[LeetCode] 399. Evaluate Division

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.
Example:
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 ].
The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.
According to the example above:
equations = [ ["a", "b"], ["b", "c"] ],
values = [2.0, 3.0],
queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

Thought process:
Build graph and do DFS.

Solution:

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class Solution {
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        Map<String, List<String>> graph = new HashMap<>();
        Map<String, Double> weights = new HashMap<>();
        
        for (int i = 0; i < equations.length; i++) {
            String s1 = equations[i][0];
            String s2 = equations[i][1];
            if (!graph.containsKey(s1)) {
                graph.put(s1, new ArrayList<>());
            }
            graph.get(s1).add(s2);
            if (!graph.containsKey(s2)) {
                graph.put(s2, new ArrayList<>());
            }
            graph.get(s2).add(s1);
            
            weights.put(s1 + "," + s2, values[i]);
            weights.put(s2 + "," + s1, 1 / values[i]);
        }
        
        double[] result = new double[queries.length];
        for (int i = 0; i < queries.length; i++) {
            String src = queries[i][0];
            if (!graph.containsKey(src)) {
                result[i] = -1;
                continue;
            }
            String dest = queries[i][1];
            
            Set<String> visited = new HashSet<>();
            dfs(graph, weights, visited, src, dest, 1, result, i);
            if (result[i] == 0) {
                result[i] = -1;
            }
        }
        return result;
    }
    
    private void dfs(Map<String, List<String>> graph, Map<String, Double> weights, Set<String> visited, String src, String dest,
                     double weight, double[] result, int index) {
        if (src.equals(dest)) {
            result[index] = weight;
            return;
        }
        
        visited.add(src);
        for (String neighbor : graph.get(src)) {
            if (!visited.contains(neighbor)) {
                dfs(graph, weights, visited, neighbor, dest, weight * weights.get(src + "," + neighbor), result, index);
            }
        }
    }
}

Time complexity:
Say there're e equations and q queries. Building the graph takes O(e). The queries take O(q) iterations and O(e) per iteration. So the overall time complexity is O(eq).
Location: Philadelphia, PA, USA

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