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Abhishek Chaudhary
Abhishek Chaudhary

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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()
            visited.add(curr)
            colors = set()
            for j in graph.get(curr, []):
                colors.add(types[j])
                if j not in visited:
                    stack.append(j)
            if types[curr] in colors:
                for i in range(1, n + 1):
                    if i not in colors:
                        types[curr] = i
                        break
        return types
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