Snapchat·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Snapchat SWE coding round, one graph problem the whole time. Pretty standard connected components stuff but the exact formulation tripped me up for a minute.

Questions Asked (1)

Given an undirected graph (number of nodes, plus two parallel arrays representing edges), where each node has an integer value, find the largest difference between the max and min node values across all connected components.

Algorithms & Data Structures

Author's notes

My first instinct was union-find but I defaulted to BFS since I could code it faster under pressure.

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AI HintsAI Generated

Suggested Approach

Model the graph as an adjacency list and traverse each connected component using BFS or DFS. For each component, track the minimum and maximum node values, then compute the difference and keep the global maximum. This yields O(V + E) time and O(V + E) space.

Pro tip: Mention that you can avoid building an adjacency list by using Union-Find to group nodes, then aggregate min/max per root—this shows awareness of alternative trade-offs and can be more memory-efficient for dense graphs.

1. Clarify input and constraints

Confirm the graph is undirected, nodes are 0-indexed or 1-indexed, and values can be negative. Ask about graph size to choose between BFS/DFS and Union-Find.

2. Choose traversal strategy

Decide between BFS/DFS with an adjacency list or Union-Find. Explain that both achieve O(V+E) time, but Union-Find may simplify component aggregation.

3. Traverse and track min/max

For each unvisited node, start a traversal. During traversal, update the component's min and max values by comparing with each visited node's value.

4. Compute and update global maximum

After finishing a component, compute max - min and update the global answer if larger. Handle isolated nodes (difference 0).

5. Analyze complexity and edge cases

State O(V+E) time and O(V+E) space for BFS/DFS, or O(V+E α(V)) for Union-Find. Discuss empty graph, single node, and negative values.

Key Points to Mention

Connected components can be found via BFS, DFS, or Union-Find.
Track min and max per component during traversal to avoid a second pass.
Time complexity is O(V + E) for graph traversal; space is O(V + E) for adjacency list.
Union-Find with path compression and union by rank gives near O(V + E) time and can be more memory-efficient.
Edge cases: empty graph, isolated nodes, all nodes in one component, negative values.
The answer is the maximum difference across all components, not the overall max minus min.

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