990. Satisfiability of Equality Equations
Description
You are given an array of strings equations
that represent relationships between variables where each string equations[i]
is of length 4
and takes one of two different forms: "xi==yi"
or "xi!=yi"
.Here, xi
and yi
are lowercase letters (not necessarily different) that represent one-letter variable names.
Return true
if it is possible to assign integers to variable names so as to satisfy all the given equations, or false
otherwise.
Example 1:
Input: equations = ["a==b","b!=a"] Output: false Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second. There is no way to assign the variables to satisfy both equations.
Example 2:
Input: equations = ["b==a","a==b"] Output: true Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
Constraints:
1 <= equations.length <= 500
equations[i].length == 4
equations[i][0]
is a lowercase letter.equations[i][1]
is either'='
or'!'
.equations[i][2]
is'='
.equations[i][3]
is a lowercase letter.
Solution
satisfiability-of-equality-equations.py
class UnionFind:
def __init__(self):
self._parent = {}
self._size = {}
def union(self, a, b):
a, b = self.find(a), self.find(b)
if a == b:
return
if self._size[a] < self._size[b]:
a, b = b, a
self._parent[b] = a
self._size[a] += self._size[b]
def find(self, x):
if x not in self._parent:
self._parent[x] = x
self._size[x] = 1
while self._parent[x] != x:
self._parent[x] = self._parent[self._parent[x]]
x = self._parent[x]
return x
class Solution:
def equationsPossible(self, equations: List[str]) -> bool:
N = len(equations)
uf = UnionFind()
for x, e, _ , y in equations:
if e == "=":
uf.union(x, y)
return not any(e == "!" and uf.find(x) == uf.find(y) for x, e, _, y in equations)