Answer Posted / ashekur rahman

Time Complexity T(n) <= 3n/2
Proof:
We will calculate comparison in three steps.
1) compare pairwise will divide two sub array.
2) linear search of first sub array
3) linear search of second sub array


For even number of elements,
step 1 needs n/2 comparisons where both sub array contains
n/2 number of elements. Step 2 or 3 needs (n/2 - 1)
comparisons by linear search.

T(n) = n/2 + 2(n/2 - 1)
= 3n/2 - 1

Again,

For odd number of elements,
step 1 needs (n/2 + 1) comparison.
Now if first sub array contains n/2 number of elements then
second sub array contains (n/2 + 1) number of elements

or if first sub array contains (n/2 + 1) number of elements
then second sub array contains n/2 number of elements

Therefore,
T(n) = (n/2 + 1) + (n/2 + 1 - 1) + (n/2 - 1)
= 3n/2

So, T(n) = 3n/2 - 1, if n is even number
and T(n) = 3n/2, if n is odd number

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