(b) option says logn , you mean to say option (c).

3 votes

Any decision tree that sorts $n$ elements has height

- $\Omega(n)$
- $\Omega(\text{lg}n)$
- $\Omega(n \text{lg} n)$
- $\Omega(n^2)$

3 votes

For any comparison based sorting algorithm the decision tree will have n! leaf nodes

so height will be log(n!)= Ώ(nlogn)

so height will be log(n!)= Ώ(nlogn)

0 votes

**Answer:****B**

**Explanation:**

We know that the maximum nodes in a binary tree of height, $\color {red}{\mathrm h = 2^{\mathrm h + 1} - 1} $ nodes.

Also, for $\mathrm n$ elements the decision would be taken from $\color {red}{\mathrm n!}$ elements.

$\Rightarrow 2^{\mathrm h+1} - 1 \ge\mathrm n!\\\Rightarrow \mathrm h \ge \log_2 \mathrm n! + 1\\\Rightarrow \mathrm h \ge \log_2\mathrm n!\\\Rightarrow \mathrm h \ge\mathrm n\log\mathrm n \;\;\;\;\;\;\text{[Using Stirling's approximation, $\mathrm n! = \frac{(\mathrm n^{\mathrm n})}{e}]$}\\\Rightarrow \Omega( \mathrm n \log\mathrm n)$

$\therefore$ **B** is the correct option.