Answer:
Because it is too dense for that to happen
Explanation:
Answer:
Explanation:
Firstly of all we have to construct the min-heap of the k-sub list and each sub list which is a node in the constructed min-heap.
We have several steps to follows:
Step-1. When we compare the two sub lists, at the starting we can compare their first elements which is actually their minimum elements.
Step-2. The min-heap formation will cost be O(k) time.
Step-3. After the step 1 & step 2 we can run the minimum algorithm which can be extracted from the minimum element in the root list.
Step-4. Then Update the root list in the heap and after that simplify the min-heap as maintained by the new minimum element in the root list.
Step-5. If any root sub-list becomes empty in the step 4 then we can take any leaf sub-list from the root and simplify it.
Step-6. At every Extraction of the element it can take up to O(log k) time.
Hence, We can say that the extract of n element in the total whose
Running time will be O(n log k + k) which can be equal to the O(n log k+ k) (since k < n).
