Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Salesforce SWE interview with a pretty meaty algorithm problem. The constraints were wild (n up to 2^61-1) and they wanted a full justification, not just code.

Questions Asked (1)

Given a positive integer n (up to 2^61-1), you can halve it if even, or increment/decrement by 1 if odd. What is the minimum number of steps to reach 1, and how do you decide whether to go n+1 or n-1 for odd values?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The core insight I kept second-guessing myself on was the n+1 vs n-1 choice.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by explaining the greedy strategy: for even n, always halve; for odd n, decide based on n mod 4 (except n=3). Then justify why this works using bit patterns and provide a step-by-step algorithm with examples.

Pro tip: Mention that the greedy choice is optimal because it minimizes the number of trailing ones in the binary representation, which directly reduces future operations. Also, handle the edge case n=3 separately.

1. Clarify the problem and constraints

Restate the problem: given n up to 2^61-1, find the minimum steps to reach 1 using halving (if even) and ±1 (if odd). Note that n can be very large, so an O(log n) solution is needed.

2. Derive the greedy rule for odd numbers

For odd n > 3, if n % 4 == 1, decrement (n-1); if n % 4 == 3, increment (n+1). For n=3, decrement to 2 then halve to 1 (2 steps). Explain that this minimizes the number of trailing ones in binary, leading to more halving opportunities.

3. Prove optimality with binary representation

Show that halving removes trailing zeros, and ±1 adjusts the least significant bits. The greedy choice reduces the number of consecutive 1s at the end, which would otherwise require multiple operations to clear. This ensures the minimum number of steps.

4. Outline the algorithm and complexity

While n > 1: if n even, n /= 2; else if n == 3 or n % 4 == 1, n -= 1; else n += 1. Count steps. Time complexity is O(log n) since each step reduces n by at least half every two operations.

5. Test with examples and edge cases

Walk through examples: n=7 (7→8→4→2→1, 4 steps), n=15 (15→16→8→4→2→1, 5 steps), n=3 (3→2→1, 2 steps). Highlight that n=1 requires 0 steps.

Key Points to Mention

Greedy strategy: halve when even; for odd, use n mod 4 to decide increment or decrement.
Special case n=3: decrement to 2 then halve (2 steps) is optimal.
Binary representation: trailing ones indicate inefficiency; greedy minimizes them.
Optimality proof: each operation reduces the number of bits or trailing ones, ensuring minimal steps.
Time complexity: O(log n) because n halves every two steps on average.
Edge cases: n=1 (0 steps), n=2 (1 step), n=3 (2 steps).

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