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Abhishek Chaudhary

Posted on

# Flower Planting With No Adjacent

You have `n` gardens, labeled from `1` to `n`, and an array `paths` where `paths[i] = [xi, yi]` describes a bidirectional path between garden `xi` to garden `yi`. In each garden, you want to plant one of 4 types of flowers.

All gardens have at most 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]
Explanation:
Gardens 1 and 2 have different types.
Gardens 2 and 3 have different types.
Gardens 3 and 1 have different types.
Hence, [1,2,3] is a valid answer. Other valid answers include [1,2,4], [1,4,2], and [3,2,1].

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]

Constraints:

• `1 <= n <= 104`
• `0 <= paths.length <= 2 * 104`
• `paths[i].length == 2`
• `1 <= xi, yi <= n`
• `xi != yi`
• Every garden has at most 3 paths coming into or leaving it.

SOLUTION:

``````class Solution:
def gardenNoAdj(self, n: int, paths: List[List[int]]) -> List[int]:
types = [1] * n
graph = {}
for a, b in paths:
graph[a - 1] = graph.get(a - 1, []) + [b - 1]
graph[b - 1] = graph.get(b - 1, []) + [a - 1]
visited = set()
stack = list(range(n))
while len(stack) > 0:
curr = stack.pop()
colors = set()
for j in graph.get(curr, []):