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Approach: Represent the given variable equations as a graph. Each variable is a node, and an equation between two variables becomes a directed edge with a weight from one node to another representing the ratio. To find the result of a query, search for a path from the dividend node to the divisor node using Depth First Search (DFS). If a path is found, multiply the weights along the path to get the result. If no path is found, return -1.0 as the result cannot be determined.
Time Complexity: O(V + E) for each query, where V is the number of variables and E is the number of equations due to the DFS traversal. Overall complexity is O(Q * (V + E)) for Q queries.
Space Complexity: O(V + E) for storing the graph.
1from collections import defaultdict
2
3def calcEquation(equations, values, queries):
4 graph = defaultdict(dict)
5
6 # Build the graph
7 for (dividend, divisor), value in zip(equations, values):
8 graph[dividend][divisor] = value
9 graph[divisor][dividend] = 1 / value
10
11 def dfs(start, end, visited, value):
12 if start == end:
13 return value
14 visited.add(start)
15
16 for neighbor, weight in graph[start].items():
17 if neighbor not in visited:
18 result = dfs(neighbor, end, visited, value * weight)
19 if result is not None:
20 return result
21
22 return None
23
24 results = []
25 for dividend, divisor in queries:
26 if dividend not in graph or divisor not in graph:
27 results.append(-1.0)
28 else:
29 visited = set()
30 result = dfs(dividend, divisor, visited, 1)
31 results.append(result if result is not None else -1.0)
32
33 return resultsThe Python implementation uses a recursive DFS function to find a path from the start variable to the end variable. The graph is built using a dictionary of dictionaries, allowing easy access to directly connected nodes and the associated division value. If a direct or indirect path exists, the function calculates the result by multiplying the weights along the path. If no path can be found, it returns -1.0.
Approach: Use Union-Find (Disjoint Set Union, DSU) with path compression to represent connected components of variables where each component can efficiently find the "parent" or representative of a set. Each variable is associated with a weight that represents its relative scale with respect to its parent. This method efficiently handles union operations and queries to determine the result by finding the root of each variable and combining their scales.
Time Complexity: O(E + Q * α(V)), where E is the number of equations, Q is the number of queries, V is the number of variables, and α is the inverse Ackermann function, which grows very slowly.
Space Complexity: O(V) for storing parent information and weights.
1#include <unordered_map>
2#include <vector>
3#include <string>
4
using namespace std;
class Solution {
public:
vector<double> calcEquation(
vector<vector<string>>& equations,
vector<double>& values,
vector<vector<string>>& queries) {
unordered_map<string, pair<string, double>> parents;
// Find function with path compression
function<pair<string, double>(string)> find = [&](string v) {
if (!parents.count(v)) return make_pair(v, 1.0);
auto& [p, w] = parents[v];
if (v != p) {
auto result = find(p);
parents[v] = {result.first, w * result.second};
}
return parents[v];
};
// Union function
auto unite = [&](string v1, string v2, double ratio) {
auto [root1, weight1] = find(v1);
auto [root2, weight2] = find(v2);
if (root1 != root2) {
parents[root2] = {root1, ratio * weight1 / weight2};
}
};
for (int i = 0; i < equations.size(); ++i) {
const auto& eq = equations[i];
unite(eq[0], eq[1], values[i]);
}
vector<double> results;
for (const auto& query : queries) {
if (!parents.count(query[0]) || !parents.count(query[1])) {
results.push_back(-1.0);
continue;
}
auto [root1, weight1] = find(query[0]);
auto [root2, weight2] = find(query[1]);
if (root1 != root2) {
results.push_back(-1.0);
} else {
results.push_back(weight1 / weight2);
}
}
return results;
}
};This C++ solution uses a Union-Find data structure with path compression. For each equation, it unifies the variables into a single component, maintaining their relative ratios. For each query, it determines whether the variables are in the same component and calculates the result using the stored weights (which represent ratios to their respective roots).