Ans: minimum number of comparison require to find minimum and maximum is: Approach is divide and conquer ....
T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2
T(2) = 1 // if two element then compare both and return max and min
T(1) = 0 // if one element then return both max and min same
If n is a power of 2, then we can write T(n) as:
T(n) = 2T(n/2) + 2
After solving above recursion, we get
T(n) = (3/2)n -2
Thus, the approach does 3/2n -2 comparisons if n is a power of 2. And it does more than (3/2)n -2 comparisons if n is not a power of 2.
