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by Sumit Chourasia | Oct 15, 2020 | Category :coding | Tags : algorithm data-structure easy graph leetcode

Flower Planting With No Adjacent - Graph - Easy - LeetCode

Flower Planting With No Adjacent - Graph - Easy - LeetCode

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.

 

Contributed By: Sumit Chourasia
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