Evaluate Division Problem
Description
LeetCode Problem 399.
You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [A_i, B_i] and values[i] represent the equation A_i / B_i = values[i]. Each A_i or B_i is a string that represents a single variable.
You are also given some queries, where queries[j] = [C_j, D_j] represents the j^th query where you must find the answer for C_j / D_j = ?.
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:
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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:
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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:
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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 <= A_i.length, B_i.length <= 5
- values.length == equations.length
- 0.0 < values[i] <= 20.0
- 1 <= queries.length <= 20
- queries[i].length == 2
- 1 <= C_j.length, D_j.length <= 5
- A_i, B_i, C_j, D_j consist of lower case English letters and digits.
Sample C++ Code
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class Solution {
public:
unordered_map<string, vector<pair<string, double>>> graph;
vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
int n = equations.size();
for (int i = 0; i < n; i++) {
string a = equations[i][0];
string b = equations[i][1];
double val = values[i];
graph[a].push_back({b, val});
graph[b].push_back({a, (double)1/val});
}
vector<double> result;
for (auto query : queries) {
unordered_set<string> visited;
result.push_back(dfs(query[0], query[1], visited));
}
return result;
}
double dfs(string src, string dst, unordered_set<string> &visited) {
if (graph.find(src) == graph.end()) return -1;
if (src == dst) return 1;
for (auto node : graph[src]) {
if (visited.count(node.first)) continue;
visited.insert(node.first);
double res = dfs(node.first, dst, visited);
if (res != -1) return res * node.second;
}
return -1;
}
};