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.

Examples

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]

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 <= 20
  • equations[i].length == 2
  • 1 <= Ai.length, Bi.length <= 5
  • values.length == equations.length
  • 0.0 < values[i] <= 20.0
  • 1 <= queries.length <= 20
  • queries[i].length == 2
  • 1 <= Cj.length, Dj.length <= 5
  • Ai, Bi, Cj, Dj consist of lower case English letters and digits.

Thinking Process

  1. Weighted Union-Find: Maintains ratios relative to root
  • Model entities as nodes and relationships as edges.
  • Pick traversal (BFS/DFS) or shortest-path (Dijkstra) based on weights.
  • Union-Find helps when connectivity updates are frequent.
Graph BFS layers S a b t BFS: expand by layers (queue)

Common Approaches

Typical techniques for this pattern:

Approach Time Space Notes
Recursive DFS (this problem) O(n) O(h) stack Natural for trees and graphs
Iterative DFS (stack) O(n) O(n) Avoid recursion depth limits
DFS with memoization O(n) O(n) Overlapping subproblems on graphs
Backtracking DFS O(2^n) typical O(n) Enumerate choices with pruning

Solution

Time Complexity: O((E + Q) × α(n)) where E = equations, Q = queries, n = variables
Space Complexity: O(n) - For the union-find structure

This solution uses Union-Find with path compression and union by weight to maintain ratios between variables.

class Solution:
    def __init__(self):
        self.parent = {}
        self.weight = {}

    def find(self, x):
        if x not in self.parent:
            self.parent[x] = x
            self.weight[x] = 1.0
            return x

        if self.parent[x] != x:
            orig_parent = self.parent[x]
            root = self.find(orig_parent)

            self.weight[x] *= self.weight[orig_parent]
            self.parent[x] = root

        return self.parent[x]

    def union(self, a, b, value):
        if a not in self.parent:
            self.parent[a] = a
            self.weight[a] = 1.0
        if b not in self.parent:
            self.parent[b] = b
            self.weight[b] = 1.0

        rootA = self.find(a)
        rootB = self.find(b)

        if rootA != rootB:
            self.parent[rootA] = rootB
            self.weight[rootA] = value * self.weight[b] / self.weight[a]

    def calcEquation(self, equations, values, queries):
        for (a, b), val in zip(equations, values):
            self.union(a, b, val)

        res = []

        for a, b in queries:
            if a not in self.parent or b not in self.parent:
                res.append(-1.0)
                continue

            rootA = self.find(a)
            rootB = self.find(b)

            if rootA != rootB:
                res.append(-1.0)
            else:
                res.append(self.weight[a] / self.weight[b])

        return res

Solution Explanation

Approach: Recursive DFS (this problem)

Key idea: 1. Weighted Union-Find: Maintains ratios relative to root

How the code works:

  1. Weighted Union-Find: Maintains ratios relative to root
    • Model entities as nodes and relationships as edges.
    • Pick traversal (BFS/DFS) or shortest-path (Dijkstra) based on weights.
    • Union-Find helps when connectivity updates are frequent.

Walkthrough — input equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]], expected output [6.00000,0.50000,-1.00000,1.00000,-1.00000]:

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]

Solution Time Space Notes
Union-Find O((E+Q)×α(n)) O(n) Optimal for many queries
Graph DFS O(E + Q×V) O(E+V) Simple, good for few queries

How Solution 1 Works

  1. Union-Find Structure:
    • weights[node] = {parent, weight} where weight = node / parent
    • Example: If a / b = 2.0, then weights[a] = {b, 2.0}
  2. Find with Path Compression:
    • Finds root and compresses path
    • Updates weight along path: node_weight = node_weight × parent_weight
    • Returns {root, node_weight}
  3. Union Operation:
    • Connects two components with a ratio
    • If a / b = value, connects roots with appropriate weight
    • Weight formula: root_weight = (divisor_weight × value) / dividend_weight
  4. Query Processing:
    • If both variables exist and have same root, return dividend_weight / divisor_weight
    • Otherwise return -1.0

Key Insight

The Union-Find structure maintains ratios relative to the root:

  • weights[node] = {root, node/root}
  • To find a / b: If both have same root, (a/root) / (b/root) = a/b

    Example Walkthrough

Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"]]

Solution 1 (Union-Find):

Step 1: Process "a / b = 2.0"
  unite(a, b, 2.0):
    find(a) → {a, 1.0}
    find(b) → {b, 1.0}
    weights[a] = {b, 2.0}

Step 2: Process "b / c = 3.0"
  unite(b, c, 3.0):
    find(b) → {b, 1.0}
    find(c) → {c, 1.0}
    weights[b] = {c, 3.0}

Step 3: Query "a / c = ?"
  find(a) → find(b) → {c, 1.0}
    Path: a → b → c
    weights[a] = {b, 2.0}
    find(b) → {c, 3.0}
    weights[a] = {c, 2.0 × 3.0 = 6.0}
  
  find(c) → {c, 1.0}
  
  Result: 6.0 / 1.0 = 6.0

Solution 2 (Graph DFS):

Graph:
  a → [(b, 2.0)]
  b → [(a, 0.5), (c, 3.0)]
  c → [(b, 0.333)]

Query "a / c":
  DFS(a, c):
    a → b (product = 2.0)
    b → c (product = 2.0 × 3.0 = 6.0)
    Found! Return 6.0

Complexity

| Solution | Time | Space | Notes | |———-|——|——-|——-| | Union-Find | O((E+Q)×α(n)) | O(n) | Optimal for many queries | | Graph DFS | O(E + Q×V) | O(E+V) | Simple, good for few queries |

Common Mistakes

  1. Variable not in equations: Return -1.0
  2. Same variable query: a / a = 1.0
  3. Disconnected components: Variables in different components return -1.0
  4. Single equation: Handle minimal input

  5. Transitive relationships: a/b=2, b/c=3 → a/c=6

  6. Path compression: Not updating weights during path compression
  7. Union weight calculation: Wrong formula for connecting roots
  8. Division by zero: Should not occur per problem constraints
  9. Missing variables: Not checking if variables exist before querying

Optimization Tips

  1. Path compression: Essential for O(α(n)) amortized time
  2. Lazy initialization: Only create entries when needed
  3. Early termination: Check if variables exist before processing

Pattern Recognition

This problem demonstrates the “Weighted Union-Find” pattern:

1. Maintain parent and weight in union-find structure
2. Path compression updates weights along path
3. Union operation connects with correct weight
4. Query uses weight ratio when same root

Similar problems:

Key Takeaways

  1. Weighted Union-Find: Maintains ratios relative to root
  2. Path compression: Updates weights along path for efficiency
  3. Union by weight: Connects roots with correct ratio
  4. Query formula: dividend_weight / divisor_weight when same root