LeetCode-in-Java

399. Evaluate Division

Medium

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [A_i, B_i] and values[i] represent the equation A_i / B_i = values[i]. Each A_i or B_i is a string that represents a single variable.

You are also given some queries, where queries[j] = [C_j, D_j] represents the j^th query where you must find the answer for C_j / D_j = ?.

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 <= A_i.length, B_i.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= C_j.length, D_j.length <= 5
A_i, B_i, C_j, D_j consist of lower case English letters and digits.

Solution

import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
    private Map<String, String> root;
    private Map<String, Double> rate;

    public double[] calcEquation(
            List<List<String>> equations, double[] values, List<List<String>> queries) {
        root = new HashMap<>();
        rate = new HashMap<>();
        int n = equations.size();
        for (List<String> equation : equations) {
            String x = equation.get(0);
            String y = equation.get(1);
            root.put(x, x);
            root.put(y, y);
            rate.put(x, 1.0);
            rate.put(y, 1.0);
        }
        for (int i = 0; i < n; ++i) {
            String x = equations.get(i).get(0);
            String y = equations.get(i).get(1);
            union(x, y, values[i]);
        }
        double[] result = new double[queries.size()];
        for (int i = 0; i < queries.size(); ++i) {
            String x = queries.get(i).get(0);
            String y = queries.get(i).get(1);
            if (!root.containsKey(x) || !root.containsKey(y)) {
                result[i] = -1;
                continue;
            }
            String rootX = findRoot(x, x, 1.0);
            String rootY = findRoot(y, y, 1.0);
            result[i] = rootX.equals(rootY) ? rate.get(x) / rate.get(y) : -1.0;
        }
        return result;
    }

    private void union(String x, String y, double v) {
        String rootX = findRoot(x, x, 1.0);
        String rootY = findRoot(y, y, 1.0);
        root.put(rootX, rootY);
        double r1 = rate.get(x);
        double r2 = rate.get(y);
        rate.put(rootX, v * r2 / r1);
    }

    private String findRoot(String originalX, String x, double r) {
        if (root.get(x).equals(x)) {
            root.put(originalX, x);
            rate.put(originalX, r * rate.get(x));
            return x;
        }
        return findRoot(originalX, root.get(x), r * rate.get(x));
    }
}

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