399. Evaluate Division

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399. Evaluate Division

Description

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [A<sub>i</sub>, B<sub>i</sub>] and values[i] represent the equation A<sub>i</sub> / B<sub>i</sub> = values[i]. Each A<sub>i</sub> or B<sub>i</sub> is a string that represents a single variable.

You are also given some queries, where queries[j] = [C<sub>j</sub>, D<sub>j</sub>] represents the j^th query where you must find the answer for C<sub>j</sub> / D<sub>j</sub> = ?.

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.

Note: The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.

Example 1:

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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 ]
note: x is undefined => -1.0

Example 2:

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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:

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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.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    unordered_map<string, unordered_map<string, double>> graph;
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        // build the graph and union find roots
        int n = equations.size();
        for (int i = 0; i < n; i++) {
            auto& equation = equations[i];
            string dividend = equation[0], divisor = equation[1];
            double val = values[i];
            graph[dividend][divisor] = val;
            graph[divisor][dividend] = 1.0 / val;
        }

        vector<double> res;
        unordered_set<string> visited;
        for (auto& q : queries) {
            string dividend = q[0], divisor = q[1];
            if (!graph.contains(dividend) || !graph.contains(divisor)) {
                res.push_back(-1.0);
            } else {
                res.push_back(dfs(dividend, divisor, visited));
            }
        }
        return res;
    }

    double dfs(string root, string target, unordered_set<string>& visited) {
        if (root == target) {
            return 1.0;
        }
        visited.insert(root);
        double mx = -1.0;
        for (auto [u, v]: graph[root]) {
            if (!visited.contains(u)) {
                mx = max(mx, dfs(u, target, visited) * v);
            }
        }
        visited.erase(root);
        return mx < 0 ? -1 : mx;
    }
};