Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Bytedance SWE interview with a graph/union-find style problem. Pretty standard stuff for a big tech coding round, though the transitive similarity angle made me think harder than I expected.

Questions Asked (1)

Given n photos and an n x n binary matrix indicating which photos are directly similar, find the total number of photo groups, where similarity is transitive (if A is similar to B and B is similar to C, then A and C are in the same group).

Algorithms & Data Structures

Author's notes

Classic connected components problem dressed up with a photo theme.

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AI HintsAI Generated

Suggested Approach

Model the photos as nodes in a graph where edges represent direct similarity, then find the number of connected components. Use Union-Find (Disjoint Set Union) for near-linear time, or BFS/DFS for O(n^2) time. Clearly explain the transitive property and why connected components solve the problem.

Pro tip: Mention that Union-Find with path compression and union by rank is optimal for dynamic connectivity, but since the graph is static, BFS/DFS is equally efficient and simpler. Also, clarify that the matrix is symmetric and the diagonal represents self-similarity, which doesn't affect the count.

1. Understand the problem

Restate that similarity is transitive, so groups are connected components in the graph defined by the matrix. Confirm that the matrix is symmetric and diagonal entries are 1 (or 0) but irrelevant.

2. Choose an algorithm

Decide between Union-Find and graph traversal (BFS/DFS). Union-Find is efficient for merging sets; BFS/DFS is straightforward for static graphs.

3. Implement the solution

For Union-Find: initialize parent array, union all pairs (i,j) where matrix[i][j]==1, then count unique roots. For BFS/DFS: build adjacency list or use matrix directly, traverse unvisited nodes, increment count for each new component.

4. Analyze complexity

Time: O(n^2) for both approaches (Union-Find with path compression is nearly O(n^2 α(n))). Space: O(n) for Union-Find or visited array, O(n^2) if adjacency list built.

5. Test with examples

Walk through a small example (e.g., n=3 with matrix [[1,1,0],[1,1,0],[0,0,1]]) to verify the count is 2. Discuss edge cases like n=0 or all zeros.

Key Points to Mention

Graph representation: nodes as photos, edges as direct similarity.
Transitive closure leads to connected components.
Union-Find (Disjoint Set Union) with path compression and union by rank.
BFS/DFS traversal for connected components.
Time complexity O(n^2) and space complexity O(n).
Handling edge cases: empty input, all photos similar, no similarities.

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