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 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:
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