# LeetCode 399. Evaluate Division

## Description

https://leetcode.com/problems/most-common-word/

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.

dfs

## Python Solution

``````class Solution:
def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
graph = collections.defaultdict(list)

for equation, value  in zip(equations, values):
numerator, denominator = equation, equation
graph[numerator].append((denominator, value))
graph[denominator].append((numerator, 1 / value))

result = []

for numerator, denominator in queries:
visited = set()
value = self.helper(graph, numerator, denominator, visited)
result.append(value)

return result

def helper(self, graph, numerator, denominator, visited):
if numerator not in graph:
return -1.0
if numerator == denominator:
return 1.0

for neighbor, value in graph[numerator]:
if neighbor == denominator:
return value
if neighbor not in visited:
quotient = self.helper(graph, neighbor, denominator, visited)

if quotient != -1:
return value * quotient

return -1.0
``````
• Time Complexity: ~NM
• Space Complexity: ~N

Let N be the number of input equations and M be the number of queries.