Leetcode daily # 30.04.2023

[1579. Remove Max Number of Edges to Keep Graph Fully Traversable]

30.04.2023

1579. Remove Max Number of Edges to Keep Graph Fully Traversable hard

fun IntArray.root(a: Int): Int {
    var x = a
    while (this[x] != x) x = this[x]
    this[a] = x
    return x
}
fun IntArray.union(a: Int, b: Int): Boolean {
    val rootA = root(a)
    val rootB = root(b)
    if (rootA != rootB) this[rootB] = rootA
    return rootA != rootB
}
fun IntArray.connected(a: Int, b: Int) = root(a) == root(b)
fun maxNumEdgesToRemove(n: Int, edges: Array<IntArray>): Int {
    val uf1 = IntArray(n + 1) { it }
    val uf2 = IntArray(n + 1) { it }
    var skipped = 0
    edges.forEach { (type, a, b) ->
        if (type == 3) {
            uf1.union(a, b)
            if (!uf2.union(a, b)) skipped++
        }
    }
    edges.forEach { (type, a, b) ->
        if (type == 1 && !uf1.union(a, b)) skipped++
    }
    edges.forEach { (type, a, b) ->
        if (type == 2 && !uf2.union(a, b)) skipped++
    }
    for (i in 2..n)
    if (!uf1.connected(i - 1, i) || !uf2.connected(i - 1, i)) return -1
    return skipped
}

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Intuition

After connecting all type 3 nodes, we can skip already connected nodes for Alice and for Bob. To detect if all the nodes are connected, we can just check if all nodes connected to one particular node.

Approach

Use separate Union-Find objects for Alice and for Bob

Complexity

• Time complexity:
  O(n), as root and union operations take < 5 for any n <= Int.MAX.

• Space complexity:
  O(n)