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886. Possible Bipartition

Description

We want to split a group of n people (labeled from 1 to n) into two groups of any size. Each person may dislike some other people, and they should not go into the same group.

Given the integer n and the array dislikes where dislikes[i] = [ai, bi] indicates that the person labeled ai does not like the person labeled bi, return true if it is possible to split everyone into two groups in this way.

 

Example 1:

Input: n = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: The first group has [1,4], and the second group has [2,3].

Example 2:

Input: n = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false
Explanation: We need at least 3 groups to divide them. We cannot put them in two groups.

 

Constraints:

  • 1 <= n <= 2000
  • 0 <= dislikes.length <= 104
  • dislikes[i].length == 2
  • 1 <= ai < bi <= n
  • All the pairs of dislikes are unique.

Solutions

Union find.

  • class Solution {
    private int[] p;

    public boolean possibleBipartition(int n, int[][] dislikes) {
        p = new int[n];
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (var e : dislikes) {
            int a = e[0] - 1, b = e[1] - 1;
            g[a].add(b);
            g[b].add(a);
        }
        for (int i = 0; i < n; ++i) {
            for (int j : g[i]) {
                if (find(i) == find(j)) {
                    return false;
                }
                p[find(j)] = find(g[i].get(0));
            }
        }
        return true;
    }

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

  • class Solution {
public:
    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        unordered_map<int, vector<int>> g;
        for (auto& e : dislikes) {
            int a = e[0] - 1, b = e[1] - 1;
            g[a].push_back(b);
            g[b].push_back(a);
        }
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) p[x] = find(p[x]);
            return p[x];
        };
        for (int i = 0; i < n; ++i) {
            for (int j : g[i]) {
                if (find(i) == find(j)) return false;
                p[find(j)] = find(g[i][0]);
            }
        }
        return true;
    }
};

  • class Solution:
    def possibleBipartition(self, n: int, dislikes: List[List[int]]) -> bool:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        g = defaultdict(list)
        for a, b in dislikes:
            a, b = a - 1, b - 1
            g[a].append(b)
            g[b].append(a)
        p = list(range(n))
        for i in range(n):
            for j in g[i]:
                if find(i) == find(j):
                    return False
                p[find(j)] = find(g[i][0])
        return True


  • func possibleBipartition(n int, dislikes [][]int) bool {
	p := make([]int, n)
	g := make([][]int, n)
	for i := range p {
		p[i] = i
	}
	for _, e := range dislikes {
		a, b := e[0]-1, e[1]-1
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for i := 0; i < n; i++ {
		for _, j := range g[i] {
			if find(i) == find(j) {
				return false
			}
			p[find(j)] = find(g[i][0])
		}
	}
	return true
}

  • function possibleBipartition(n: number, dislikes: number[][]): boolean {
    const color = new Array(n + 1).fill(0);
    const g = Array.from({ length: n + 1 }, () => []);
    const dfs = (i: number, v: number) => {
        color[i] = v;
        for (const j of g[i]) {
            if (color[j] === color[i] || (color[j] === 0 && dfs(j, 3 ^ v))) {
                return true;
            }
        }
        return false;
    };
    for (const [a, b] of dislikes) {
        g[a].push(b);
        g[b].push(a);
    }
    for (let i = 1; i <= n; i++) {
        if (color[i] === 0 && dfs(i, 1)) {
            return false;
        }
    }
    return true;
}


  • impl Solution {
    fn dfs(i: usize, v: usize, color: &mut Vec<usize>, g: &Vec<Vec<usize>>) -> bool {
        color[i] = v;
        for &j in (*g[i]).iter() {
            if color[j] == color[i] || (color[j] == 0 && Self::dfs(j, v ^ 3, color, g)) {
                return true;
            }
        }
        false
    }

    pub fn possible_bipartition(n: i32, dislikes: Vec<Vec<i32>>) -> bool {
        let n = n as usize;
        let mut color = vec![0; n + 1];
        let mut g = vec![Vec::new(); n + 1];
        for d in dislikes.iter() {
            let (i, j) = (d[0] as usize, d[1] as usize);
            g[i].push(j);
            g[j].push(i);
        }
        for i in 1..=n {
            if color[i] == 0 && Self::dfs(i, 1, &mut color, &g) {
                return false;
            }
        }
        true
    }
}

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