You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i]. Each Ai or Bi is a string that represents a single variable.
You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.
Return the answers to all queries. If a single answer cannot be determined, return -1.0.
Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.
Note: The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.
Example 1:
Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]] Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000] Explanation: Given: a / b = 2.0, b / c = 3.0 queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? return: [6.0, 0.5, -1.0, 1.0, -1.0 ] note: x is undefined => -1.0
Example 2:
Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]] Output: [3.75000,0.40000,5.00000,0.20000]
Example 3:
Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]] Output: [0.50000,2.00000,-1.00000,-1.00000]
Constraints:
1 <= equations.length <= 20equations[i].length == 21 <= Ai.length, Bi.length <= 5values.length == equations.length0.0 < values[i] <= 20.01 <= queries.length <= 20queries[i].length == 21 <= Cj.length, Dj.length <= 5Ai, Bi, Cj, Dj consist of lower case English letters and digits.Problem Overview: You receive equations like a / b = 2.0 and must answer queries such as a / c. If the relationships exist through other variables, compute the result using those connections. If no path exists between two variables, return -1.0.
Approach 1: Graph Representation and DFS/BFS (Time: O(Q * (V + E)), Space: O(V + E))
Model the equations as a weighted graph. Each variable becomes a node, and every equation a / b = k creates two edges: a → b with weight k and b → a with weight 1/k. To answer a query like x / y, run a Depth-First Search or Breadth-First Search starting from x. Multiply edge weights along the path until you reach y. If a path exists, the product of weights gives the answer; otherwise return -1.0.
This approach works because ratios compose multiplicatively along paths in the graph. During traversal, maintain a visited set to avoid cycles. DFS is commonly used because the graph is usually small, but BFS works equally well. Building the adjacency list takes O(E) time where E is the number of equations, and each query may traverse the graph once.
Problems like this are classic graph traversal tasks and commonly appear alongside Depth-First Search interview questions.
Approach 2: Union-Find with Path Compression (Time: ~O((E + Q) α(N)), Space: O(N))
Another approach treats each variable as part of a disjoint set. Instead of storing only connectivity, store a weight representing the ratio between a node and its parent. When processing an equation a / b = k, union the sets while preserving the ratio relationship. The parent links maintain multiplicative weights so any node can compute its value relative to the set root.
When answering a query x / y, check whether both variables share the same root. If they do, compute the ratio using their stored weights relative to that root. If they belong to different sets, the relationship is undefined and the answer is -1.0. Path compression keeps the structure shallow and maintains correct weight adjustments.
This weighted Union-Find technique is common in problems involving relative relationships or ratios. It combines Union-Find with algebraic weight propagation to answer queries efficiently.
Recommended for interviews: Start with the graph + DFS explanation because it directly models the problem and demonstrates strong understanding of graph traversal. Many candidates implement this first. The Union-Find approach is more advanced and often viewed as the optimized design because it handles multiple queries efficiently with near-constant time operations.
Approach: Represent the given variable equations as a graph. Each variable is a node, and an equation between two variables becomes a directed edge with a weight from one node to another representing the ratio. To find the result of a query, search for a path from the dividend node to the divisor node using Depth First Search (DFS). If a path is found, multiply the weights along the path to get the result. If no path is found, return -1.0 as the result cannot be determined.
The Python implementation uses a recursive DFS function to find a path from the start variable to the end variable. The graph is built using a dictionary of dictionaries, allowing easy access to directly connected nodes and the associated division value. If a direct or indirect path exists, the function calculates the result by multiplying the weights along the path. If no path can be found, it returns -1.0.
Python
JavaScript
Time Complexity: O(V + E) for each query, where V is the number of variables and E is the number of equations due to the DFS traversal. Overall complexity is O(Q * (V + E)) for Q queries.
Space Complexity: O(V + E) for storing the graph.
Approach: Use Union-Find (Disjoint Set Union, DSU) with path compression to represent connected components of variables where each component can efficiently find the "parent" or representative of a set. Each variable is associated with a weight that represents its relative scale with respect to its parent. This method efficiently handles union operations and queries to determine the result by finding the root of each variable and combining their scales.
This C++ solution uses a Union-Find data structure with path compression. For each equation, it unifies the variables into a single component, maintaining their relative ratios. For each query, it determines whether the variables are in the same component and calculates the result using the stored weights (which represent ratios to their respective roots).
Time Complexity: O(E + Q * α(V)), where E is the number of equations, Q is the number of queries, V is the number of variables, and α is the inverse Ackermann function, which grows very slowly.
Space Complexity: O(V) for storing parent information and weights.
| Approach | Complexity |
|---|---|
| Graph Representation and DFS | Time Complexity: O(V + E) for each query, where V is the number of variables and E is the number of equations due to the DFS traversal. Overall complexity is O(Q * (V + E)) for Q queries. |
| Union-Find with Path Compression | Time Complexity: O(E + Q * α(V)), where E is the number of equations, Q is the number of queries, V is the number of variables, and α is the inverse Ackermann function, which grows very slowly. |
| Default Approach | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Graph + DFS | O(Q * (V + E)) | O(V + E) | General solution; easy to implement and ideal when query count is small |
| Graph + BFS | O(Q * (V + E)) | O(V + E) | Preferred when iterative traversal is easier than recursion |
| Union-Find with Weights | O((E + Q) α(N)) | O(N) | Best when many queries must be answered after building relationships |
Evaluate Division - Leetcode 399 - Python • NeetCodeIO • 59,147 views views
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