Snapchat·Backend Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Snapchat backend screen, one algorithmic problem the whole time. Pretty standard union-find territory but I fumbled the counting step at the end and had to backtrack.

Questions Asked (1)

Given n employees and a list of pairs indicating which employees belong to the same group (with transitivity), count the total number of unordered pairs of employees that come from different groups.

Algorithms & Data Structures

Author's notes

I recognized the connected components angle pretty fast and went with union-find.

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AI HintsAI Generated

Suggested Approach

Model employees as nodes and pairs as edges in an undirected graph, then use Union-Find (Disjoint Set Union) to group employees into connected components. After finding the size of each group, compute the total number of unordered pairs across different groups as (n*(n-1)/2) minus the sum of (size*(size-1)/2) for each group.

Pro tip: Mention that you can compute the answer during the union operations by maintaining the total number of same-group pairs, avoiding a second pass over the groups. Also, clarify that the input pairs are 0-indexed or 1-indexed and handle edge cases like n=0 or n=1.

1. Clarify the problem and assumptions

Confirm that the pairs represent a transitive relation (if A-B and B-C, then A, B, C are in the same group) and that we need unordered pairs of employees from different groups. Ask about input size, indexing, and whether the graph is guaranteed to be connected or not.

2. Choose Union-Find for grouping

Explain that Union-Find with path compression and union by rank/size efficiently groups employees into connected components. Each union merges two groups, and we can track the size of each group.

3. Compute group sizes

After processing all pairs, iterate over all employees to find the root of each and count the size of each group. Alternatively, maintain a size array during unions.

4. Calculate total cross-group pairs

Compute total possible pairs as n*(n-1)/2. Subtract the sum of within-group pairs, which is sum over groups of size*(size-1)/2. The result is the number of unordered pairs from different groups.

5. Analyze complexity and edge cases

State that the time complexity is nearly O(n + m) with inverse Ackermann for Union-Find, where m is the number of pairs. Discuss edge cases: no pairs (all singleton groups), all employees in one group, and large n requiring efficient memory usage.

Key Points to Mention

Union-Find (Disjoint Set Union) with path compression and union by rank/size for near-constant time operations.
Transitivity of the 'same group' relation means connected components in an undirected graph.
Formula for total unordered pairs: n*(n-1)/2, and within-group pairs: sum(size*(size-1)/2).
Time complexity: O((n + m) α(n)) where α is the inverse Ackermann function, effectively linear.
Space complexity: O(n) for parent and size arrays.
Edge cases: n=0, n=1, no pairs, all pairs forming a single group, and disconnected components.

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