886. Possible Bipartition

886. Possible Bipartition (Medium)

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group. 

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only 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: group1 [1,4], group2 [2,3]

Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

 

Note:

1. 1 <= N <= 2000
2. 0 <= dislikes.length <= 10000
3. 1 <= dislikes[i][j] <= N
4. dislikes[i][0] < dislikes[i][1]
5. There does not exist i != j for which dislikes[i] == dislikes[j].

Related Topics:
Depth-first Search

Solution 1. DFS

// OJ: https://leetcode.com/problems/possible-bipartition/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
    vector<unordered_set<int>> G;
    vector<int> id;
    bool dfs(int u, int prev = 1) {
        if (id[u]) return id[u] != prev;
        id[u] = -prev;
        for (int v : G[u]) {
            if (!dfs(v, id[u])) return false;
        }
        return true;
    }
public:
    bool possibleBipartition(int N, vector<vector<int>>& A) {
        G.assign(N + 1, {});
        id.assign(N + 1, 0);
        for (auto &v : A) {
            G[v[0]].insert(v[1]);
            G[v[1]].insert(v[0]);
        }
        for (int i = 1; i <= N; ++i) {
            if (id[i]) continue;
            if (!dfs(i)) return false;
        }
        return true;
    }
};

Solution 2. BFS

// OJ: https://leetcode.com/problems/possible-bipartition/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
    bool possibleBipartition(int N, vector<vector<int>>& A) {
        vector<vector<int>> G(N + 1);
        vector<int> id(N + 1, 0);
        for (auto &v : A) {
            G[v[0]].push_back(v[1]);
            G[v[1]].push_back(v[0]);
        }
        queue<int> q;
        for (int i = 1; i <= N; ++i) {
            if (id[i]) continue;
            q.push(i);
            id[i] = 1;
            while (q.size()) {
                int u = q.front();
                q.pop();
                for (int v : G[u]) {
                    if (id[v]) {
                        if (id[v] != -id[u]) return false;
                        else continue;
                    }
                    id[v] = -id[u];
                    q.push(v);
                }
            }
        }
        return true;
    }
};