Formatted question description: https://leetcode.ca/all/1042.html

1042. Flower Planting With No Adjacent (Easy)

You have N gardens, labelled 1 to N.  In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 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]

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]

 

Note:

  • 1 <= N <= 10000
  • 0 <= paths.size <= 20000
  • No garden has 4 or more paths coming into or leaving it.
  • It is guaranteed an answer exists.

Related Topics:
Graph

Solution 1.

// OJ: https://leetcode.com/problems/flower-planting-with-no-adjacent/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
        vector<vector<int>> G(N);
        for (auto & e : paths) {
            G[e[0] - 1].push_back(e[1] - 1);
            G[e[1] - 1].push_back(e[0] - 1);
        }
        vector<int> ans(N);
        for (int i = 0; i < N; ++i) {
            int color[5] = {};
            for (int j : G[i]) color[ans[j]] = 1;
            for (int j = 4; j > 0; --j) {
                if (color[j]) continue;
                ans[i] = j;
                break;
            }
        }
        return ans;
    }
};

Java

  • class Solution {
        public int[] gardenNoAdj(int N, int[][] paths) {
            int[] flowers = new int[N];
            Map<Integer, List<Integer>> adjacentMap = new HashMap<Integer, List<Integer>>();
            for (int[] path : paths) {
                int garden1 = path[0] - 1, garden2 = path[1] - 1;
                List<Integer> pathList1 = adjacentMap.getOrDefault(garden1, new ArrayList<Integer>());
                pathList1.add(garden2);
                adjacentMap.put(garden1, pathList1);
                List<Integer> pathList2 = adjacentMap.getOrDefault(garden2, new ArrayList<Integer>());
                pathList2.add(garden1);
                adjacentMap.put(garden2, pathList2);
            }
            flowers[0] = 1;
            for (int i = 1; i < N; i++) {
                List<Integer> adjacent = adjacentMap.getOrDefault(i, new ArrayList<Integer>());
                boolean[] used = new boolean[5];
                for (int garden : adjacent) {
                    int flower = flowers[garden];
                    if (flower > 0)
                        used[flower] = true;
                }
                for (int j = 1; j < 5; j++) {
                    if (!used[j]) {
                        flowers[i] = j;
                        break;
                    }
                }
            }
            return flowers;
        }
    }
    
  • // OJ: https://leetcode.com/problems/flower-planting-with-no-adjacent/
    // Time: O(N)
    // Space: O(N)
    class Solution {
    public:
        vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
            vector<vector<int>> G(N);
            for (auto & e : paths) {
                G[e[0] - 1].push_back(e[1] - 1);
                G[e[1] - 1].push_back(e[0] - 1);
            }
            vector<int> ans(N);
            for (int i = 0; i < N; ++i) {
                int color[5] = {};
                for (int j : G[i]) color[ans[j]] = 1;
                for (int j = 4; j > 0; --j) {
                    if (color[j]) continue;
                    ans[i] = j;
                    break;
                }
            }
            return ans;
        }
    };
    
  • class Solution(object):
        def gardenNoAdj(self, N, paths):
            """
            :type N: int
            :type paths: List[List[int]]
            :rtype: List[int]
            """
            res = [0] * N
            graph = [[] for i in range(N)]
            for path in paths:
                graph[path[0] - 1].append(path[1] - 1)
                graph[path[1] - 1].append(path[0] - 1)
            for i in range(N):
                neighbor_colors = []
                for neighbor in graph[i]:
                    neighbor_colors.append(res[neighbor])
                for color in range(1, 5):
                    if color in neighbor_colors:
                        continue
                    res[i] = color
                    break
            return res
    

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