# Leetcode Algorithms

1042. Flower Planting With No Adjacent

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.

Logic:

create an empty res list, a graph list, and a neighbor_colors list.

This is the graph problem. I am not very familiar with this.

Solution:

`class Solution:`

def gardenNoAdj(self, N: int, paths: List[List[int]]) -> 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): # only 4 colors

if color in neighbor_colors:

continue

res[i] = color

break

return res