Problem Description

This problem asks you to evaluate division expressions based on given equation relationships between variables.

You're given a set of equations where each equation tells you the result of dividing one variable by another. For example, if you have a / b = 2.0, this means variable a divided by variable b equals 2.0.

Input:

• equations: An array of variable pairs like [["a", "b"], ["b", "c"]]
• values: An array of corresponding division results like [2.0, 3.0]
  • Together these mean: a / b = 2.0 and b / c = 3.0
• queries: An array of variable pairs you need to evaluate, like [["a", "c"], ["b", "a"]]

Task: For each query pair [C, D], you need to find the value of C / D using the given equations.

Key Rules:

1. If you can derive the answer from the given equations (even indirectly through chaining), return that value
2. If either variable in the query doesn't appear in any equation, return -1.0
3. If both variables exist but there's no path connecting them through the equations, return -1.0

Example: If you know a / b = 2.0 and b / c = 3.0, you can derive:

• a / c = (a / b) * (b / c) = 2.0 * 3.0 = 6.0
• b / a = 1 / (a / b) = 1 / 2.0 = 0.5
• a / a = 1.0 (any variable divided by itself equals 1)

The solution uses a Union-Find (Disjoint Set Union) data structure with path compression and weight tracking to efficiently find these relationships and calculate the division results.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem can be modeled as a graph where variables are nodes and equations represent weighted edges between them. For example, if a / b = 2.0, we have an edge from a to b with weight 2.0 and an edge from b to a with weight 0.5.

Is it a tree?

• No: The graph may contain cycles. For instance, we could have a / b = 2.0, b / c = 3.0, and c / a = 0.167, forming a cycle.

Is the problem related to directed acyclic graphs?

• No: As mentioned, the graph can have cycles, so it's not a DAG.

Is the problem related to shortest paths?

• No: We're not finding shortest paths. We need to find if there's a path between two variables and calculate the product of weights along that path.

Does the problem involve connectivity?

• Yes: The core challenge is determining if two variables are connected through the equation relationships. If they're in the same connected component, we can calculate the division result.

Disjoint Set Union

• Yes: This is the approach used in the solution. DSU (Union-Find) efficiently handles connectivity queries and maintains the multiplication factors (weights) between variables.

Conclusion: While the flowchart leads us to DSU, it's worth noting that this problem can also be solved using DFS. With DFS, we would traverse from the source variable to the target variable, multiplying the edge weights along the path. The DSU approach is more efficient for multiple queries as it preprocesses the relationships, making each query O(α(n)) time complexity after the initial setup.

Intuition

The key insight is recognizing that division relationships form a network where we need to find paths and calculate products along those paths.

Think about it this way: if a / b = 2 and b / c = 3, then a / c = 6 because a = 2b and b = 3c, so a = 2 * 3c = 6c. This is essentially path multiplication - we multiply the weights along the path from a to c.

The challenge is efficiently handling multiple queries. We could use DFS for each query, but that would involve repeated graph traversals. Instead, we can preprocess the relationships to make queries faster.

The Union-Find (DSU) approach is clever here. It groups variables that are connected through equations into the same set. But regular Union-Find only tells us if elements are connected - we need more. We need to know the multiplication factor between any variable and its root.

Here's the brilliant part: we maintain a weight w[x] for each variable x that represents x / root(x). When we query c / d, if they're in the same connected component (same root), then:

• c / d = (c / root) / (d / root) = w[c] / w[d]

During the union operation, when connecting two components with roots pa and pb, we need to maintain consistency. If we're processing a / b = v, and a has root pa with weight w[a] (meaning a / pa = w[a]), and similarly for b, then:

• a / b = v
• (pa * w[a]) / (pb * w[b]) = v
• pa / pb = v * w[b] / w[a]

This gives us the weight for connecting pa to pb.

The path compression optimization in find() is crucial - when we compress the path, we update weights by multiplying them along the chain, ensuring each variable directly knows its ratio to the root.

Solution Approach

The solution implements a weighted Union-Find (Disjoint Set Union) data structure with path compression to efficiently handle division queries.

