LeetCode in Kotlin

1042. Flower Planting With No Adjacent

Medium

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.

Solution

class Solution {
    private lateinit var graph: Array<ArrayList<Int>?>
    private lateinit var color: IntArray
    private lateinit var visited: BooleanArray

    fun gardenNoAdj(n: Int, paths: Array<IntArray>): IntArray {
        buildGraph(n, paths)
        color = IntArray(n)
        visited = BooleanArray(n)
        for (i in 0 until n) {
            if (!visited[i]) {
                dfs(i)
            }
        }
        return color
    }

    private fun dfs(at: Int) {
        visited[at] = true
        var used = 0
        for (to in graph[at]!!) {
            if (color[to] != 0) {
                used = used or (1 shl color[to] - 1)
            }
        }

        // use available color
        for (i in 0..3) {
            if (used and (1 shl i) == 0) {
                color[at] = i + 1
                break
            }
        }
    }

    private fun buildGraph(n: Int, paths: Array<IntArray>) {
        graph = arrayOfNulls(n)
        for (i in 0 until n) {
            graph[i] = ArrayList()
        }
        for (path in paths) {
            val u = path[0] - 1
            val v = path[1] - 1
            graph[u]!!.add(v)
            graph[v]!!.add(u)
        }
    }
}

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