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Formatted question description: https://leetcode.ca/all/990.html

# 990. Satisfiability of Equality Equations (Medium)

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. 1 <= equations.length <= 500
2. equations[i].length == 4
3. equations[i][0] and equations[i][3] are lowercase letters
4. equations[i][1] is either '=' or '!'
5. equations[i][2] is '='

Related Topics:
Union Find, Graph

## Solution 1. Union Find

// OJ: https://leetcode.com/problems/satisfiability-of-equality-equations/
// Time: O(N)
// Space: O(1)
class UnionFind {
unordered_map<char, char> id;
unordered_map<char, int> rank;
public:
void connect(char a, char b) {
int x = find(a), y = find(b);
if (x == y) return;
if (rank[x] <= rank[y]) {
id[x] = y;
if (rank[x] == rank[y]) rank[y]++;
} else id[y] = x;
}
char find(char a) {
if (!id.count(a)) {
rank[a] = 1;
id[a] = a;
}
return id[a] == a ? a : (id[a] = find(id[a]));
}
};
class Solution {
public:
bool equationsPossible(vector<string>& equations) {
UnionFind uf;
for (auto e : equations) {
if (e[1] == '=') uf.connect(e[0], e[3]);
}
for (auto e : equations) {
if (e[1] == '!' && uf.find(e[0]) == uf.find(e[3])) return false;
}
return true;
}
};


## Solution 2. Union Find

// OJ: https://leetcode.com/problems/satisfiability-of-equality-equations/
// Time: O(N)
// Space: O(1)
class Solution {
int uf[26];
int find(int x) {
return uf[x] == x ? x : (uf[x] = find(uf[x]));
}
public:
bool equationsPossible(vector<string>& equations) {
for (int i = 0; i < 26; ++i) uf[i] = i;
for (auto e : equations) {
if (e[1] == '=') uf[find(e[0] - 'a')] = find(e[3] - 'a');
}
for (auto e : equations) {
if (e[1] == '!' && find(e[0] - 'a') == find(e[3] - 'a')) return false;
}
return true;
}
};


## Solution 3. Graph Coloring (DFS)

// OJ: https://leetcode.com/problems/satisfiability-of-equality-equations/
// Time: O(N)
// Space: O(N)
class Solution {
vector<bool> visited = vector<bool>(26, false);
int color[26];
void dfs(int u, int c) {
visited[u] = true;
color[u] = c;
for (auto v : adj[u]) {
if (!visited[v]) dfs(v, c);
}
}
public:
bool equationsPossible(vector<string>& equations) {
for (auto e : equations) {
if (e[1] != '=') continue;
}
for (int i = 0, c = 0; i < 26; ++i) {
if (!visited[i]) dfs(i, c++);
}
for (auto e : equations) {
if (e[1] == '!' && color[e[0] - 'a'] == color[e[3] - 'a']) return false;
}
return true;
}
};

• class Solution {
public boolean equationsPossible(String[] equations) {
Map<Character, Set<Character>> equalMap = new HashMap<Character, Set<Character>>();
Map<Character, Set<Character>> unequalMap = new HashMap<Character, Set<Character>>();
for (String equation : equations) {
char var1 = equation.charAt(0), var2 = equation.charAt(3);
if (equation.charAt(1) == '=') {
Set<Character> equalSet1 = equalMap.getOrDefault(var1, new HashSet<Character>());
Set<Character> equalSet2 = equalMap.getOrDefault(var2, new HashSet<Character>());
for (char var : equalSet1)
equalMap.put(var, equalSet1);
for (char var : equalSet2)
equalMap.put(var, equalSet2);
} else if (var1 == var2)
return false;
}
for (String equation : equations) {
if (equation.charAt(1) == '!') {
char var1 = equation.charAt(0), var2 = equation.charAt(3);
Set<Character> equalSet = equalMap.getOrDefault(var1, new HashSet<Character>());
if (equalSet.contains(var2))
return false;
}
}
return true;
}
}

############

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];
}
}

• // OJ: https://leetcode.com/problems/satisfiability-of-equality-equations/
// Time: O(N)
// Space: O(1)
class UnionFind {
unordered_map<char, char> id;
unordered_map<char, int> rank;
public:
void connect(char a, char b) {
int x = find(a), y = find(b);
if (x == y) return;
if (rank[x] <= rank[y]) {
id[x] = y;
if (rank[x] == rank[y]) rank[y]++;
} else id[y] = x;
}
char find(char a) {
if (!id.count(a)) {
rank[a] = 1;
id[a] = a;
}
return id[a] == a ? a : (id[a] = find(id[a]));
}
};
class Solution {
public:
bool equationsPossible(vector<string>& equations) {
UnionFind uf;
for (auto e : equations) {
if (e[1] == '=') uf.connect(e[0], e[3]);
}
for (auto e : equations) {
if (e[1] == '!' && uf.find(e[0]) == uf.find(e[3])) return false;
}
return true;
}
};

• 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

############

class Solution(object):
def equationsPossible(self, equations):
"""
:type equations: List[str]
:rtype: bool
"""
self.m = collections.defaultdict(list)
for eq in equations:
if eq[1] == '=':
self.m[eq[0]].append(eq[3])
self.m[eq[3]].append(eq[0])
for eq in equations:
if eq[1] == '!':
if self.find(set(), eq[0], eq[3]) or self.find(set(), eq[3], eq[0]):
return False
return True

def find(self, visited, begin, end):
if begin in visited:
return False
if begin == end:
return True
for n in self.m[begin]:
if self.find(visited, n, end):
return True
return False

• 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;
}