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 '='
class Solution {
Map<Character, Character> parent;
public boolean equationsPossible(String[] equations) {
parent = new HashMap<>();
for (String edge : equations) {
char var1 = edge.charAt(0);
char var2 = edge.charAt(3);
if (!parent.containsKey(var1))
parent.put(var1, var1);
if (!parent.containsKey(var2))
parent.put(var2, var2);
if (edge.charAt(1) == '=')
if (var1 != var2)
union(var1, var2);
}
for (String edge : equations) {
if (edge.charAt(1) == '!') {
char p1 = findParent(edge.charAt(0));
char p2 = findParent(edge.charAt(3));
if (p1 == p2)
return false;
}
}
return true;
}
public void union(char v1, char v2) {
char p1 = findParent(v1);
char p2 = findParent(v2);
if (p1 != p2)
parent.put(p1, p2);
}
public char findParent(char c) {
if (c != parent.get(c))
parent.put(c, findParent(parent.get(c)));
return parent.get(c);
}
}