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:
class Solution:
def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
graph = {}
n = len(equations)
for i in range(n):
a, b = equations[i]
graph[a] = graph.get(a, []) + [(b, values[i])]
graph[b] = graph.get(b, []) + [(a, 1 / values[i])]
op = []
for c, d in queries:
found = False
paths = [(c, 1, {c})]
while len(paths) > 0:
curr, prod, visited = paths.pop()
if curr not in graph:
break
if curr == d:
op.append(prod)
found = True
break
for j, div in graph.get(curr, []):
if j not in visited:
paths.append((j, prod * div, visited.union({j})))
if not found:
op.append(-1.0)
return op
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