Evaluate Division (Union-Find)
Given a set of equations $A/B = k$ and queries $C/D = ?$, calculate the result of the queries. Use a Graph (or Union-Find) to represent the division relationships.
Why Interviewers Ask This
Google interviewers use this problem to assess a candidate's ability to model real-world relationships as graph structures and select the optimal algorithmic approach. They specifically evaluate your understanding of Disjoint Set Union (Union-Find) with path compression and rank optimization, testing if you can handle dynamic connectivity problems efficiently rather than just solving isolated equations.
How to Answer This Question
1. Clarify requirements: Confirm edge cases like division by zero or disconnected variables immediately.
2. Model the problem: Explain that each variable is a node and an equation A/B=k creates a weighted directed edge from A to B with weight k, while B/A has weight 1/k.
3. Choose the strategy: Propose using Union-Find where each node tracks its parent and the ratio relative to that parent, allowing O(α(n)) amortized time for operations.
4. Implement logic: Detail how 'find' compresses paths and updates ratios, and how 'union' merges sets by attaching one root to another while calculating the new scaling factor.
5. Handle queries: Describe traversing the path to find roots; if roots differ, return -1, otherwise compute the result by dividing the accumulated weights.
6. Analyze complexity: Explicitly state O(N + Q * α(N)) time and O(N) space complexity to demonstrate depth of understanding expected at Google.
Key Points to Cover
- Modeling the problem as a weighted graph where edges carry multiplicative values
- Implementing Union-Find with path compression that maintains cumulative ratios
- Handling disconnected components by returning -1 when roots differ
- Achieving O(N + Q * α(N)) time complexity through optimized union-find
- Demonstrating awareness of edge cases like division by zero or missing variables
Sample Answer
To solve the Evaluate Division problem, I would model the variables as nodes in a weighted graph where edges represent the division results. For instance, if we have A/B = 2.0, we create a directed edge from A to B with weight 2.0 and an inverse edge from B to A with weight 0.5. This transforms the query C/D into finding a path between two nodes and multiplying the edge weights along that path.
While a DFS approach works, I believe a Union-Find data structure with path compression is more efficient for handling multiple queries dynamically. In this implementation, each node stores its parent and a value representing the ratio to that parent. When performing a 'find' operation, we recursively locate the root while updating the current node's value to be the product of all intermediate ratios, effectively compressing the path. The 'union' operation connects two disjoint sets by making one root a child of the other, carefully calculating the new weight based on the known ratio between the two input variables.
For any query C/D, I first check if both variables exist in our map. If they do, I find their roots. If the roots are different, the variables are disconnected, and I return -1. If they share a root, the answer is simply the ratio of C to its root divided by the ratio of D to its root. This approach ensures near-constant time complexity per query after linear preprocessing, which aligns well with Google's emphasis on scalable, efficient solutions for large datasets.
Common Mistakes to Avoid
- Failing to update the ratio value during path compression, leading to incorrect query results
- Ignoring the need for bidirectional edges (A/B vs B/A) when building the initial structure
- Not handling disconnected variables, causing runtime errors or infinite loops in traversal
- Using simple BFS/DFS without memoization, resulting in TLE for large numbers of queries
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