Algorithm
Problem Name: 1042. Flower Planting With No Adjacent
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 <= 1040 <= paths.length <= 2 * 104paths[i].length == 21 <= xi, yi <= nxi != yi- Every garden has at most 3 paths coming into or leaving it.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const gardenNoAdj = function(N, paths) {
const map = {};
for (let i = 0; i < N; i++) {
map[i] = [];
}
for (let path of paths) {
let l = path[0] - 1;
let r = path[1] - 1;
map[l].push(r);
map[r].push(l);
}
const result = new Array(N).fill(-1);
for (let i = 0; i < N; i++) {
let colors = new Array(4).fill(false);
for (let neighbor of map[i]) {
if (result[neighbor] !== -1) {
colors[result[neighbor]] = true;
}
}
for (let j = 0; j < colors.length; j++) {
if (!colors[j]) {
result[i] = j;
break;
}
}
}
return result.map(i => ++i);
};
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def gardenNoAdj(self, N: int, paths: List[List[int]]) -> List[int]:
res = [0] * N
G = [[] for i in range(N)]
for x, y in paths:
G[x - 1].append(y - 1)
G[y - 1].append(x - 1)
for i in range(N):
res[i] = ({1, 2, 3, 4} - {res[j] for j in G[i]}).pop()
return res
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _1042_FlowerPlantingWithNoAdjacent
{
public int[] GardenNoAdj(int N, int[][] paths)
{
var edges = new Dictionary < int, IList<int>>();
for (int i = 0; i < N; i++)
edges[i] = new List<int>();
foreach (var path in paths)
{
edges[path[0] - 1].Add(path[1] - 1);
edges[path[1] - 1].Add(path[0] - 1);
}
var results = new int[N];
for (int i = 0; i < N; i++)
{
var usedColor = new bool[5];
foreach (var v in edges[i])
usedColor[results[v]] = true;
for (int c = 1; c < = 4; c++)
if (!usedColor[c])
{
results[i] = c;
break;
}
}
return results;
}
}
}
Input
Output