LeetCode #785 — MEDIUM

Is Graph Bipartite?

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

The Problem

Problem Statement

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:

There are no self-edges (graph[u] does not contain u).
There are no parallel edges (graph[u] does not contain duplicate values).
If v is in graph[u], then u is in graph[v] (the graph is undirected).
The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.

Return true if and only if it is bipartite.

Example 1:

Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.

Example 2:

Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.

Constraints:

graph.length == n
1 <= n <= 100
0 <= graph[u].length < n
0 <= graph[u][i] <= n - 1
graph[u] does not contain u.
All the values of graph[u] are unique.
If graph[u] contains v, then graph[v] contains u.

Step 01

Brute Force Baseline

Problem summary: There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties: There are no self-edges (graph[u] does not contain u). There are no parallel edges (graph[u] does not contain duplicate values). If v is in graph[u], then u is in graph[v] (the graph is undirected). The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them. A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B. Return true if and only if it is bipartite.

Baseline thinking

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

Pattern signal: Union-Find

Example 1

[[1,2,3],[0,2],[0,1,3],[0,2]]

Example 2

[[1,3],[0,2],[1,3],[0,2]]

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 #785: Is Graph Bipartite?
class Solution {
    private int[] color;
    private int[][] g;

    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        color = new int[n];
        g = graph;
        for (int i = 0; i < n; ++i) {
            if (color[i] == 0 && !dfs(i, 1)) {
                return false;
            }
        }
        return true;
    }

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

// Accepted solution for LeetCode #785: Is Graph Bipartite?
func isBipartite(graph [][]int) bool {
	n := len(graph)
	color := make([]int, n)
	var dfs func(int, int) bool
	dfs = func(a, c int) bool {
		color[a] = c
		for _, b := range graph[a] {
			if color[b] == c || (color[b] == 0 && !dfs(b, -c)) {
				return false
			}
		}
		return true
	}
	for i := range graph {
		if color[i] == 0 && !dfs(i, 1) {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #785: Is Graph Bipartite?
class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        def dfs(a: int, c: int) -> bool:
            color[a] = c
            for b in graph[a]:
                if color[b] == c or (color[b] == 0 and not dfs(b, -c)):
                    return False
            return True

        n = len(graph)
        color = [0] * n
        for i in range(n):
            if color[i] == 0 and not dfs(i, 1):
                return False
        return True

// Accepted solution for LeetCode #785: Is Graph Bipartite?
impl Solution {
    pub fn is_bipartite(graph: Vec<Vec<i32>>) -> bool {
        let n = graph.len();
        let mut color = vec![0; n];

        fn dfs(a: usize, c: i32, graph: &Vec<Vec<i32>>, color: &mut Vec<i32>) -> bool {
            color[a] = c;
            for &b in &graph[a] {
                if color[b as usize] == c
                    || (color[b as usize] == 0 && !dfs(b as usize, -c, graph, color))
                {
                    return false;
                }
            }
            true
        }

        for i in 0..n {
            if color[i] == 0 && !dfs(i, 1, &graph, &mut color) {
                return false;
            }
        }
        true
    }
}

// Accepted solution for LeetCode #785: Is Graph Bipartite?
function isBipartite(graph: number[][]): boolean {
    const n = graph.length;
    const color: number[] = Array(n).fill(0);
    const dfs = (a: number, c: number): boolean => {
        color[a] = c;
        for (const b of graph[a]) {
            if (color[b] === c || (color[b] === 0 && !dfs(b, -c))) {
                return false;
            }
        }
        return true;
    };
    for (let i = 0; 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(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.