SoFi·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

SoFi software engineer interview that went deep on a classic jump game problem. They didn't just want code, they wanted the full treatment: proof of correctness, complexity analysis, edge cases, the works.

Questions Asked (1)

Given an array of non-negative integers where each value represents the max steps you can jump forward from that index, first determine if you can reach the last index from index 0, and then find the minimum number of jumps to get there. They wanted an O(n) time, O(1) space solution, plus a correctness proof, complexity analysis, and a full edge case walkthrough.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This is jump game 1 and 2 stapled together, which I didn't expect.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then present the greedy BFS-layer approach: track the current jump's reachable window and the farthest reachable index. After confirming reachability, extend the window layer by layer, incrementing jumps when the current window is exhausted. Conclude with a correctness proof, complexity analysis, and edge case walkthrough.

Pro tip: Emphasize that the greedy approach is essentially a BFS over jump layers, and explicitly state the invariant that after k jumps you can reach any index up to the k-th layer's farthest point. This shows deep understanding and makes the proof trivial.

1. Clarify and handle edge cases

Confirm input constraints (non-negative integers, array length, values). Discuss edge cases: empty array, single element, zero values, unreachable last index, and large jumps.

2. Present the greedy O(n) O(1) algorithm

Maintain variables: jumps, current_end (end of current jump layer), farthest (max reach so far). Iterate through array; update farthest; when i == current_end, increment jumps and set current_end = farthest. Stop if current_end >= last index.

3. Prove correctness

Use induction: after k jumps, current_end is the farthest index reachable in exactly k jumps. Show that the greedy choice of extending to farthest is optimal because any index within the current layer can be reached with the same number of jumps.

4. Analyze complexity

Time: O(n) because each element is visited once. Space: O(1) because only a few variables are used. Mention that this is optimal for the problem.

5. Walk through edge cases and examples

Demonstrate with examples: [2,3,1,1,4] (reachable, min jumps 2), [3,2,1,0,4] (unreachable), [0] (already at last index, 0 jumps), [1,0] (unreachable). Explain how the algorithm handles each.

Key Points to Mention

Greedy BFS-layer approach: each jump expands the reachable window.
Invariant: after k jumps, all indices up to current_end are reachable.
Reachability check: if farthest never reaches last index, return false/-1.
Minimum jumps: increment when the current layer is exhausted.
Time O(n) and space O(1) optimality.
Edge cases: empty array, single element, zeros, unreachable last index.

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