785. Is Graph Bipartite?

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785. Is Graph Bipartite?

Description

Difficulty: Medium

Related Topics: Depth-First Search, Breadth-First Search, Union Find, Graph

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:

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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:

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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.

Hints/Notes

  • Traverse the graph, color the next node if not visited, or check if the color match

Solution

Language: C++

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class Solution {
public:
    vector<bool> color;
    vector<bool> visited;
    bool ok = true;

    bool isBipartite(vector<vector<int>>& graph) {
        int numNodes = graph.size();
        color = vector<bool>(numNodes, false);
        visited = vector<bool>(numNodes, false);

        for (int i = 0; i < numNodes; i++) {
            traverse(i, graph);
        }

        return ok;
    }

    void traverse(int node, vector<vector<int>>& graph) {
        visited[node] = true;

        for (auto w : graph[node]) {
            if (visited[w]) {
                // visited, should check the color
                if (color[w] == color[node]) {
                    ok = false;
                    break;
                }
            } else {
                color[w] = !color[node];
                traverse(w, graph);
            }
        }
    }
};