# 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 =  * 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): # only 4 colors                if color in neighbor_colors:                    continue                res[i] = color                break                        return res`