990. Satisfiability of Equality Equations

Problem

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.

Solution (Java)

class Solution {
    private int[] par;

    public boolean equationsPossible(String[] equations) {
        int counter = 0;
        HashMap<Character, Integer> map = new HashMap<>();
        for (String str : equations) {
            char ch = str.charAt(0);
            if (!map.containsKey(ch)) {
                map.put(ch, counter);
                counter++;
            }
            ch = str.charAt(3);
            if (!map.containsKey(ch)) {
                map.put(ch, counter);
                counter++;
            }
        }
        par = new int[counter];
        for (int i = 0; i < par.length; i++) {
            par[i] = i;
        }
        for (String str : equations) {
            String oper = str.substring(1, 3);
            if (oper.equals("==")) {
                int px = find(map.get(str.charAt(0)));
                int py = find(map.get(str.charAt(3)));
                if (px != py) {
                    par[px] = py;
                }
            }
        }
        for (String str : equations) {
            String oper = str.substring(1, 3);
            if (oper.equals("!=")) {
                int px = find(map.get(str.charAt(0)));
                int py = find(map.get(str.charAt(3)));
                if (px == py) {
                    return false;
                }
            }
        }
        return true;
    }

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).