0399--evaluate-division.md

399. Evaluate Division

Difficulty: Medium

https://leetcode.com/problems/evaluate-division/

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.

 

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 ]

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]

 

Constraints:

• 1 <= equations.length <= 20
• equations[i].length == 2
• 1 <= Ai.length, Bi.length <= 5
• values.length == equations.length
• 0.0 < values[i] <= 20.0
• 1 <= queries.length <= 20
• queries[i].length == 2
• 1 <= Cj.length, Dj.length <= 5
• Ai, Bi, Cj, Dj consist of lower case English letters and digits.

My Solution(s):

JavaScript

var calcEquation = function (equations, values, queries) {
    const graph = {};
    const result = [];

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

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

        graph[a][b] = value;
        graph[b][a] = 1 / value;
    }

    for (let i = 0; i < queries.length; i++) {
        const [a, b] = queries[i];
        const value = dfs(graph, a, b, {});
        result.push(value);
    }

    return result;

};

function dfs(graph, a, b, visited) {
    if (!graph[a]) return -1;
    if (a === b) return 1;

    visited[a] = true;

    const neighbors = Object.keys(graph[a]);
    for (let i = 0; i < neighbors.length; i++) {
        const neighbor = neighbors[i];
        if (visited[neighbor]) continue;

        const value = dfs(graph, neighbor, b, visited);
        if (value !== -1) {
            return value * graph[a][neighbor];
        }
    }

    return -1;
}

PHP

class Solution
{
    /**
     * @param String[][] $equations
     * @param Float[] $values
     * @param String[][] $queries
     * @return Float[]
     */
function calcEquation(array $equations, array $values, array $queries): array {
        $graph = [];
        $result = [];

        // Build Graph
        foreach ($equations as $key => $equation) {
            $graph[$equation[0]][$equation[1]] = $values[$key];
            $graph[$equation[1]][$equation[0]] = 1 / $values[$key];
        }

        foreach ($queries as $query) {
            $visited = [];
            $tmp = $this->dfs($graph, $query[0], $query[1], 1, $visited);
            $result[] = $tmp == null ? -1 : $tmp;
        }

        return $result;
    }

    function dfs(array $graph, $node, $end, $count, &$visited) {
        if (!isset($graph[$node]) || !isset($graph[$end])) {
            return -1;
        }

        if ($node == $end) {
            return $count;
        }

        if (isset($visited[$node])) {
            return null;
        }

        $visited[$node] = true;
        foreach ($graph[$node] as $key => $value) {
            $result = $this->dfs($graph, $key, $end, $count * $value, $visited);
            if ($result) {
                return $result;
            }
        }

        return null;
    }
}