Lazy binomial heap
Idea: 


Advantages:  while insert and meld now take only \(O(1)\) time, extractmin is still \(O(\log n)\) amortized 
Laziness
A standard binomial heap may contain only a single binomial tree of each order. This guarantees that a heap with \(n\) elements consists of at most \(\log n\) binomial trees. Every time we add a new node or merge in new trees, we need to do a cleanup and link the trees with the same order.
In lazy binomial heap, we drop this invariant. When merging two heaps, we simply link the two lists in constant time and leave the cleanup for later. This lazy merging means that if we insert
\(n\) new elements, we will create a chain of \(n\) roots. We only do cleanups during the extractmin
operation. When searching for a new minimum, we need to traverse the whole list of roots anyway, so we might just as well use this time to clean up.
This way merge
and insert
will take only constant time, while extractmin
will still take \(O(\log n)\) amortized time.
merge 
insert 
extractmin 


binomial heap  \(O(\log n)\)  \(O(\log n)\)  \(O(\log n)\) 
lazy binomial heap  \(O(1)\)  \(O(1)\)  \(O(\log n)\) amortized 
Analysis
We will use the accounting method of amortized analysis, so imagine we have to pay for all computation and 1$ can buy us \(O(1)\) time. We prove:
Theorem: If we get 2$ for every insert
, 1$ for every merge
and \(2\log n+1\)$ for every extractmin
operation, we can pay for all the computation.
Invariant: Each root will have 1$ saved for cleanup in future.
Proof:
merge
takes \(O(1)\) time  just link two lists and choose the new minimum; we can pay for it with 1$ (and preserve the invariant).insert
: creating and linking a new root node takes \(O(1)\) time  we pay for it with 1$ and save the other 1$ for future (to preserve the invariant). for
extractmin
, we need to delete the min root and merge back its children (1$); each child becomes a new root, so it gets 1$ (\(\leq \log n\)$ in total); then we need to do a cleanup  the cleanup of \(t\) trees takes \(O(t)\) time; let \(\ell\) be the number of trees that are linked under some other trees and let \(r\) be the number of roots which remain roots even after cleanup (\(t=\ell+r\))
 each one of the \(\ell\) roots has 1$ saved and since they will not be roots anymore after the cleanup, they are free to spend the money; they will pay for the \(\ell\) comparisons and linking operations
 on the other hand, \(r\leq \log n\) so we can pay \(\log n\)$ from the money allocated for the operation (this includes finding the new minimum)
In the potential method, we would define potential \(\Phi(H)=\)number of binomial trees = number of roots. The extractmin
operation takes \(O(t+\log n)=O(\ell+\log n)\) time. On the other hand, the change in potential is at most \(\Delta\Phi \leq \log n  \ell\). The extra \(\leq \log n\) term corresponds to the children of the deleted root becoming new roots and the \(\ell\) term cancels out with the \(O(\ell)\) term in the actual time: \[ T_{amort} = T_{actual} + c\cdot \Delta\Phi = O(\ell + \log n) +c\cdot (\log n  \ell) = O(\log n).\]