Possible Bipartition

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- Tag: #graph, #classic, #bipartite

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

// https://code.dennyzhang.com/possible-bipartition // Basic Ideas: // Complexity: Time O(n), Space O(n) func possibleBipartition(N int, dislikes [][]int) bool { m := map[int][]int{} for _, pair := range dislikes { n1, n2 := pair[0], pair[1] m[n1] = append(m[n1], n2) m[n2] = append(m[n2], n1) } types := make([]int, N+1) for i:=1; i<=N; i++ { if types[i] != 0 { continue } // put current node to one set types[i] = 1 queue := []int{i} for len(queue) != 0 { l := []int{} for _, node := range queue { // get dislikes for _, node2 := range m[node] { if types[node2] == types[node] { return false } else { if types[node2] == 0 { types[node2] = -types[node] l = append(l, node2) } } } } queue = l } } return true }