 # LeetCode Evaluate Division Solution

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.

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.

## Solution

To solve the problem, we observe that

• given an equation `a/b = n > 0` we also know `b/a = 1/n`.
• given two equations `a/b=n` `b/c=m`, by transitivity, `a/c=(a/b)(b/c)=nm`.

For each query (`a/b`), we wish to find a path from `dividend`(`a`) to `divisor`(`b`) and multiply the quotients along the way to find the final answer. If no such path exists, then we can return `-1.0`.

We will use aggregated backtracking to find the answer to each query.

#### Implementation

``````1  def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
2      graph = defaultdict(list)
3      for i, [dividend, divisor] in enumerate(equations):
4          graph[dividend].append((divisor, values[i]))
5          graph[divisor].append((dividend, 1/values[i]))
6
7      def divide(dividend, divisor, visited):
8          if dividend == divisor:
9              return 1
10          for new_dividend, multiplier in graph[dividend]:
11              if new_dividend not in visited:
13                  ans = divide(new_dividend, divisor, visited)
14                  if ans > 0:
15                      return ans*multiplier
16                  visited.remove(new_dividend)
17          return -1
18
19      res = []
20      for query in queries:
21          if graph[query] == [] or graph[query] == []:
22              res.append(-1)
23          else:
24              visited = set()