1042. Flower Planting With No Adjacent

Problem

You have n gardens, labeled from 1 to n, and an array paths where paths[i] = [xi, yi] describes a bidirectional path between garden xi to garden yi. In each garden, you want to plant one of 4 types of flowers.

All gardens have at most 3 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]
Explanation:
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].

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 <= 10^4

• 0 <= paths.length <= 2 * 10^4

• paths[i].length == 2

• 1 <= xi, yi <= n

• xi != yi

• Every garden has at most 3 paths coming into or leaving it.

Solution (Java)

class Solution {
    private List<Integer>[] graph;
    private int[] color;
    private boolean[] visited;

    public int[] gardenNoAdj(int n, int[][] paths) {
        buildGraph(n, paths);
        this.color = new int[n];
        this.visited = new boolean[n];

        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                dfs(i);
            }
        }

        return color;
    }

    private void dfs(int at) {
        visited[at] = true;
        int used = 0;

        for (int to : graph[at]) {
            if (color[to] != 0) {
                used |= 1 << color[to] - 1;
            }
        }

        // use available color
        for (int i = 0; i < 4; i++) {
            if ((used & 1 << i) == 0) {
                color[at] = i + 1;
                break;
            }
        }
    }

    private void buildGraph(int n, int[][] paths) {
        graph = new ArrayList[n];

        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }

        for (int[] path : paths) {
            int u = path[0] - 1;
            int v = path[1] - 1;
            graph[u].add(v);
            graph[v].add(u);
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).