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Leetcode Evaluate Division problem solution

YASH PAL, 31 July 202422 January 2026

In this Leetcode Evaluate Division problem 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.

Leetcode Evaluate Division problem solution

Leetcode Evaluate Division problem solution in Python.

class Solution(object):
    def calcEquation(self, equations, values, queries):
        """
        :type equations: List[List[str]]
        :type values: List[float]
        :type queries: List[List[str]]
        :rtype: List[float]
        """
        dic = {}
        for i,e in enumerate(equations):
            if not e[0] in dic:dic[e[0]]={}
            if not e[1] in dic:dic[e[1]]={}
            dic[e[0]][e[1]] = values[i]
            dic[e[1]][e[0]] = 1/values[i]
        res = []
        for q in queries:
            start  = q[0]
            target = q[1]
            res.append(self.dfs(start,target,dic,1,set()))
        return res
            
    def dfs(self,start,target,dic,tmp,visited):
        if not start in dic or not target in dic:return -1
        if start==target:return tmp
        for child in dic[start]:
            if not (start,child) in visited:
                visited.add((start,child))
                coco = self.dfs(child,target,dic,tmp*dic[start][child],visited)
                if coco!=-1:return coco
                visited.remove((start,child))
        return -1

Evaluate Division problem solution in Java.

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
         Map<String,Map<String, Double>> map = new HashMap<>();
         for(int i=0;i<equations.size();i++){
             List<String> list = equations.get(i);
             String a = list.get(0);
             String b = list.get(1);
             if(map.containsKey(a)==false){
                 map.put(a, new HashMap<>());
             }
             map.get(a).put(b, values[i]);
             
             if(map.containsKey(b)==false){
                 map.put(b, new HashMap<>());
             }
             map.get(b).put(a, 1/values[i]);
         }
        
         int index = 0;
         double[] res = new double[queries.size()];
         for(List<String> ele: queries){
                if(map.containsKey(ele.get(0))==false || 
                   map.containsKey(ele.get(1))==false){
                    res[index++] = -1;
                    continue;
                }
                double temp = dfs(map, new HashSet<>(), ele.get(0), ele.get(1));     
                res[index++] = (temp==0?-1:temp);
         }
        
         return res;
    }
    
    public double dfs(Map<String, Map<String, Double>> map, Set<String> seen, 
                   String start, String end){
        
            if(start.equals(end)){
                return 1;
            }
            if(seen.contains(start)){
                return 0;
            }
            
            
            seen.add(start);
            for(String next: map.get(start).keySet()){
                  double temp = dfs(map, seen, next, end);
                  if(temp!=0){
                      return temp*map.get(start).get(next);
                  }
            }
        
            seen.remove(start);
            return 0;
        
    }
}

Problem solution in C++.

class Solution {
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        for(int i = 0; i < equations.size(); i++) {
            string o = equations[i][0];
            string d = equations[i][1];
            double w = values[i];
            double w_reciprocal = 1/w;
            
            adj_[o].emplace(d, w);
            adj_[d].emplace(o, w_reciprocal);
        }
        
        vector<double> res{};
        for(const auto &q : queries) {
            bc_.clear();
            string origin = q[0];
            string dest = q[1];
            res.emplace_back(dfs(origin, dest));
        }
        
        return res;
    }
private:
    unordered_map<string, unordered_map<string, double>> adj_{};
    unordered_set<string> bc_{};
    
    double dfs(string origin, string dest) {        
        if(adj_.find(origin) == adj_.end()) {
            return -1;
        }
        
        if(origin == dest) {
            return 1;
        }
        
        if(bc_.find(origin) != bc_.end()) {
            return -1;
        }
        
        bc_.emplace(origin);
        for(const auto &p : adj_[origin]) {
            string next = p.first;
            double weight = p.second;
            
            double res = dfs(next, dest);
            if(res != -1) {
                return res * weight;
            }
        }
        
        return -1;
    }
};

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