785. Is Graph Bipartite?

Problem

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.

Solution (Java)

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

    private boolean helper(int[][] graph, int curr, int c, int[] color) {
        if (color[curr] == c) {
            return true;
        }
        color[curr] = c;
        for (int x : graph[curr]) {
            if (color[x] == c || !helper(graph, x, c * -1, color)) {
                return false;
            }
        }
        return true;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).