title
Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: “a==b” or “a!=b”. Here, a and b are lowercase letters (not necessarily different) that represent one-letter variable names.
Return true if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.
Example 1:
Input: ["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: ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
Example 3:
Input: ["a==b","b==c","a==c"]
Output: true
Example 4:
Input: ["a==b","b!=c","c==a"]
Output: false
Example 5:
Input: ["c==c","b==d","x!=z"]
Output: true
Note:
1 <= equations.length <= 500
equations[i].length == 4
equations[i][0] and equations[i][3] are lowercase letters
equations[i][1] is either '=' or '!'
equations[i][2] is '='
solution
class Solution:
def equationsPossible(self, equations: List[str]) -> bool:
uf = {}
def find(x):
if uf[x] != x:
uf[x] = find(uf[x])
return uf[x]
def union(a, b):
uf.setdefault(a, a)
uf.setdefault(b, b)
uf[find(a)] = find(b)
def is_connect(a, b):
uf.setdefault(a, a)
uf.setdefault(b, b)
return find(a) == find(b)
equals = [x for x in equations if x[1:3] == "=="]
for ie in equals:
a, b = ie[0], ie[3]
union(a, b)
not_equals = [x for x in equations if x[1:3] == "!="]
for ie in not_equals:
a, b = ie[0], ie[3]
if is_connect(a, b):
return False
return True