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Formatted question description: https://leetcode.ca/all/886.html

# 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/
// 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/
// 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;
}
};

• class Solution {
public boolean possibleBipartition(int N, int[][] dislikes) {
int[] degrees = new int[N];
Map<Integer, Set<Integer>> dislikesMap = new HashMap<Integer, Set<Integer>>();
for (int[] dislike : dislikes) {
int person1 = dislike[0] - 1, person2 = dislike[1] - 1;
degrees[person1]++;
degrees[person2]++;
Set<Integer> dislikesSet1 = dislikesMap.getOrDefault(person1, new HashSet<Integer>());
Set<Integer> dislikesSet2 = dislikesMap.getOrDefault(person2, new HashSet<Integer>());
dislikesMap.put(person1, dislikesSet1);
dislikesMap.put(person2, dislikesSet2);
}
int[] groups = new int[N];
for (int i = 0; i < N; i++) {
if (degrees[i] > 0 && groups[i] == 0) {
groups[i] = 1;
boolean possible = breadthFirstSearch(dislikesMap, groups, i);
if (!possible)
return false;
}
}
return true;
}

public boolean breadthFirstSearch(Map<Integer, Set<Integer>> dislikesMap, int[] groups, int start) {
queue.offer(start);
while (!queue.isEmpty()) {
int person = queue.poll();
int group = groups[person];
int newGroup = 3 - group;
Set<Integer> dislikes = dislikesMap.getOrDefault(person, new HashSet<Integer>());
for (int dislike : dislikes) {
if (groups[dislike] == group)
return false;
else if (groups[dislike] == 0) {
groups[dislike] = newGroup;
queue.offer(dislike);
}
}
}
return true;
}
}

############

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;
}
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];
}
}

• // OJ: https://leetcode.com/problems/possible-bipartition/
// 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;
}
};

• 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

############

class Solution(object):
def possibleBipartition(self, N, dislikes):
"""
:type N: int
:type dislikes: List[List[int]]
:rtype: bool
"""
graph = collections.defaultdict(list)
for dislike in dislikes:
graph[dislike[0] - 1].append(dislike[1] - 1)
graph[dislike[1] - 1].append(dislike[0] - 1)
color = [0] * N
for i in range(N):
if color[i] != 0: continue
bfs = collections.deque()
bfs.append(i)
color[i] = 1
while bfs:
cur = bfs.popleft()
for e in graph[cur]:
if color[e] != 0:
if color[cur] == color[e]:
return False
else:
color[e] = -color[cur]
bfs.append(e)
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
}
}