LeetCode #886 — MEDIUM

Possible Bipartition

Move from brute-force thinking to an efficient approach using union-find strategy.

The Problem

Problem Statement

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] = [a_i, b_i] indicates that the person labeled a_i does not like the person labeled b_i, 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 <= a_i < b_i <= n
All the pairs of dislikes are unique.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Union-Find

Example 1

4
[[1,2],[1,3],[2,4]]

Example 2

3
[[1,2],[1,3],[2,3]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #886: Possible Bipartition
class Solution {
    private List<Integer>[] g;
    private int[] color;

    public boolean possibleBipartition(int n, int[][] dislikes) {
        g = new List[n];
        color = new int[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : dislikes) {
            int a = e[0] - 1, b = e[1] - 1;
            g[a].add(b);
            g[b].add(a);
        }
        for (int i = 0; i < n; ++i) {
            if (color[i] == 0) {
                if (!dfs(i, 1)) {
                    return false;
                }
            }
        }
        return true;
    }

    private boolean dfs(int i, int c) {
        color[i] = c;
        for (int j : g[i]) {
            if (color[j] == c) {
                return false;
            }
            if (color[j] == 0 && !dfs(j, 3 - c)) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #886: Possible Bipartition
func possibleBipartition(n int, dislikes [][]int) bool {
	g := make([][]int, n)
	for _, e := range dislikes {
		a, b := e[0]-1, e[1]-1
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	color := make([]int, n)
	var dfs func(int, int) bool
	dfs = func(i, c int) bool {
		color[i] = c
		for _, j := range g[i] {
			if color[j] == c {
				return false
			}
			if color[j] == 0 && !dfs(j, 3-c) {
				return false
			}
		}
		return true
	}
	for i, c := range color {
		if c == 0 && !dfs(i, 1) {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #886: Possible Bipartition
class Solution:
    def possibleBipartition(self, n: int, dislikes: List[List[int]]) -> bool:
        def dfs(i, c):
            color[i] = c
            for j in g[i]:
                if color[j] == c:
                    return False
                if color[j] == 0 and not dfs(j, 3 - c):
                    return False
            return True

        g = defaultdict(list)
        color = [0] * n
        for a, b in dislikes:
            a, b = a - 1, b - 1
            g[a].append(b)
            g[b].append(a)
        return all(c or dfs(i, 1) for i, c in enumerate(color))

// Accepted solution for LeetCode #886: Possible Bipartition
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
    }
}

// Accepted solution for LeetCode #886: Possible Bipartition
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(α(n))

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.