# 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.

## Solutions

Union find.

• class Solution {
private int[] p;

public boolean equationsPossible(String[] equations) {
p = new int[26];
for (int i = 0; i < 26; ++i) {
p[i] = i;
}
for (String e : equations) {
int a = e.charAt(0) - 'a', b = e.charAt(3) - 'a';
if (e.charAt(1) == '=') {
p[find(a)] = find(b);
}
}
for (String e : equations) {
int a = e.charAt(0) - 'a', b = e.charAt(3) - 'a';
if (e.charAt(1) == '!' && find(a) == find(b)) {
return false;
}
}
return true;
}

private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}

• class Solution {
public:
vector<int> p;

bool equationsPossible(vector<string>& equations) {
p.resize(26);
for (int i = 0; i < 26; ++i) p[i] = i;
for (auto& e : equations) {
int a = e[0] - 'a', b = e[3] - 'a';
if (e[1] == '=') p[find(a)] = find(b);
}
for (auto& e : equations) {
int a = e[0] - 'a', b = e[3] - 'a';
if (e[1] == '!' && find(a) == find(b)) return false;
}
return true;
}

int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};

• class Solution:
def equationsPossible(self, equations: List[str]) -> bool:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

p = list(range(26))
for e in equations:
a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
if e[1] == '=':
p[find(a)] = find(b)
for e in equations:
a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
if e[1] == '!' and find(a) == find(b):
return False
return True


• func equationsPossible(equations []string) bool {
p := make([]int, 26)
for i := 1; i < 26; i++ {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for _, e := range equations {
a, b := int(e[0]-'a'), int(e[3]-'a')
if e[1] == '=' {
p[find(a)] = find(b)
}
}
for _, e := range equations {
a, b := int(e[0]-'a'), int(e[3]-'a')
if e[1] == '!' && find(a) == find(b) {
return false
}
}
return true
}

• class UnionFind {
private parent: number[];

constructor() {
this.parent = Array.from({ length: 26 }).map((_, i) => i);
}

find(index: number) {
if (this.parent[index] === index) {
return index;
}
this.parent[index] = this.find(this.parent[index]);
return this.parent[index];
}

union(index1: number, index2: number) {
this.parent[this.find(index1)] = this.find(index2);
}
}

function equationsPossible(equations: string[]): boolean {
const uf = new UnionFind();
for (const [a, s, _, b] of equations) {
if (s === '=') {
const index1 = a.charCodeAt(0) - 'a'.charCodeAt(0);
const index2 = b.charCodeAt(0) - 'a'.charCodeAt(0);
uf.union(index1, index2);
}
}
for (const [a, s, _, b] of equations) {
if (s === '!') {
const index1 = a.charCodeAt(0) - 'a'.charCodeAt(0);
const index2 = b.charCodeAt(0) - 'a'.charCodeAt(0);
if (uf.find(index1) === uf.find(index2)) {
return false;
}
}
}
return true;
}