You have n gardens, labeled from 1 to n, and an array paths where paths[i] = [xi, yi] describes the existence of a bidirectional path from garden xi to garden yi. In each garden, you want to plant one of 4 types of flowers.
There is no garden that has more than three paths coming into or leaving it.
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.
Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.
Example 1:
Input: n = 3, paths = [[1,2],[2,3],[3,1]]
Output: [1,2,3]
Example 2:
Input: n = 4, paths = [[1,2],[3,4]]
Output: [1,2,1,2]
Example 3:
Input: n = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
Output: [1,2,3,4]
Constraints:
1 <= n <= 104
0 <= paths.length <= 2 * 104
paths[i].length == 2
1 <= xi, yi <= n
xi != yi
No garden has four or more paths coming into or leaving it.
public class Solution {
Dictionary<int,List<int>> graph = new Dictionary<int,List<int>>();
HashSet<int> visited = new HashSet<int>();
private void AddVertex(int v){
graph.Add(v,new List<int>());
}
private void AddEdge(int s, int d){
var sEdges = graph[s];
var dEdges = graph[d];
sEdges.Add(d);
dEdges.Add(s);
graph[s]=sEdges;
graph[d]=dEdges;
}
public int[] GardenNoAdj(int n, int[][] paths) {
for(int i=1;i<=n;i++){
AddVertex(i);
}
for(int i=0;i<paths.Length;i++){
AddEdge(paths[i][0],paths[i][1]);
}
var ans = new int[n];
for(int i=1;i<=n;i++){
var colors = new int[5];
foreach(var item in graph[i]){
colors[ans[item-1]]=1;
}
for(int j=4;j>=1;j--){
if(colors[j]!=1){
ans[i-1]=j;
}
}
}
return ans;
}
}
t
Time Complexity: O(V.E)
Space Complexity: O(V)
Where V and E are the vertices and Edges of graph.