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 - 1].push_back(e - 1);
G[e - 1].push_back(e - 1);
}
vector<int> ans(N);
for (int i = 0; i < N; ++i) {
int color = {};
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 - 1, garden2 = path - 1;
List<Integer> pathList1 = adjacentMap.getOrDefault(garden1, new ArrayList<Integer>());
List<Integer> pathList2 = adjacentMap.getOrDefault(garden2, new ArrayList<Integer>());
}
flowers = 1;
for (int i = 1; i < N; i++) {
boolean[] used = new boolean;
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 - 1].push_back(e - 1);
G[e - 1].push_back(e - 1);
}
vector<int> ans(N);
for (int i = 0; i < N; ++i) {
int color = {};
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):
"""
:type N: int
:type paths: List[List[int]]
:rtype: List[int]
"""
res =  * N
graph = [[] for i in range(N)]
for path in paths:
graph[path - 1].append(path - 1)
graph[path - 1].append(path - 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