# 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;
}
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
}
}