In a min-heap with $n$ elements with the smallest element at the root, the $log n^{th}$ smallest element can be found in time.
Is this approach correct?
For $1$ st smallest - root node is the element - takes $0$ comparison
For $2$ nd smallest - $2$ elements are to be compared - takes $1$ comparison
For $3$ rd smallest - $3$ elements are to be compared - takes $2$ comparison
For $4$ th smallest - $4$ elements are to be compared - takes $3$ comparison
.
.
For $logn^{th}$ smallest - $logn$ elements are to be compared - takes $logn-1$ comparison
So total comparisons would be = $1 + 2 + 3 + 4 + . . . . . . + (logn-1)$
= $\frac{(logn)(logn+1)}{2}$
= $O$$(logn)$$2$
Please suggest is this correct approach??