Data Structures:

• p: A dictionary storing the parent of each variable (for Union-Find)
• w: A dictionary storing the weight of each variable, representing the division ratio to its parent

Implementation Walkthrough:

1. Find Function with Path Compression:

   def find(x):
       if p[x] != x:
           origin = p[x]
           p[x] = find(p[x])
           w[x] *= w[origin]
       return p[x]

   • Recursively finds the root of variable x
   • During path compression, updates the weight by multiplying along the path
   • After compression, w[x] represents x / root(x)

2. Initialization:

   w = defaultdict(lambda: 1)
   p = defaultdict()
   for a, b in equations:
       p[a], p[b] = a, b

   • Initialize weights to 1 (each variable equals itself)
   • Set each variable as its own parent initially

3. Union Operation:

   for i, v in enumerate(values):
       a, b = equations[i]
       pa, pb = find(a), find(b)
       if pa == pb:
           continue
       p[pa] = pb
       w[pa] = w[b] * v / w[a]

   • For each equation a / b = v, find roots of both variables
   • If already in same component, skip (avoid redundant unions)
   • Connect component of a to component of b by making pb the parent of pa
   • Calculate the weight: Since a / b = v, and we know a / pa = w[a] and b / pb = w[b], we need pa / pb = (a / w[a]) / (b / w[b]) = (a / b) * (w[b] / w[a]) = v * w[b] / w[a]

4. Query Processing:

   return [
       -1 if c not in p or d not in p or find(c) != find(d) else w[c] / w[d]
       for c, d in queries
   ]

   • For each query c / d:
     • Return -1 if either variable doesn't exist or they're in different components
     • Otherwise, since both have the same root after find(), calculate c / d = (c / root) / (d / root) = w[c] / w[d]

Time Complexity:

• Building the union-find: O(N * α(N)) where N is the number of equations and α is the inverse Ackermann function
• Processing queries: O(M * α(N)) where M is the number of queries
• Overall: O((M + N) * α(N)) ≈ O(M + N) for practical purposes

Space Complexity: O(N) for storing the parent and weight dictionaries

Example Walkthrough

Let's walk through a concrete example to understand how the weighted Union-Find solution works.

Given:

• equations: [["a","b"], ["b","c"]]
• values: [2.0, 3.0] (meaning a/b = 2.0 and b/c = 3.0)
• queries: [["a","c"], ["b","a"], ["x","y"]]

Step 1: Initialization

p = {a: a, b: b, c: c}  # Each variable is its own parent
w = {a: 1, b: 1, c: 1}  # Initial weights are 1

Step 2: Process First Equation (a/b = 2.0)

• Find roots: find(a) = a, find(b) = b
• Different roots, so union them:
  • Make b the parent of a: p[a] = b
  • Calculate weight: w[a] = w[b] * 2.0 / w[a] = 1 * 2.0 / 1 = 2.0
• State after:

  p = {a: b, b: b, c: c}
  w = {a: 2.0, b: 1, c: 1}

  This means a/b = 2.0 (a is 2 times its parent b)

Step 3: Process Second Equation (b/c = 3.0)

• Find roots: find(b) = b, find(c) = c
• Different roots, so union them:
  • Make c the parent of b: p[b] = c
  • Calculate weight: w[b] = w[c] * 3.0 / w[b] = 1 * 3.0 / 1 = 3.0
• State after:

  p = {a: b, b: c, c: c}
  w = {a: 2.0, b: 3.0, c: 1}

Step 4: Process Queries

Query 1: a/c

• Call find(a):
  • p[a] = b ≠ a, so compress path
  • origin = b
  • Recursively find(b): returns c (and updates b's parent to c directly)
  • Update w[a] = w[a] * w[origin] = 2.0 * 3.0 = 6.0
  • Set p[a] = c
• Call find(c): returns c
• Both have root c, so calculate: a/c = w[a]/w[c] = 6.0/1 = 6.0

Query 2: b/a

• find(b) = c, find(a) = c (already compressed)
• Both have root c, so calculate: b/a = w[b]/w[a] = 3.0/6.0 = 0.5

Query 3: x/y

• x not in p, return -1.0

Final Answer: [6.0, 0.5, -1.0]

The key insight is that after path compression, each variable stores its division ratio directly to the root, making queries O(1) after the find operation.

