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
Top comments (0)