if all edges are already sorted then this problem will reduced to union-find problem on a graph with $E$ edges and $V$ vertices.
for each edge (u,v) in E
if(FIND-SET(u) != FIND-SET(v))
FIND-SET$(v)$ and UNION$(u,v)$ runs in $α(|V|)$
where $α(n)$ is inverse ackermann function i.e $\log*(n)$
So overall complexity becomes $O(|E|.α(|V|))$
Check these !!
I think the answer should simply be O(E)
According to wikipedia amortized time complexity of unoin find is O(log*V), as Arjun said in one other answer log*n is practically a constant. For E unoin find operations, the answer would simply be E*O(1), Therefore O(E)
Complexity of kruskal is O(ElogE + E) .... ElogE for sorting the edges and E for union operation of E edges ...
So here it is sorted ... so O(E) ... correct me if i am wrong ... Refer this ...
On a lighter ...