Union Find
Union-Find represent each subset as backward trees ! 0 ! 1 ! 2 ! ! 1 ! 2 ! 0 !
package algorithmdesignmanualbook.graph
import _utils.UseCommentAsDocumentation
import java.util.*
import kotlin.test.assertFalse
import kotlin.test.assertTrue
/**
* Union-Find represent each subset as backward trees
* ! 0 ! 1 ! 2 !
* ! 1 ! 2 ! 0 !
*
* [Video1](https://www.youtube.com/watch?v=ayW5B2W9hfo) - [Video2](https://www.youtube.com/watch?v=eTaWFhPXPz4) - [Solution](https://www.geeksforgeeks.org/union-find/)
*/
@UseCommentAsDocumentation
private class UnionFindSimpleGraph(private val vertexCount: Int, private val edgesCount: Int) {
class Edge(var src: Int, var dest: Int)
val edge = Array<Edge?>(edgesCount) {
null
}
fun find(parent: Array<Int>, i: Int): Int {
// you've reached the root/parent of the "tree"/relationship
if (parent[i] == -1 || parent[i] == i) {
return i
}
// visit its parent if unionized
return find(parent, parent[i])
}
fun hasCycle(): Boolean {
// disjoint sets
val parent = Array(vertexCount) { -1 }
for (i in 0 until edgesCount) {
val root1 = find(parent, edge[i]!!.src)
val root2 = find(parent, edge[i]!!.dest)
// since SimpleGraph#find returns itself if not initialized, root1 & root2 cant be equal
if (root1 == root2)
return true
// make r2 the parent of r1 i.e put in same set
union(parent, root1, root2)
}
return false
}
private fun union(parent: Array<Int>, root1: Int, root2: Int) {
parent[root1] = root2
}
}
fun main() {
run {
val graph = UnionFindSimpleGraph(3, 3)
graph.edge[0] = UnionFindSimpleGraph.Edge(0, 1)
graph.edge[1] = UnionFindSimpleGraph.Edge(1, 2)
graph.edge[2] = UnionFindSimpleGraph.Edge(2, 0)
assertTrue { graph.hasCycle() }
}
run {
val graph = UnionFindSimpleGraph(4, 3)
graph.edge[0] = UnionFindSimpleGraph.Edge(0, 1)
graph.edge[1] = UnionFindSimpleGraph.Edge(1, 2)
graph.edge[2] = UnionFindSimpleGraph.Edge(2, 3)
assertFalse { graph.hasCycle() }
}
}
Updated on 2021-04-03