g0301_0400.s0399_evaluate_division.Solution Maven / Gradle / Ivy
Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation
Show all versions of leetcode-in-java21 Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
package g0301_0400.s0399_evaluate_division;
// #Medium #Array #Depth_First_Search #Breadth_First_Search #Graph #Union_Find #Shortest_Path
// #2022_07_15_Time_1_ms_(99.52%)_Space_43_MB_(20.05%)
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
* 399 - Evaluate Division\.
*
* Medium
*
* 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.
**/
public class Solution {
private Map root;
private Map rate;
public double[] calcEquation(
List> equations, double[] values, List> queries) {
root = new HashMap<>();
rate = new HashMap<>();
int n = equations.size();
for (List 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));
}
}
© 2015 - 2025 Weber Informatics LLC | Privacy Policy