# 399. Evaluate Division

## Description

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.

Note: The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.

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 ]
note: x is undefined => -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.

## Solutions

Union find.

• class Solution {
private Map<String, String> p;
private Map<String, Double> w;

public double[] calcEquation(
List<List<String>> equations, double[] values, List<List<String>> queries) {
int n = equations.size();
p = new HashMap<>();
w = new HashMap<>();
for (List<String> e : equations) {
p.put(e.get(0), e.get(0));
p.put(e.get(1), e.get(1));
w.put(e.get(0), 1.0);
w.put(e.get(1), 1.0);
}
for (int i = 0; i < n; ++i) {
List<String> e = equations.get(i);
String a = e.get(0), b = e.get(1);
String pa = find(a), pb = find(b);
if (Objects.equals(pa, pb)) {
continue;
}
p.put(pa, pb);
w.put(pa, w.get(b) * values[i] / w.get(a));
}
int m = queries.size();
double[] ans = new double[m];
for (int i = 0; i < m; ++i) {
String c = queries.get(i).get(0), d = queries.get(i).get(1);
ans[i] = !p.containsKey(c) || !p.containsKey(d) || !Objects.equals(find(c), find(d))
? -1.0
: w.get(c) / w.get(d);
}
return ans;
}

private String find(String x) {
if (!Objects.equals(p.get(x), x)) {
String origin = p.get(x);
p.put(x, find(p.get(x)));
w.put(x, w.get(x) * w.get(origin));
}
return p.get(x);
}
}

• class Solution {
public:
unordered_map<string, string> p;
unordered_map<string, double> w;

vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
int n = equations.size();
for (auto e : equations) {
p[e[0]] = e[0];
p[e[1]] = e[1];
w[e[0]] = 1.0;
w[e[1]] = 1.0;
}
for (int i = 0; i < n; ++i) {
vector<string> e = equations[i];
string a = e[0], b = e[1];
string pa = find(a), pb = find(b);
if (pa == pb) continue;
p[pa] = pb;
w[pa] = w[b] * values[i] / w[a];
}
int m = queries.size();
vector<double> ans(m);
for (int i = 0; i < m; ++i) {
string c = queries[i][0], d = queries[i][1];
ans[i] = p.find(c) == p.end() || p.find(d) == p.end() || find(c) != find(d) ? -1.0 : w[c] / w[d];
}
return ans;
}

string find(string x) {
if (p[x] != x) {
string origin = p[x];
p[x] = find(p[x]);
w[x] *= w[origin];
}
return p[x];
}
};

• class Solution:
def calcEquation(
self, equations: List[List[str]], values: List[float], queries: List[List[str]]
) -> List[float]:
def find(x):
if p[x] != x:
origin = p[x]
p[x] = find(p[x])
w[x] *= w[origin]
return p[x]

w = defaultdict(lambda: 1)
p = defaultdict()
for a, b in equations:
p[a], p[b] = a, b
for i, v in enumerate(values):
a, b = equations[i]
pa, pb = find(a), find(b)
if pa == pb:
continue
p[pa] = pb
w[pa] = w[b] * v / w[a]
return [
-1 if c not in p or d not in p or find(c) != find(d) else w[c] / w[d]
for c, d in queries
]


• func calcEquation(equations [][]string, values []float64, queries [][]string) []float64 {
p := make(map[string]string)
w := make(map[string]float64)
for _, e := range equations {
a, b := e[0], e[1]
p[a], p[b] = a, b
w[a], w[b] = 1.0, 1.0
}

var find func(x string) string
find = func(x string) string {
if p[x] != x {
origin := p[x]
p[x] = find(p[x])
w[x] *= w[origin]
}
return p[x]
}

for i, v := range values {
a, b := equations[i][0], equations[i][1]
pa, pb := find(a), find(b)
if pa == pb {
continue
}
p[pa] = pb
w[pa] = w[b] * v / w[a]
}
var ans []float64
for _, e := range queries {
c, d := e[0], e[1]
if p[c] == "" || p[d] == "" || find(c) != find(d) {
ans = append(ans, -1.0)
} else {
ans = append(ans, w[c]/w[d])
}
}
return ans
}

• use std::collections::HashMap;

#[derive(Debug)]
pub struct DSUNode {
parent: String,
weight: f64,
}

pub struct DisjointSetUnion {
nodes: HashMap<String, DSUNode>,
}

impl DisjointSetUnion {
pub fn new(equations: &Vec<Vec<String>>) -> DisjointSetUnion {
let mut nodes = HashMap::new();
for equation in equations.iter() {
for iter in equation.iter() {
nodes.insert(iter.clone(), DSUNode {
parent: iter.clone(),
weight: 1.0,
});
}
}
DisjointSetUnion { nodes }
}

pub fn find(&mut self, v: &String) -> String {
let origin = self.nodes[v].parent.clone();
if origin == *v {
return origin;
}

let root = self.find(&origin);
self.nodes.get_mut(v).unwrap().parent = root.clone();
self.nodes.get_mut(v).unwrap().weight *= self.nodes[&origin].weight;
root
}

pub fn union(&mut self, a: &String, b: &String, v: f64) {
let pa = self.find(a);
let pb = self.find(b);
if pa == pb {
return;
}
let (wa, wb) = (self.nodes[a].weight, self.nodes[b].weight);
self.nodes.get_mut(&pa).unwrap().parent = pb;
self.nodes.get_mut(&pa).unwrap().weight = (wb * v) / wa;
}

pub fn exist(&mut self, k: &String) -> bool {
self.nodes.contains_key(k)
}

pub fn calc_value(&mut self, a: &String, b: &String) -> f64 {
if !self.exist(a) || !self.exist(b) || self.find(a) != self.find(b) {
-1.0
} else {
let wa = self.nodes[a].weight;
let wb = self.nodes[b].weight;
wa / wb
}
}
}

impl Solution {
pub fn calc_equation(
equations: Vec<Vec<String>>,
values: Vec<f64>,
queries: Vec<Vec<String>>
) -> Vec<f64> {
let mut dsu = DisjointSetUnion::new(&equations);
for (i, &v) in values.iter().enumerate() {
let (a, b) = (&equations[i][0], &equations[i][1]);
dsu.union(a, b, v);
}

let mut ans = vec![];
for querie in queries {
let (c, d) = (&querie[0], &querie[1]);
ans.push(dsu.calc_value(c, d));
}
ans
}
}


• function calcEquation(equations: string[][], values: number[], queries: string[][]): number[] {
const g: Record<string, [string, number][]> = {};
const ans = Array.from({ length: queries.length }, () => -1);

for (let i = 0; i < equations.length; i++) {
const [a, b] = equations[i];
(g[a] ??= []).push([b, values[i]]);
(g[b] ??= []).push([a, 1 / values[i]]);
}

for (let i = 0; i < queries.length; i++) {
const [c, d] = queries[i];
const vis = new Set<string>();
const q: [string, number][] = [[c, 1]];

if (!g[c] || !g[d]) continue;
if (c === d) {
ans[i] = 1;
continue;
}

for (const [current, v] of q) {
if (vis.has(current)) continue;

for (const [intermediate, multiplier] of g[current]) {
if (vis.has(intermediate)) continue;

if (intermediate === d) {
ans[i] = v * multiplier;
break;
}

q.push([intermediate, v * multiplier]);
}

if (ans[i] !== -1) break;
}
}

return ans;
}