Skip to content
Programmingoneonone
Programmingoneonone
  • Engineering Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
    • 100+ C++ Programs
  • Solutions
    • HackerRank
      • Algorithms Solutions
      • C solutions
      • C++ solutions
      • Java solutions
      • Python solutions
    • Leetcode Solutions
    • HackerEarth Solutions
  • Work with US
Programmingoneonone
Programmingoneonone

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;
    }
};

coding problems solutions Leetcode Problems Solutions Leetcode

Post navigation

Previous post
Next post

Leave a Reply

Your email address will not be published. Required fields are marked *

Programmingoneonone

We at Programmingoneonone, also known as Programming101 is a learning hub of programming and other related stuff. We provide free learning tutorials/articles related to programming and other technical stuff to people who are eager to learn about it.

Pages

  • About US
  • Contact US
  • Privacy Policy

Practice

  • Java
  • C++
  • C

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes