LeetCode in Kotlin

684. Redundant Connection

Medium

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input.

Example 1:

Input: edges = [[1,2],[1,3],[2,3]]

Output: [2,3]

Example 2:

Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]

Output: [1,4]

Constraints:

• n == edges.length
• 3 <= n <= 1000
• edges[i].length == 2
• 1 <= ai < bi <= edges.length
• ai != bi
• There are no repeated edges.
• The given graph is connected.

Solution

class Solution {
    private lateinit var par: IntArray

    fun findRedundantConnection(edges: Array<IntArray>): IntArray {
        val ans = IntArray(2)
        val n = edges.size
        par = IntArray(n + 1)
        for (i in 0 until n) {
            par[i] = i
        }
        for (edge in edges) {
            val lx = find(edge[0])
            val ly = find(edge[1])
            if (lx != ly) {
                par[lx] = ly
            } else {
                ans[0] = edge[0]
                ans[1] = edge[1]
            }
        }
        return ans
    }

    private fun find(x: Int): Int {
        return if (par[x] == x) {
            x
        } else find(par[x])
    }
}

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