Flower Planting With No Adjacent

Problem

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 , whereanswer[i]is the type of flower planted in the(i+1)th garden. The flower types are denoted1 ,2 ,3 , or4 . It is guaranteed an answer exists.

Examples

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 <= 10^4
• 0 <= paths.length <= 2 * 10^4
• paths[i].length == 2
• 1 <= xi, yi <= n
• xi != yi
• Every garden has at most 3 paths coming into or leaving it.

Solution

Method 1 – Greedy Coloring (Graph Coloring)

Intuition

Since each garden has at most 3 neighbors and there are 4 flower types, we can always assign a flower type to each garden such that no two adjacent gardens have the same type. This is a classic greedy coloring problem.

Approach

1. Build an adjacency list for all gardens.
2. For each garden from 1 to n:
   • Check the flower types used by its neighbors.
   • Assign the smallest flower type (1-4) not used by its neighbors.
3. Return the flower types for all gardens.

Code

C++

class Solution {
public:
    vector<int> gardenNoAdj(int n, vector<vector<int>>& paths) {
        vector<vector<int>> g(n);
        for (auto& p : paths) {
            g[p[0]-1].push_back(p[1]-1);
            g[p[1]-1].push_back(p[0]-1);
        }
        vector<int> ans(n, 0);
        for (int i = 0; i < n; ++i) {
            bool used[5] = {};
            for (int v : g[i]) used[ans[v]] = true;
            for (int c = 1; c <= 4; ++c) {
                if (!used[c]) {
                    ans[i] = c;
                    break;
                }
            }
        }
        return ans;
    }
};

Go

type Solution struct{}
func (Solution) gardenNoAdj(n int, paths [][]int) []int {
    g := make([][]int, n)
    for _, p := range paths {
        g[p[0]-1] = append(g[p[0]-1], p[1]-1)
        g[p[1]-1] = append(g[p[1]-1], p[0]-1)
    }
    ans := make([]int, n)
    for i := 0; i < n; i++ {
        used := [5]bool{}
        for _, v := range g[i] {
            used[ans[v]] = true
        }
        for c := 1; c <= 4; c++ {
            if !used[c] {
                ans[i] = c
                break
            }
        }
    }
    return ans
}

Java

class Solution {
    public int[] gardenNoAdj(int n, int[][] paths) {
        List<Integer>[] g = new List[n];
        for (int i = 0; i < n; i++) g[i] = new ArrayList<>();
        for (int[] p : paths) {
            g[p[0]-1].add(p[1]-1);
            g[p[1]-1].add(p[0]-1);
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            boolean[] used = new boolean[5];
            for (int v : g[i]) used[ans[v]] = true;
            for (int c = 1; c <= 4; c++) {
                if (!used[c]) {
                    ans[i] = c;
                    break;
                }
            }
        }
        return ans;
    }
}

Kotlin

class Solution {
    fun gardenNoAdj(n: Int, paths: Array<IntArray>): IntArray {
        val g = Array(n) { mutableListOf<Int>() }
        for (p in paths) {
            g[p[0]-1].add(p[1]-1)
            g[p[1]-1].add(p[0]-1)
        }
        val ans = IntArray(n)
        for (i in 0 until n) {
            val used = BooleanArray(5)
            for (v in g[i]) used[ans[v]] = true
            for (c in 1..4) {
                if (!used[c]) {
                    ans[i] = c
                    break
                }
            }
        }
        return ans
    }
}

Python

class Solution:
    def gardenNoAdj(self, n: int, paths: list[list[int]]) -> list[int]:
        g = [[] for _ in range(n)]
        for x, y in paths:
            g[x-1].append(y-1)
            g[y-1].append(x-1)
        ans = [0] * n
        for i in range(n):
            used = {ans[v] for v in g[i]}
            for c in range(1, 5):
                if c not in used:
                    ans[i] = c
                    break
        return ans

Rust

impl Solution {
    pub fn garden_no_adj(n: i32, paths: Vec<Vec<i32>>) -> Vec<i32> {
        let n = n as usize;
        let mut g = vec![vec![]; n];
        for p in &paths {
            g[(p[0]-1) as usize].push((p[1]-1) as usize);
            g[(p[1]-1) as usize].push((p[0]-1) as usize);
        }
        let mut ans = vec![0; n];
        for i in 0..n {
            let mut used = [false; 5];
            for &v in &g[i] {
                used[ans[v] as usize] = true;
            }
            for c in 1..=4 {
                if !used[c] {
                    ans[i] = c as i32;
                    break;
                }
            }
        }
        ans
    }
}

TypeScript

class Solution {
  gardenNoAdj(n: number, paths: number[][]): number[] {
    const g: number[][] = Array.from({length: n}, () => []);
    for (const [x, y] of paths) {
      g[x-1].push(y-1);
      g[y-1].push(x-1);
    }
    const ans: number[] = Array(n).fill(0);
    for (let i = 0; i < n; i++) {
      const used = new Set<number>();
      for (const v of g[i]) used.add(ans[v]);
      for (let c = 1; c <= 4; c++) {
        if (!used.has(c)) {
          ans[i] = c;
          break;
        }
      }
    }
    return ans;
  }
}

Complexity

• ⏰ Time complexity: O(n), since each garden and its neighbors are processed once.
• 🧺 Space complexity: O(n + m), where m is the number of paths, for the adjacency list and answer array.