399. Evaluate Division.java

/*
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.
  */
  class Solution {
  public double[] calcEquation(List<List<String>> equations, double[] values,
                               List<List<String>> queries) {
    double[] ans = new double[queries.size()];
    // graph.get(A).get(B) := A / B
    Map<String, Map<String, Double>> graph = new HashMap<>();

    // construct the graph
    for (int i = 0; i < equations.size(); ++i) {
      final String A = equations.get(i).get(0);
      final String B = equations.get(i).get(1);
      graph.putIfAbsent(A, new HashMap<>());
      graph.putIfAbsent(B, new HashMap<>());
      graph.get(A).put(B, values[i]);
      graph.get(B).put(A, 1.0 / values[i]);
    }

    for (int i = 0; i < queries.size(); ++i) {
      final String A = queries.get(i).get(0);
      final String C = queries.get(i).get(1);
      if (!graph.containsKey(A) || !graph.containsKey(C))
        ans[i] = -1.0;
      else
        ans[i] = divide(graph, A, C, new HashSet<>());
    }

    return ans;
  }

  // return A / C
  private double divide(Map<String, Map<String, Double>> graph, final String A, final String C,
                        Set<String> seen) {
    if (A.equals(C))
      return 1.0;

    seen.add(A);

    for (final String B : graph.get(A).keySet()) {
      if (seen.contains(B))
        continue;
      final double res = divide(graph, B, C, seen); // B / C
      if (res > 0)                                  // valid result
        return graph.get(A).get(B) * res;           // A / C = (A / B) * (B / C)
    }

    return -1.0; // invalid result
  }
}