990. Satisfiability of Equality Equations #

Problem #

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 '='

Problem Summary #

Given an array of string equations representing relationships between variables, each string equation 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 only if it is possible to assign integers to variable names so that all the given equations are satisfied; otherwise return false. 

Constraints:

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 ‘=’

Solution Approach #

• Given a string array, the array contains relationships between letters, with only two types of relationships: '==' and '! ='. Determine whether there is a contradiction among these given relationships.
• This is a simple union-find problem. First union() all letters with '==' relationships, then check each '! =' relationship to see whether there is a combination that has a '==' relationship. If so, return false; if none are found after traversal, return true.

Code #

package leetcode

import (
	"github.com/halfrost/leetcode-go/template"
)

func equationsPossible(equations []string) bool {
	if len(equations) == 0 {
		return false
	}
	uf := template.UnionFind{}
	uf.Init(26)
	for _, equ := range equations {
		if equ[1] == '=' && equ[2] == '=' {
			uf.Union(int(equ[0]-'a'), int(equ[3]-'a'))
		}
	}
	for _, equ := range equations {
		if equ[1] == '!' && equ[2] == '=' {
			if uf.Find(int(equ[0]-'a')) == uf.Find(int(equ[3]-'a')) {
				return false
			}
		}
	}
	return true
}