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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/
// Author: github.com/lzl124631x
// 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;
    }
};