花花酱 LeetCode 886. Possible Bipartition

zxi — Sun, 12 Aug 2018 08:40:59 +0000

Problem

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

Solution: Graph Coloring

Color a node with one color, and color all it’s disliked nodes with another color, if can not finish return false.

Time complexity: O(V+E)

Space complexity: O(V+E)

C++ / DFS

// Author: Huahua
// Running time: 96 ms
class Solution {
public:
    bool possibleBipartition(int N, vector>& dislikes) {
      g_ = vector>(N);
      for (const auto& d : dislikes) {
        g_[d[0] - 1].push_back(d[1] - 1);
        g_[d[1] - 1].push_back(d[0] - 1);
      }
      colors_ = vector(N, 0);  // 0: unknown, 1: red, -1: blue
      for (int i = 0; i < N; ++i)
        if (colors_[i] == 0 && !dfs(i, 1)) return false;
      return true;      
    }
private:
  vector> g_;
  vector colors_;
  bool dfs(int cur, int color) {
    colors_[cur] = color;
    for (int nxt : g_[cur]) {
      if (colors_[nxt] == color) return false;      
      if (colors_[nxt] == 0 && !dfs(nxt, -color)) return false;
    }
    return true;
  }
};

C++ / BFS

// Author: Huahua
// Running time: 100 ms
class Solution {
public:
  bool possibleBipartition(int N, vector>& dislikes) {
    vector> g(N);
    for (const auto& d : dislikes) {
      g[d[0] - 1].push_back(d[1] - 1);
      g[d[1] - 1].push_back(d[0] - 1);
    }
    queue q;
    vector colors(N, 0);  // 0: unknown, 1: red, -1: blue
    for (int i = 0; i < N; ++i) {
      if (colors[i] != 0) continue;
      q.push(i);
      colors[i] = 1;
      while (!q.empty()) {
        int cur = q.front(); q.pop();
        for (int nxt : g[cur]) {
          if (colors[nxt] == colors[cur]) return false;
          if (colors[nxt] != 0) continue;
          colors[nxt] = -colors[cur];
          q.push(nxt);
        }
      }
    }    
    return true;
  }
};

bipartite Archives - Huahua's Tech Road

花花酱 LeetCode 886. Possible Bipartition

Problem

Solution: Graph Coloring

C++ / DFS

C++ / BFS

Related Problem