Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Bytedance SWE interview with a graph traversal problem that looks deceptively simple but has a specific constraint that changes the approach entirely. Pretty standard coding round vibe.

Questions Asked (1)

You have an integer array and a target value T. Starting at index 0, you can jump forward 1 to 10 positions at a time. Each index has a cost equal to its value. Find the minimum number of jumps needed to land on an index whose value equals T. Return -1 if no reachable index has that value.

Algorithms & Data Structures

Author's notes

My first instinct was dynamic programming because of the 'minimum steps' framing and the jump constraint, which felt like a classic staircase problem.

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AI HintsAI Generated

Suggested Approach

Model the problem as a shortest path on a directed graph where each index is a node and edges go from i to i+1 through i+10. Use BFS to find the minimum number of jumps to any index with value T, since all edges have unit weight. If no such index is reachable, return -1.

Pro tip: Clarify with the interviewer whether the cost of an index affects the jump count or is just a value to match; the problem statement says 'cost equal to its value' but asks for minimum jumps, so the cost is likely irrelevant except for matching T. Also, mention that BFS is optimal here because each jump counts as one step, and early termination when T is found ensures efficiency.

1. Understand the problem and constraints

Confirm that you start at index 0, can jump 1 to 10 positions forward, and need the minimum number of jumps to land on an index with value T. Ask about edge cases like T at index 0, array length, and negative values.

2. Choose the right algorithm

Recognize this as a shortest path problem on an unweighted graph, so BFS is ideal. Alternatively, dynamic programming can be used, but BFS naturally finds the minimum jumps.

3. Implement BFS with a queue

Initialize a queue with index 0 and a jumps count of 0. For each index, check if its value equals T; if so, return jumps. Otherwise, enqueue all reachable indices (i+1 to i+10) that are within bounds and not visited.

4. Handle termination and edge cases

If the queue empties without finding T, return -1. Also handle cases where T is at index 0 (return 0 jumps) and ensure you don't revisit indices to avoid cycles.

5. Analyze complexity and optimize

Time complexity is O(N * 10) = O(N) since each index is enqueued at most once and we check up to 10 neighbors. Space complexity is O(N) for the queue and visited set. Mention that early termination can improve average performance.

Key Points to Mention

BFS guarantees minimum jumps because all edges have unit weight.
Use a visited array or set to avoid reprocessing indices and infinite loops.
The cost of an index is only used to check if it equals T; it does not affect jump count.
Edge cases: T at index 0 (return 0), T not present, array length less than jump range.
Time complexity O(N) and space complexity O(N) due to BFS.
Alternative DP approach: dp[i] = min jumps to reach i, but BFS is more intuitive for minimum steps.

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