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990. Satisfiability of Equality Equations

Description

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.

Solutions

Union find.

  • class Solution {
    private int[] p;

    public boolean equationsPossible(String[] equations) {
        p = new int[26];
        for (int i = 0; i < 26; ++i) {
            p[i] = i;
        }
        for (String e : equations) {
            int a = e.charAt(0) - 'a', b = e.charAt(3) - 'a';
            if (e.charAt(1) == '=') {
                p[find(a)] = find(b);
            }
        }
        for (String e : equations) {
            int a = e.charAt(0) - 'a', b = e.charAt(3) - 'a';
            if (e.charAt(1) == '!' && find(a) == find(b)) {
                return false;
            }
        }
        return true;
    }

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

  • class Solution {
public:
    vector<int> p;

    bool equationsPossible(vector<string>& equations) {
        p.resize(26);
        for (int i = 0; i < 26; ++i) p[i] = i;
        for (auto& e : equations) {
            int a = e[0] - 'a', b = e[3] - 'a';
            if (e[1] == '=') p[find(a)] = find(b);
        }
        for (auto& e : equations) {
            int a = e[0] - 'a', b = e[3] - 'a';
            if (e[1] == '!' && find(a) == find(b)) return false;
        }
        return true;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

  • class Solution:
    def equationsPossible(self, equations: List[str]) -> bool:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(26))
        for e in equations:
            a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
            if e[1] == '=':
                p[find(a)] = find(b)
        for e in equations:
            a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
            if e[1] == '!' and find(a) == find(b):
                return False
        return True


  • func equationsPossible(equations []string) bool {
	p := make([]int, 26)
	for i := 1; i < 26; i++ {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, e := range equations {
		a, b := int(e[0]-'a'), int(e[3]-'a')
		if e[1] == '=' {
			p[find(a)] = find(b)
		}
	}
	for _, e := range equations {
		a, b := int(e[0]-'a'), int(e[3]-'a')
		if e[1] == '!' && find(a) == find(b) {
			return false
		}
	}
	return true
}

  • class UnionFind {
    private parent: number[];

    constructor() {
        this.parent = Array.from({ length: 26 }).map((_, i) => i);
    }

    find(index: number) {
        if (this.parent[index] === index) {
            return index;
        }
        this.parent[index] = this.find(this.parent[index]);
        return this.parent[index];
    }

    union(index1: number, index2: number) {
        this.parent[this.find(index1)] = this.find(index2);
    }
}

function equationsPossible(equations: string[]): boolean {
    const uf = new UnionFind();
    for (const [a, s, _, b] of equations) {
        if (s === '=') {
            const index1 = a.charCodeAt(0) - 'a'.charCodeAt(0);
            const index2 = b.charCodeAt(0) - 'a'.charCodeAt(0);
            uf.union(index1, index2);
        }
    }
    for (const [a, s, _, b] of equations) {
        if (s === '!') {
            const index1 = a.charCodeAt(0) - 'a'.charCodeAt(0);
            const index2 = b.charCodeAt(0) - 'a'.charCodeAt(0);
            if (uf.find(index1) === uf.find(index2)) {
                return false;
            }
        }
    }
    return true;
}

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