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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.
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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); } }
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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]; } };
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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 ]
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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 }
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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 } }
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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; vis.add(current); 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; }