399. 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.

Rust Solution

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);
}

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