LC 990. Satisfiability of Equality Equations

Posted 2021-02-08 ethanhong

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 ‘=‘

 

Runtime: 12 ms, faster than 100.00% of C++ online submissions for Satisfiability of Equality Equations.

Memory Usage: 7.1 MB, less than 100.00% of C++ online submissions for Satisfiability of Equality Equations.

class Solution {

private:
  int arr[26];
public:

  void unionab(int a, int b) {
    arr[parent(a)] = arr[parent(b)];
  }
  int parent(int a) {
    if(arr[a] != a) return parent(arr[a]);
    return a;
  }
  bool uninit(int a) {
    return arr[a] == a ? true : false;
  }
  bool hassameroot(int a, int b) {
    return parent(a) == parent(b);
  }

  bool equationsPossible(vector<string>& equations) {
    for(int i=0; i<26; i++) arr[i] = i;
    for(int i=0; i<equations.size(); i++) {
      int a = ((int)equations[i][0] - (int)‘a‘);
      int b = ((int)equations[i][3] - (int)‘a‘);
      if ((int)equations[i][1] == (int)‘=‘) {
        if(!hassameroot(a,b)) unionab(a,b);
      }
    }
    for(int i=0; i<equations.size(); i++) {
      int a = ((int)equations[i][0] - (int)‘a‘);
      int b = ((int)equations[i][3] - (int)‘a‘);
      if((int)equations[i][1] == (int)‘!‘) {
        if(hassameroot(a,b)) return false;
      }
    }
    return true;
  }
};