You are given a 2D array of strings equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] means that Ai / Bi = values[i].
Determine if there exists a contradiction in the equations. Return true if there is a contradiction, or false otherwise.
Note:
When checking if two numbers are equal, check that their absolute difference is less than 10-5.
The testcases are generated such that there are no cases targeting precision, i.e. using double is enough to solve the problem.
Example 1:
Input: equations = [["a","b"],["b","c"],["a","c"]], values = [3,0.5,1.5]
Output: false
Explanation:
The given equations are: a / b = 3, b / c = 0.5, a / c = 1.5
There are no contradictions in the equations. One possible assignment to satisfy all equations is:
a = 3, b = 1 and c = 2.
Example 2:
Input: equations = [["le","et"],["le","code"],["code","et"]], values = [2,5,0.5]
Output: true
Explanation:
The given equations are: le / et = 2, le / code = 5, code / et = 0.5
Based on the first two equations, we get code / et = 0.4.
Since the third equation is code / et = 0.5, we get a contradiction.
Constraints:
1 <= equations.length <= 100
equations[i].length == 2
1 <= Ai.length, Bi.length <= 5
Ai, Bi consist of lowercase English letters.
equations.length == values.length
0.0 < values[i] <= 10.0
values[i] has a maximum of 2 decimal places.
Solutions
Solution 1: Weighted Union-Find
First, we convert the strings into integers starting from $0$. Then, we traverse all the equations, map the two strings in each equation to the corresponding integers $a$ and $b$. If these two integers are not in the same set, we merge them into the same set and record the weights of the two integers, which is the ratio of $a$ to $b$. If these two integers are in the same set, we check whether their weights satisfy the equation. If not, we return true.
The time complexity is $O(n \times \log n)$ or $O(n \times \alpha(n))$, and the space complexity is $O(n)$. Here, $n$ is the number of equations.