Problem Description

The problem presented involves understanding and calculating the ratios between various variables given some base equations and then using this information to solve a series of queries for the ratios between other pairs of variables. Specifically, we are given an array of equations where each element is a pair [A_i, B_i] that represents a single equation A_i / B_i. Alongside, there is an array of values where values[i] represents the value of the ith equation, A_i / B_i = values[i].

Our task is to answer a series of queries where each query [C_j, D_j] asks for the value of C_j / D_j. If we can determine the answer based on the given equations and values, we should include it in our output. If we cannot determine the answer (for example, if one or both of the variables don't appear in the given equations, or if there is no connection between them through the equations), we should return -1.0 for that query.

It is guaranteed that the input will be valid, meaning that no division by zero will occur, all equations are possible, and there are no contradictory equations.

Flowchart Walkthrough

To analyze Leetcode 399: Evaluate Division, let's employ the algorithm flowchart for guidance. The problem involves equations given as fractions, which are essentially ratios between values, and you need to compute results for queries based on these equations. It can be thought of as a graph problem where each variable represents a node and each equation represents an edge with a certain weight.

Follow along the Flowchart per the steps:

Is it a graph?

• Yes: Each variable can be represented as a node. Each division equation represents a directed edge between nodes. The weight of the edge is the result of the division.

Is it a tree?

• No: The graph is not necessarily a hierarchy without cycles. In fact, cycles are allowed because you can go from one variable to another and back using different division results.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The graph could have cycles because you can derive multiple paths that reflect different derived equations.

Is the problem related to shortest paths?

• No: The problem isn't about finding shortest paths; it's about finding specific relationships (or weights) between connected nodes.

Does the problem involve connectivity?

• Yes: The problem involves checking if two variables are connected and finding the product of the weights (division results or ratios) along the path between these variables.

Is the graph weighted?

• Yes: Each edge in the graph has a weight, which is the division result between the two connected nodes.

Since this weighted connectivity problem doesn't have anything to do with minimum paths or similar shortest-path conditions, and the graph may include cycles, the Depth-First Search (DFS) algorithm is suitable. DFS can be efficiently used to explore all possible connections between nodes, recursively multiplying the weights of the edges to compute the result for each query.

Conclusion: Based on the journey through the flowchart where the path involves determining connectivity in a weighted graph with cycles, DFS is the appropriate algorithm for efficiently solving the Evaluate Division problem.

Intuition

The intuition behind solving this problem lies in recognizing it as a graph problem. Each variable can be thought of as a node in a graph, and each equation as an edge that connects two variables. The value associated with each equation is the weight of the corresponding edge.

Using this interpretation, the problem of finding the ratio of two variables becomes a problem of finding a path between two nodes in the graph. If there's a direct path between the nodes (variables) representing the numerator and denominator, we can calculate the result; if not, or if one of the variables doesn't exist in the graph, the result is -1.0.

To efficiently find the connections and calculate the ratios between variables, we can use the Union-Find algorithm, which is suitable for handling such dynamic connectivity problems. It helps track which variables are connected and combines groups of connected components efficiently.

Here's how we arrive at the solution approach:

1. Initialize Union-Find data structures.
2. Union the nodes based on the given equations, effectively connecting the variables with edges weighted by the given values.
3. Iterate over the queries, for each:
   • If both variables are in the graph and belong to the same connected component, calculate the ratio using the weights along the path that connects the variables.
   • If the variables are not in the same connected component or at least one is not in the graph, return -1.0 for that query.

The solution uses path compression in the Union-Find find function, which optimizes the structure by making nodes in the path point directly to the root. As we call find, we simultaneously update the weights (w) to ensure direct access to the relative weight of any node to its root. This optimization allows for faster queries following the construction of the graph.

In summary, the solution approach employs a graph representation and Union-Find algorithm with path compression to efficiently solve the ratio queries.

Learn more about Depth-First Search, Breadth-First Search, Union Find, Graph and Shortest Path patterns.

Solution Approach

The reference solution approach mentions "Union find," which is a data structure that is particularly adept at handling the "disjoint-set" class of problems. In this problem, we are working with variables that are members of certain groups—where each group represents the variables that have been equated through the given equations.

The Union-Find data structure has two primary operations:

• find: This function determines the root representative of a set that an element belongs to, and it also applies path compression here.
• union: This function merges two sets together.

In the provided Python code, we have a find function that handles finding the root representative and path compression, but instead of a separate union function, the union operation is embedded within the main logic of the calcEquation method:

1. We initialize two dictionaries w and p. w holds the relative weight of a variable to its root, and p holds the parent (or root representative) of each variable.
2. We iterate over the list of equations to set up the initial connections. Each variable is initially set as a root of itself.
3. As we go through the values alongside equations, we find the root representatives of a and b (pa and pb respectively). If they are already connected (i.e., have the same root), we skip to the next equation.
4. If they are not connected, we connect them by setting the parent of pa to be pb and adjusting the weight of pa to maintain the correct ratio as given by the current equation. The weight of pa is updated to be the weight of b times the given value divided by the weight of a. This ensures that the weight reflects the ratio between these two connected components up to their common root.

