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.
Problem Link: https://leetcode.com/problems/evaluate-division/
Solution
To solve the problem, we observe that
- given an equation
a/b = n > 0
we also knowb/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:
12 visited.add(new_dividend)
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[0]] == [] or graph[query[1]] == []:
22 res.append(-1)
23 else:
24 visited = set()
25 visited.add(query[0])
26 res.append(divide(query[0], query[1], visited))
27 return res