Leetcode 399

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        // to solve this question
        // we first wanna to think this as a graph and using dfs find the path from a to b
        // in order to do so ,
        // we must construct a weighted graph
        HashMap<String,HashMap<String,Double>> graph = new HashMap<>();
        
        // we need to get equations to gain our two string node
        // we also need to have a index to keep track 
        int index = 0;
        for(List<String> e : equations){
            String a = e.get(0);
            String b = e.get(1);
            
            graph.putIfAbsent(a,new HashMap<>());
            graph.putIfAbsent(b,new HashMap<>());
            
            
            graph.get(a).put(b,values[index]);
            graph.get(b).put(a,1 / values[index]);
            index ++;
            
            graph.get(a).put(a,1.0);
            graph.get(b).put(b,1.0);
            
        }
        // we have sucessfully build our graph
        // lets look at our queiries
        double [] ans = new double[queries.size()];
        Arrays.fill(ans,-1.0);
        for(int i = 0; i < queries.size(); i ++){
            List<String> q = queries.get(i);
            String start = q.get(0);
            String end = q.get(1);
            
            // check if start and end in the graph at all
            if(!graph.containsKey(start) || !graph.containsKey(end)){
                continue;
            }else {
                dfs(graph,start,end,new HashSet<String>(),1.0,ans,i);
            
                continue;
            }
        }
        return ans;
    }
    // let's make our dfs method
    private void dfs (HashMap<String,HashMap<String,Double>> graph, String start, String end, Set<String> visited, double pre, double[] ans, int index){
        visited.add(start);
        if(graph.get(start).containsKey(end)){
            ans[index] = graph.get(start).get(end) * pre;
        }
        
        for(String next : graph.get(start).keySet()){
            if (visited.contains(next)) continue;
            dfs(graph,next,end,visited,graph.get(start).get(next) * pre ,ans,index);
        }
    }
}