<p>You have <code>n</code> gardens, labeled from <code>1</code> to <code>n</code>, and an array <code>paths</code> where <code>paths[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> describes a bidirectional path between garden <code>x<sub>i</sub></code> to garden <code>y<sub>i</sub></code>. In each garden, you want to plant one of 4 types of flowers.</p>

<p>All gardens have <strong>at most 3</strong> paths coming into or leaving it.</p>

<p>Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.</p>

<p>Return <em><strong>any</strong> such a choice as an array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> is the type of flower planted in the </em><code>(i+1)<sup>th</sup></code><em> garden. The flower types are denoted </em><code>1</code><em>, </em><code>2</code><em>, </em><code>3</code><em>, or </em><code>4</code><em>. It is guaranteed an answer exists.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, paths = [[1,2],[2,3],[3,1]]
<strong>Output:</strong> [1,2,3]
<strong>Explanation:</strong>
Gardens 1 and 2 have different types.
Gardens 2 and 3 have different types.
Gardens 3 and 1 have different types.
Hence, [1,2,3] is a valid answer. Other valid answers include [1,2,4], [1,4,2], and [3,2,1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, paths = [[1,2],[3,4]]
<strong>Output:</strong> [1,2,1,2]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
<strong>Output:</strong> [1,2,3,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= paths.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>paths[i].length == 2</code></li>
	<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= n</code></li>
	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
	<li>Every garden has <strong>at most 3</strong> paths coming into or leaving it.</li>
</ul>

class Solution {
public:
    void bfs(int strt, vector<int> &ans, queue<int> &q,vector<vector<int>> &graph){
        // 3. push starting
        q.push(strt);

        //4. start bfs
        while(!q.empty()){
            int curr=q.front();
            q.pop();

            int flowers_used[5]={0}; //to check for flowers already used
            for(auto adjnodes: graph[curr]){
                if(ans[adjnodes]!=0){
                    flowers_used[ans[adjnodes]]=1;
                }
                else{
                    q.push(adjnodes); //if no flower push them for further use
                }
            }

            for(int i=1;i<5;i++){
                //if this flower not used use it
                //always we will find atleast one not used as maximum 3 connections 
                //hence max 3 will be used so one will be left
                if(flowers_used[i]==0){
                    ans[curr]=i;
                }
            }

        }
        return;
    }
    vector<int> gardenNoAdj(int n, vector<vector<int>>& paths) {
        //cann use bfs or dfs, 
        //pahle se placed kisi k equal nhi hona chahiye bas 

        //1. make graph
        vector<vector<int>> graph(n+1);
        for(auto it: paths){
            graph[it[0]].push_back(it[1]);
            graph[it[1]].push_back(it[0]);
        }

        //2. make required components
        queue<int> q;
        vector<int> ans(n+1,0); //it will work as visited also

        // graph can be disconnected hence do bfs on all
        for(int i=1;i<n+1;i++){
            if(ans[i]==0){
                bfs(i,ans,q,graph);
            }
        }
       
        ans.erase(ans.begin());
        return ans;


    }
};
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