves.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.