Leetcode•Mar 28, 2026

Evaluate Division

Hazrat Ali

Leetcode

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.

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:

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:

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:

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]

Solution
var calcEquation = function(equations, values, queries) {
    let graph = {};

    for (let i = 0; i < equations.length; i++) {
        let [a, b] = equations[i];
        let val = values[i];

        if (!graph[a]) graph[a] = [];
        if (!graph[b]) graph[b] = [];

        graph[a].push([b, val]);
        graph[b].push([a, 1 / val]);
    }

    function dfs(src, dest, visited) {
        if (!(src in graph)) return -1.0;
        if (src === dest) return 1.0;

        visited.add(src);

        for (let [next, weight] of graph[src]) {
            if (!visited.has(next)) {
                let res = dfs(next, dest, visited);
                if (res !== -1.0) {
                    return res * weight;
                }
            }
        }

        return -1.0;
    }

    let result = [];

    for (let [src, dest] of queries) {
        let visited = new Set();
        result.push(dfs(src, dest, visited));
    }

    return result;
};



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