Solution Implementation

Time and Space Complexity

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 unique variables, and α is the inverse Ackermann function.

• Initializing parent pointers for all variables in equations: O(E)
• Processing each equation involves two find operations and possibly one union operation: O(E * α(V))
• For each query, we perform at most two find operations: O(Q * α(V))
• The find operation with path compression has an amortized time complexity of O(α(V)), where α is the inverse Ackermann function, which is effectively constant for all practical values

Space Complexity: O(V) where V is the number of unique variables.

• The parent dictionary p stores at most V entries: O(V)
• The weight dictionary w stores at most V entries: O(V)
• The recursion stack for find operation can go up to O(V) in the worst case before path compression, but after path compression, the depth is reduced to O(α(V))
• The output list stores Q results: O(Q)

Overall space complexity is O(V + Q), but since we typically consider auxiliary space excluding the output, the auxiliary space complexity is O(V).

Common Pitfalls

1. Not Handling the Union Weight Calculation Correctly

The Pitfall: The most common mistake is incorrectly calculating the weight when performing the union operation. Many developers mistakenly use:

# WRONG!
w[pa] = v * w[a] / w[b]

Why it's wrong: The weight relationship needs to maintain the invariant that w[x] represents x / root(x). When connecting two components, we need to ensure that the division relationship a / b = v is preserved through the root nodes.

The Correct Approach:

w[pa] = w[b] * v / w[a]

This ensures that:

• a / b = v (given)
• a / pa = w[a] and b / pb = w[b] (after find operations)
• pa / pb = (a / w[a]) / (b / w[b]) = v * w[b] / w[a]

2. Forgetting to Compress Paths During Find Operation

The Pitfall: Implementing find without updating weights during path compression:

# WRONG - doesn't update weights!
def find(x):
    if p[x] != x:
        p[x] = find(p[x])  # Missing weight update
    return p[x]

Why it's wrong: Path compression changes the parent relationships, so weights must be updated to maintain the correct division ratios. Without updating weights, the accumulated product from node to root becomes incorrect.

The Correct Approach:

def find(x):
    if p[x] != x:
        origin = p[x]
        p[x] = find(p[x])
        w[x] *= w[origin]  # Update weight to maintain correct ratio
    return p[x]

3. Not Checking if Variables Are Already Connected

The Pitfall: Processing redundant unions without checking if nodes are already in the same component:

# WRONG - might create inconsistencies
for i, v in enumerate(values):
    a, b = equations[i]
    p[find(a)] = find(b)
    # ... weight calculation

Why it's wrong: If two variables are already connected, adding another edge between them could create inconsistencies or override existing correct relationships. While it might work in some cases, it's inefficient and can lead to subtle bugs.

The Correct Approach:

pa, pb = find(a), find(b)
if pa == pb:
    continue  # Skip if already connected
p[pa] = pb
w[pa] = w[b] * v / w[a]

4. Incorrect Query Result Calculation

The Pitfall: Calculating the query result without ensuring both variables have been compressed to their roots:

# WRONG - might use stale weights
if c in p and d in p and p[c] == p[d]:
    return w[c] / w[d]

Why it's wrong: The weights are only guaranteed to be correct after calling find(), which performs path compression and updates the weights. Without calling find(), you might be using outdated parent relationships and weights.

The Correct Approach:

if c not in p or d not in p or find(c) != find(d):
    return -1.0
else:
    return w[c] / w[d]  # Weights are now updated after find()

5. Not Initializing Variables Properly

The Pitfall: Only initializing variables when they appear as numerators:

# WRONG - might miss some variables
for eq in equations:
    p[eq[0]] = eq[0]  # Only initializing first variable

Why it's wrong: Both variables in each equation need to be initialized as their own parent. Missing initialization can cause KeyError or incorrect parent relationships.

The Correct Approach:

for a, b in equations:
    p[a], p[b] = a, b  # Initialize both variables