990. Satisfiability of Equality Equations

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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
    vector<bool> visited = vector<bool>(26, false);
    int color[26];
    vector<vector<int>> adj = vector<vector<int>>(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;
            adj[e[0] - 'a'].push_back(e[3] - 'a');
            adj[e[3] - 'a'].push_back(e[0] - 'a');
        }
        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;
    }
};