LeetCode in Kotlin

3203. Find Minimum Diameter After Merging Two Trees

Hard

There exist two undirected trees with n and m nodes, numbered from 0 to n - 1 and from 0 to m - 1, respectively. You are given two 2D integer arrays edges1 and edges2 of lengths n - 1 and m - 1, respectively, where edges1[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the first tree and edges2[i] = [u_i, v_i] indicates that there is an edge between nodes u_i and v_i in the second tree.

You must connect one node from the first tree with another node from the second tree with an edge.

Return the minimum possible diameter of the resulting tree.

The diameter of a tree is the length of the longest path between any two nodes in the tree.

Example 1:

Input: edges1 = [[0,1],[0,2],[0,3]], edges2 = [[0,1]]

Output: 3

Explanation:

We can obtain a tree of diameter 3 by connecting node 0 from the first tree with any node from the second tree.

Example 2:

Input: edges1 = [[0,1],[0,2],[0,3],[2,4],[2,5],[3,6],[2,7]], edges2 = [[0,1],[0,2],[0,3],[2,4],[2,5],[3,6],[2,7]]

Output: 5

Explanation:

We can obtain a tree of diameter 5 by connecting node 0 from the first tree with node 0 from the second tree.

Constraints:

1 <= n, m <= 10⁵
edges1.length == n - 1
edges2.length == m - 1
edges1[i].length == edges2[i].length == 2
edges1[i] = [a_i, b_i]
0 <= a_i, b_i < n
edges2[i] = [u_i, v_i]
0 <= u_i, v_i < m
The input is generated such that edges1 and edges2 represent valid trees.

Solution

import kotlin.math.max

class Solution {
    fun minimumDiameterAfterMerge(edges1: Array<IntArray>, edges2: Array<IntArray>): Int {
        val n = edges1.size + 1
        val g = packU(n, edges1)
        val m = edges2.size + 1
        val h = packU(m, edges2)
        val d1 = diameter(g)
        val d2 = diameter(h)
        var ans = max(d1[0], d2[0])
        ans = max(
            ((d1[0] + 1) / 2 + ((d2[0] + 1) / 2) + 1),
            ans,
        )
        return ans
    }

    private fun diameter(g: Array<IntArray?>): IntArray {
        val n = g.size
        val f0: Int
        val f1: Int
        val d01: Int
        val q = IntArray(n)
        val ved = BooleanArray(n)
        var qp = 0
        q[qp++] = 0
        ved[0] = true
        run {
            var i = 0
            while (i < qp) {
                val cur = q[i]
                for (e in g[cur]!!) {
                    if (!ved[e]) {
                        ved[e] = true
                        q[qp++] = e
                    }
                }
                i++
            }
        }
        f0 = q[n - 1]
        val d = IntArray(n)
        qp = 0
        ved.fill(false)
        q[qp++] = f0
        ved[f0] = true
        var i = 0
        while (i < qp) {
            val cur = q[i]
            for (e in g[cur]!!) {
                if (!ved[e]) {
                    ved[e] = true
                    q[qp++] = e
                    d[e] = d[cur] + 1
                }
            }
            i++
        }
        f1 = q[n - 1]
        d01 = d[f1]
        return intArrayOf(d01, f0, f1)
    }

    private fun packU(n: Int, ft: Array<IntArray>): Array<IntArray?> {
        val g = arrayOfNulls<IntArray>(n)
        val p = IntArray(n)
        for (u in ft) {
            p[u[0]]++
            p[u[1]]++
        }
        for (i in 0 until n) {
            g[i] = IntArray(p[i])
        }
        for (u in ft) {
            g[u[0]]!![--p[u[0]]] = u[1]
            g[u[1]]!![--p[u[1]]] = u[0]
        }
        return g
    }
}

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