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

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