← Bytedance Interview Insights
Took me a beat longer than it should have to see this was just connected components.
Model the matrix as an undirected graph where each photo is a node and a 1 at [i][j] indicates an edge. The number of distinct photo groups equals the number of connected components in this graph. Use Union-Find (Disjoint Set Union) to efficiently merge similar photos and count components, or BFS/DFS if the graph is dense.
Pro tip: Clarify whether the matrix is symmetric and whether self-similarity (diagonal) is always true; this affects initialization. Also, mention that Union-Find with path compression and union by rank gives near O(N^2) time, which is optimal for reading the matrix.
Confirm that similarity is transitive, so groups are connected components. Ask if the matrix is symmetric and if diagonal entries are always 1.
Select Union-Find for efficient merging and component counting, or BFS/DFS if the graph is sparse. Discuss trade-offs.
Initialize each photo as its own group. Iterate through the upper triangle of the matrix (i < j) and union i and j when matrix[i][j] is 1.
After processing all pairs, count the number of unique roots in the Union-Find structure. This is the number of connected components.
Time complexity is O(N^2) due to matrix traversal, with near O(1) per union/find. Space is O(N). Handle edge cases like N=0 or all zeros.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.