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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>());
            pathList1.add(garden2);
            adjacentMap.put(garden1, pathList1);
            List<Integer> pathList2 = adjacentMap.getOrDefault(garden2, new ArrayList<Integer>());
            pathList2.add(garden1);
            adjacentMap.put(garden2, pathList2);
        }
        flowers[0] = 1;
        for (int i = 1; i < N; i++) {
            List<Integer> adjacent = adjacentMap.getOrDefault(i, new ArrayList<Integer>());
            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;
            g[x].add(y);
            g[y].add(x);
        }
        int[] ans = new int[n];
        for (int u = 0; u < n; ++u) {
            Set<Integer> colors = new HashSet<>();
            for (int v : g[u]) {
                colors.add(ans[v]);
            }
            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;
}

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