Why is the Kth Largest Element solution using a MinHeap O(N lgK) in complexity?
The maximum number of elements in the heap at any given time is upper bounded by $k+1$.
Since each insertion or deletion from an heap requires $O(\log \eta)$ time where $\eta$ is the number of elements in the heap at the time of the operation, and $\eta \le k+1$, we know that each heap operation can be performed in time $O(\log (k+1) ) = O( \log k )$.
How many heap operations does the algorithm perform? Each element is inserted once, so there are $n$ insertions. In addition all but the largest $k-1$ elements are extracted. This yields a total of $n + (n-(k-1)) = 2n-k+1 = O(n)$ operations.
By multiplying the number of operations with our upper bound on the time required by each operation of $O(\log k)$  we obtain an overall time complexity of $O(n \log k)$.