# 785.Is-Graph-Bipartite?

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/is-graph-bipartite/

## ้ข็ฎๆ่ฟฐ

``````Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists.  Each node is an integer between 0 and graph.length - 1.  There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation:
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

Note:
graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].``````

## ไปฃ็ 

### Approach #1 Coloring by DFS

Complexity Analysis

• Time Complexity: O(N + E), where NN is the number of nodes in the graph, and E is the number of edges. We explore each node once when we transform it from uncolored to colored, traversing all its edges in the process.

• Space Complexity: O(N), the space used to store the `color`.

``````class Solution {
public boolean isBipartite(int[][] graph) {
int n = graph.length;
int[] color = new int[n];
Arrays.fill(color, -1);

for (int start = 0; start < n; start++) {
if (color[start] == -1) {
Stack<Integer> stack = new Stack();
stack.push(start);
// color[start] = 0;    // whatever, It doesn't matter whether you add it or not

while (!stack.isEmpty()) {
Integer node = stack.pop();
for (int nei: graph[node]) {
if (color[nei] == -1) {
stack.push(nei);
color[nei] = color[node] ^ 1; // -1 ^ 1 = -2, -2 ^ 1 = -1
} else if (color[nei] == color[node]) {
return false;
}
}
}
}
}

return true;
}
}

class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
vector<int> colors(graph.size());
for (int i = 0; i < graph.size(); ++i) {
if (colors[i] == 0 && !valid(graph, 1, i, colors)) {
return false;
}
}
return true;
}
bool valid(vector<vector<int>>& graph, int color, int cur, vector<int>& colors) {
if (colors[cur] != 0) return colors[cur] == color;
colors[cur] = color;
for (int i : graph[cur]) {
if (!valid(graph, -1 * color, i, colors)) {
return false;
}
}
return true;
}
};``````

### Approach #2 BFS

``````class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
vector<int> colors(graph.size());
for (int i = 0; i < graph.size(); ++i) {
if (colors[i] != 0) continue;
colors[i] = 1;
queue<int> q{{i}};
while (!q.empty()) {
int t = q.front(); q.pop();
for (auto a : graph[t]) {
if (colors[a] == colors[t]) return false;
if (colors[a] == 0) {
colors[a] = -1 * colors[t];
q.push(a);
}
}
}
}
return true;
}
};``````

### Approch #3 Union Find

``````class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
vector<int> root(graph.size());
for (int i = 0; i < graph.size(); ++i) root[i] = i;
for (int i = 0; i < graph.size(); ++i) {
if (graph[i].empty()) continue;
int x = find(root, i), y = find(root, graph[i][0]);
if (x == y) return false;
for (int j = 1; j < graph[i].size(); ++j) {
int parent = find(root, graph[i][j]);
if (x == parent) return false;
root[parent] = y;
}
}
return true;
}
int find(vector<int>& root, int i) {
return root[i] == i ? i : find(root, root[i]);
}
};``````

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