How do you solve Flower Planting With No Adjacent on LeetCode?

Flower Planting With No Adjacent (LeetCode #1042) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each garden can have at most 3 neighbors, ensuring at least one flower type is always available.

What is the brute force solution for Flower Planting With No Adjacent?

The brute force approach for Flower Planting With No Adjacent is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we try to assign flower types to each garden one by one, checking each time if the current assignment is valid. This method is simple but inefficient, as it may require multiple attempts to find a valid configuration. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Flower Planting With No Adjacent?

Flower Planting With No Adjacent uses the following concepts: Depth-First Search, Breadth-First Search, Graph Theory. The recommended patterns to study are: Graph Traversal, Greedy Algorithms.

What are the interview tips for Flower Planting With No Adjacent?

Always start with understanding the problem constraints and how they affect your approach. Think about how you can use data structures like arrays or lists to keep track of used flower types. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Flower Planting With No Adjacent?

Overcomplicating the problem by trying to use backtracking unnecessarily. Failing to recognize that with at most 3 neighbors, there will always be an available flower type.

#1042

Flower Planting With No Adjacent

Medium

Depth-First SearchBreadth-First SearchGraph TheoryGraph TraversalGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that each garden has at most 3 neighbors, allowing us to always find an available flower type. We can simply iterate through each garden and assign a flower that is not used by its neighbors.

⚙️

Algorithm

4 steps

1Step 1: Build an adjacency list to represent the gardens and their connections.
2Step 2: Initialize an array to store the flower types for each garden.
3Step 3: For each garden, determine which flower types are already used by its neighbors.
4Step 4: Assign the first available flower type to the current garden.

solution.py18 lines

1# Full working Python code
2class Solution:
3    def gardenNoAdj(self, n: int, paths: List[List[int]]) -> List[int]:
4        graph = [[] for _ in range(n)]
5        for x, y in paths:
6            graph[x-1].append(y-1)
7            graph[y-1].append(x-1)
8        answer = [0] * n
9        for i in range(n):
10            used = [False] * 5
11            for neighbor in graph[i]:
12                if answer[neighbor] != 0:
13                    used[answer[neighbor]] = True
14            for flower in range(1, 5):
15                if not used[flower]:
16                    answer[i] = flower
17                    break
18        return answer

ℹ

Complexity note: The time complexity is O(n) because we only iterate through each garden and its neighbors once, making it efficient. The space complexity is O(n) due to the adjacency list.

1Each garden can have at most 3 neighbors, ensuring at least one flower type is always available.
2Using a greedy approach allows us to efficiently assign flower types without backtracking.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.