Once the equations are processed and the Union-Find structure is ready, we handle the queries:

1. For each queries pair (c, d), we check if c or d are not present in the p dictionary. If either is not present, it means we cannot determine the ratio and hence return -1.
2. If c and d are present, we perform find operations on both to get their root representatives.
3. If they have different roots, the variables are in different disjoint sets, and we cannot determine their ratio, so we return -1.
4. If c and d belong to the same connected component (have the same root), we return the ratio by dividing w[c] by w[d]. This works because w[c] represents the ratio of c to its root and w[d] of d to that same root, so their division gives the ratio c/d.

This code thus uses Union-Find with path compression to implement a fast and efficient solution to the problem, handling both the input equations to build the graph as well as the queries to extract the required ratios.

Example Walkthrough

Let's illustrate the solution approach with a small example.

Suppose we have the following input:

• equations: [["a", "b"], ["b", "c"]]
• values: [2.0, 3.0]
• queries: [["a", "c"], ["c", "b"], ["b", "a"], ["a", "e"]]

Here equations and values tell us that a / b = 2.0 and b / c = 3.0.

Using the solution approach:

1. We intialize two dictionaries w and p to keep track of weights and parents.

2. We start processing the equations:

   For the first equation ["a", "b"]:

   • We find the root of a (which is a itself since it's the first time we see it) and do the same for b.
   • We union a and b, by making b the parent of a's root and we set the weight of a to 2.0 (since a / b = 2.0).

   The dictionaries are now:

   • w = {"a": 2.0, "b": 1.0}
   • p = {"a": "b", "b": "b"}

   For the second equation ["b", "c"]:

   • Find the root of b (which is b itself).
   • Find the root of c (which is c itself).
   • We union b and c, making c the parent of b's root and updating the weight of b to 3.0 (b / c = 3.0).

   The dictionaries are now updated:

   • w = {"a": 2.0, "b": 3.0, "c": 1.0}
   • p = {"a": "b", "b": "c", "c": "c"}

3. Now we handle the queries one by one:

   • For ["a", "c"]:

     • Find root of a (which is c through b), and find root of c (which is c).
     • Since the roots are the same, we can find the ratio by dividing w[a] by w[c], so a / c = w[a] / w[c] = 2.0 / 1.0 = 2.0.

   • For ["c", "b"]:

     • Roots for both c and b are c, so they are connected.
     • We get the ratio c / b = w[c] / w[b] = 1.0 / 3.0 = 0.333....

   • For ["b", "a"]:

     • Roots for both b and a are c, so they are connected.
     • We get the ratio b / a = w[b] / w[a] = 3.0 / 2.0 = 1.5.

   • For ["a", "e"]:

     • e is not found in p, so there is no known connection between a and e.
     • We return -1.0 for this query.

At the end of the process, the results for the queries are [2.0, 0.333..., 1.5, -1.0]. Through the Union-Find algorithm with path compression, we were able to efficiently solve for the ratios between various variables.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python code implements a solution to evaluate division among a set of variables based on given equations. Here's the breakdown of its time complexity:

• The initialization of variables w and p: The complexity for this is constant, O(1).

• The first for-loop iterates through the list of equations. If there are E equations, this loop runs E times, contributing an O(E) complexity.

• The second for-loop iterates through the list of values, with parallel iteration over the equations again, thus running E times. Within this loop, the find function is called for each a and b, which can lead to a depth of E in the worst-case if every find leads to another find. However, due to path compression, the amortized time complexity for each find operation drops to near O(1), so the loop contributes O(E) complexity.

• The last for-loop iterates over the list of queries. Let's say there are Q queries; this loop runs Q times. Inside this loop, there is also a find call for each c and d, which again is amortized to O(1) complexity due to path compression. This loop, therefore, adds an O(Q) complexity.

Combining all the parts, the resultant time complexity is: O(E + E + Q) which simplifies to O(E + Q), given E is the number of equations, and Q is the number of queries.

Space Complexity

Here's the breakdown of space complexity for the given code:

• The w (weights) and p (parents) dictionaries will at most store every unique variable from both equations and queries. If V represents the total distinct variables, then these structures will contribute O(V) to the space complexity.

• The output list is composed of an entry for each query. If there are Q queries, this will contribute O(Q) space complexity.

Therefore, the total space complexity is O(V + Q), where V is the number of distinct variables, and Q is the number of queries.

Learn more about how to find time and space complexity quickly using problem constraints.