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Formatted question description: https://leetcode.ca/all/1042.html

# 1042. Flower Planting With No Adjacent (Easy)

You have N gardens, labelled 1 to N.  In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 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]


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]


Note:

• 1 <= N <= 10000
• 0 <= paths.size <= 20000
• No garden has 4 or more paths coming into or leaving it.
• It is guaranteed an answer exists.

Related Topics:
Graph

## Solution 1.

• class Solution {
public int[] gardenNoAdj(int N, int[][] paths) {
int[] flowers = new int[N];
Map<Integer, List<Integer>> adjacentMap = new HashMap<Integer, List<Integer>>();
for (int[] path : paths) {
int garden1 = path[0] - 1, garden2 = path[1] - 1;
List<Integer> pathList1 = adjacentMap.getOrDefault(garden1, new ArrayList<Integer>());
List<Integer> pathList2 = adjacentMap.getOrDefault(garden2, new ArrayList<Integer>());
}
flowers[0] = 1;
for (int i = 1; i < N; i++) {
boolean[] used = new boolean[5];
for (int garden : adjacent) {
int flower = flowers[garden];
if (flower > 0)
used[flower] = true;
}
for (int j = 1; j < 5; j++) {
if (!used[j]) {
flowers[i] = j;
break;
}
}
}
return flowers;
}
}

############

class Solution {
public int[] gardenNoAdj(int n, int[][] paths) {
List<Integer>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (int[] p : paths) {
int x = p[0] - 1, y = p[1] - 1;
}
int[] ans = new int[n];
for (int u = 0; u < n; ++u) {
Set<Integer> colors = new HashSet<>();
for (int v : g[u]) {
}
for (int c = 1; c < 5; ++c) {
if (!colors.contains(c)) {
ans[u] = c;
break;
}
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/flower-planting-with-no-adjacent/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
vector<vector<int>> G(N);
for (auto & e : paths) {
G[e[0] - 1].push_back(e[1] - 1);
G[e[1] - 1].push_back(e[0] - 1);
}
vector<int> ans(N);
for (int i = 0; i < N; ++i) {
int color[5] = {};
for (int j : G[i]) color[ans[j]] = 1;
for (int j = 4; j > 0; --j) {
if (color[j]) continue;
ans[i] = j;
break;
}
}
return ans;
}
};

• class Solution:
def gardenNoAdj(self, n: int, paths: List[List[int]]) -> List[int]:
g = defaultdict(list)
for x, y in paths:
x, y = x - 1, y - 1
g[x].append(y)
g[y].append(x)
ans = [0] * n
for u in range(n):
colors = set(ans[v] for v in g[u])
for c in range(1, 5):
if c not in colors:
ans[u] = c
break
return ans

############

class Solution(object):
def gardenNoAdj(self, N, paths):
"""
:type N: int
:type paths: List[List[int]]
:rtype: List[int]
"""
res = [0] * N
graph = [[] for i in range(N)]
for path in paths:
graph[path[0] - 1].append(path[1] - 1)
graph[path[1] - 1].append(path[0] - 1)
for i in range(N):
neighbor_colors = []
for neighbor in graph[i]:
neighbor_colors.append(res[neighbor])
for color in range(1, 5):
if color in neighbor_colors:
continue
res[i] = color
break
return res

• func gardenNoAdj(n int, paths [][]int) []int {
g := make([][]int, n)
for _, p := range paths {
x, y := p[0]-1, p[1]-1
g[x] = append(g[x], y)
g[y] = append(g[y], x)
}
ans := make([]int, n)
for u := 0; u < n; u++ {
colors := make(map[int]bool)
for _, v := range g[u] {
colors[ans[v]] = true
}
for c := 1; c < 5; c++ {
if !colors[c] {
ans[u] = c
break
}
}
}
return ans
}

• function gardenNoAdj(n: number, paths: number[][]): number[] {
const g: number[][] = new Array(n).fill(0).map(() => []);
for (const [x, y] of paths) {
g[x - 1].push(y - 1);
g[y - 1].push(x - 1);
}
const ans: number[] = new Array(n).fill(0);
for (let x = 0; x < n; ++x) {
const used: boolean[] = new Array(5).fill(false);
for (const y of g[x]) {
used[ans[y]] = true;
}
for (let c = 1; c < 5; ++c) {
if (!used[c]) {
ans[x] = c;
break;
}
}
}
return ans;
}