Question

399. Evaluate Division

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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

Solution 1

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
    ]

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];
  }
};

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
  }
}