# Satisfiability Of Equality Equations Problem

## Description

LeetCode Problem 990.

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: “x_i==y_i” or “x_i!=y_i”.Here, x_i and y_i 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:

``````1
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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:

``````1
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Input: equations = ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
``````

Example 3:

``````1
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Input: equations = ["a==b","b==c","a==c"]
Output: true
``````

Example 4:

``````1
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Input: equations = ["a==b","b!=c","c==a"]
Output: false
``````

Example 5:

``````1
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Input: equations = ["c==c","b==d","x!=z"]
Output: true
``````

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.

## Sample C++ Code using Union Find

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class Solution {
public:
int _find(int p, vector<int>& nums) {
while (p != nums[p])
p = nums[p];
return p;
}
void _union(int i, int j, vector<int>& nums) {
int r1 = _find(i, nums);
int r2 = _find(j, nums);
if (r1 < r2)
nums[r2] = r1;
else
nums[r1] = r2;
}

bool equationsPossible(vector<string>& equations) {
vector<int> equal(26);
for (int i = 0; i < 26; i ++) equal[i] = i;

for (int i = 0; i < equations.size(); i ++) {
if (equations[i][1] == '=') {
_union(equations[i][0]-'a', equations[i][3]-'a', equal);
}
}

for (int i = 0; i < equations.size(); i ++) {
if (equations[i][1] == '!') {
int left = equations[i][0] - 'a';
int right = equations[i][3] - 'a';
left = _find(left, equal);
right = _find(right, equal);
if (left == right)
return false;
}
}
return true;
}
};
``````