1042. Flower Planting With No Adjacent

Description

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.

Solutions

• class Solution {
public int[] gardenNoAdj(int n, int[][] paths) {
List<Integer>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (var p : paths) {
int x = p[0] - 1, y = p[1] - 1;
}
int[] ans = new int[n];
boolean[] used = new boolean[5];
for (int x = 0; x < n; ++x) {
Arrays.fill(used, false);
for (int y : g[x]) {
used[ans[y]] = true;
}
for (int c = 1; c < 5; ++c) {
if (!used[c]) {
ans[x] = c;
break;
}
}
}
return ans;
}
}

• class Solution {
public:
vector<int> gardenNoAdj(int n, vector<vector<int>>& paths) {
vector<vector<int>> g(n);
for (auto& p : paths) {
int x = p[0] - 1, y = p[1] - 1;
g[x].push_back(y);
g[y].push_back(x);
}
vector<int> ans(n);
bool used[5];
for (int x = 0; x < n; ++x) {
memset(used, false, sizeof(used));
for (int y : g[x]) {
used[ans[y]] = true;
}
for (int c = 1; c < 5; ++c) {
if (!used[c]) {
ans[x] = c;
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 x in range(n):
used = {ans[y] for y in g[x]}
for c in range(1, 5):
if c not in used:
ans[x] = c
break
return ans


• 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 x := 0; x < n; x++ {
used := [5]bool{}
for _, y := range g[x] {
used[ans[y]] = true
}
for c := 1; c < 5; c++ {
if !used[c] {
ans[x] = 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;
}