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 <= 20equations[i].length == 21 <= Ai.length, Bi.length <= 5values.length == equations.length0.0 < values[i] <= 20.01 <= queries.length <= 20queries[i].length == 21 <= Cj.length, Dj.length <= 5Ai, Bi, Cj, Dj consist of lower case English letters and digits.struct Solution;
use std::collections::BTreeSet;
use std::collections::HashMap;
type Edge = (usize, f64);
impl Solution {
fn calc_equation(
equations: Vec<Vec<String>>,
values: Vec<f64>,
queries: Vec<Vec<String>>,
) -> Vec<f64> {
let m = equations.len();
let mut symbols: BTreeSet<String> = BTreeSet::new();
let mut ids: HashMap<String, usize> = HashMap::new();
for eq in &equations {
symbols.insert(eq[0].clone());
symbols.insert(eq[1].clone());
}
for (i, s) in symbols.into_iter().enumerate() {
ids.insert(s, i);
}
let n = ids.len();
let mut graph: Vec<Vec<Edge>> = vec![vec![]; n];
for i in 0..m {
let u = ids[&equations[i][0]];
let v = ids[&equations[i][1]];
graph[u].push((v, values[i]));
graph[v].push((u, 1.0 / values[i]));
}
let mut res = vec![];
for query in queries {
if ids.contains_key(&query[0]) && ids.contains_key(&query[1]) {
let u = ids[&query[0]];
let v = ids[&query[1]];
let mut product = -1.0;
let mut visited = vec![false; n];
let mut path: Vec<f64> = vec![];
Self::dfs(u, v, &mut visited, &mut path, &mut product, &graph);
res.push(product);
} else {
res.push(-1.0);
}
}
res
}
fn dfs(
u: usize,
v: usize,
visited: &mut Vec<bool>,
path: &mut Vec<f64>,
product: &mut f64,
graph: &[Vec<Edge>],
) {
visited[u] = true;
if u == v {
*product = path.iter().fold(1.0, |a, v| a * v);
} else {
for e in &graph[u] {
if !visited[e.0] {
path.push(e.1);
Self::dfs(e.0, v, visited, path, product, graph);
path.pop();
}
}
}
visited[u] = false;
}
}
#[test]
fn test() {
let equations = vec_vec_string![["a", "b"], ["b", "c"]];
let values = vec![2.0, 3.0];
let queries = vec_vec_string![["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"]];
let res = vec![6.0, 0.5, -1.0, 1.0, -1.0];
assert_eq!(Solution::calc_equation(equations, values, queries), res);
}