Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Salesforce software engineering interview with a single algorithmic problem. Pretty focused session, just one coding question the whole time.

Questions Asked (1)

Given a positive integer n, you can add or subtract any power of 2 from n in each operation. What is the minimum number of operations needed to reduce n to 0?

Algorithms & Data Structures

Author's notes

My first instinct was greedy, just keep subtracting the largest power of 2 that fits.

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AI HintsAI Generated

Suggested Approach

Recognize that this is a shortest path problem on integers where edges represent adding or subtracting powers of 2. Use BFS from n to 0, but optimize by considering only the nearest powers of 2 to the current value, as moving to a farther power would be suboptimal. Alternatively, derive a greedy strategy based on binary representation, but verify with BFS for small n.

Pro tip: Mention that the problem can be solved in O(log n) time by observing that the optimal strategy is to either round up or down to the nearest power of 2 at each step, and this can be computed using a recursive formula. This shows you can optimize beyond brute force.

1. Clarify the problem

Confirm that in each operation you can add or subtract any power of 2 (including 2^0=1) from the current number, and you want the minimum number of operations to reach exactly 0. Ask if n can be up to 10^9 or larger to determine the required efficiency.

2. Model as a graph problem

View each integer as a node, with edges to n ± 2^k for all k. The goal is the shortest path from n to 0. This suggests BFS, but the graph is infinite, so we need to bound the search space.

3. Identify optimal moves

Argue that from any number x, the optimal move is to add or subtract the largest power of 2 less than or equal to x, or the smallest power of 2 greater than x. Moving to a farther power is never beneficial because it overshoots and requires more steps to correct.

4. Derive a recursive or greedy solution

Define f(n) as the minimum operations. If n is a power of 2, f(n)=1. Otherwise, let p be the largest power of 2 ≤ n. Then f(n) = 1 + min(f(n-p), f(p*2 - n) + 1?) Actually, careful: f(n) = 1 + min(f(n-p), f(2p - n)) where 2p is the next power of 2. This leads to an O(log n) algorithm.

5. Validate with examples and edge cases

Test small values: n=1 -> 1, n=2 -> 1, n=3 -> 2 (3-2=1, 1-1=0), n=4 -> 1, n=5 -> 2 (5-4=1, 1-1=0), n=6 -> 2 (6-4=2, 2-2=0), n=7 -> 2 (7-8=-1? Actually 7+1=8, 8-8=0 -> 2 operations: 7+1=8, 8-8=0). Check n=15: 15+1=16, 16-16=0 -> 2 operations. This matches the pattern.

Key Points to Mention

The problem is equivalent to finding the shortest path in a graph where each node connects to n ± 2^k.
BFS can be used for small n, but the state space can be pruned by only considering the nearest powers of 2.
The optimal strategy often involves rounding up or down to the nearest power of 2, which can be proven by exchange argument.
A recursive formula f(n) = 1 + min(f(n-p), f(2p-n)) where p is the largest power of 2 ≤ n, with base case f(0)=0, gives an O(log n) solution.
The binary representation of n can guide the solution: each 1 bit can be handled by subtracting the corresponding power, but carries can reduce operations (e.g., 7 = 8-1 uses 2 operations instead of 3).
Edge cases: n=0 (0 operations), n=1 (1 operation), and large n require efficient algorithm.

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