Connected Components Solution

Start with $n$ components. Each successful union reduces the count by 1:

class UnionFind:
    function init(n):
        this.parent = array of [0, 1, ..., n-1]
        this.rank = array of n zeros
        this.count = n

    function find(x):
        if this.parent[x] != x:
            this.parent[x] = this.find(this.parent[x])
        return this.parent[x]

    function union(x, y):
        rootX = this.find(x)
        rootY = this.find(y)
        if rootX == rootY:
            return false // already connected
        if this.rank[rootX] < this.rank[rootY]:
            this.parent[rootX] = rootY
        else if this.rank[rootX] > this.rank[rootY]:
            this.parent[rootY] = rootX
        else:
            this.parent[rootY] = rootX
            this.rank[rootX] += 1
        this.count -= 1
        return true

function countComponents(n, edges):
    uf = new UnionFind(n)
    for each (a, b) in edges:
        uf.union(a, b)
    return uf.count