# Solving Leetcode Interviews in Seconds with AI: Flower Planting With No Adjacent


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1042" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's a solution to the garden flower planting problem:

*   **Greedy Coloring:** Iterate through each garden and assign it the smallest available flower type (1-4) that is not already used by its neighboring gardens.
*   **Adjacency List:** Represent the connections between gardens using an adjacency list for efficient neighbor lookups.
*   **Iterative Assignment:** Assign flower types iteratively, ensuring no adjacent gardens have the same flower.

*   **Time Complexity:** O(n + m), where n is the number of gardens and m is the number of paths.
*   **Space Complexity:** O(n + m), primarily for the adjacency list and the result array.

	
	# Code
	```python
	def gardenNoAdj(n: int, paths: list[list[int]]) -> list[int]:
    """
    Assigns flower types to gardens such that no adjacent gardens have the same flower type.

    Args:
        n: The number of gardens.
        paths: A list of bidirectional paths between gardens.

    Returns:
        A list representing the flower type assigned to each garden.
    """

    adj_list = [[] for _ in range(n)]
    for u, v in paths:
        adj_list[u - 1].append(v - 1)
        adj_list[v - 1].append(u - 1)

    result = [0] * n
    for i in range(n):
        used_colors = set()
        for neighbor in adj_list[i]:
            if result[neighbor] != 0:
                used_colors.add(result[neighbor])

        for color in range(1, 5):
            if color not in used_colors:
                result[i] = color
                break

    return result
	```