Algorithm to find high/low numbers with at most 1.5n comparisons

Question

I've been thinking about this homework question for a bit now. Given an number array of size n, design an algorithm that will find the high and and low values with at most 1.5n comparisons.

My first try was

int high=0
int low= Number.MaxValue //problem statement is unclear on what type of number to use
Number numList[0 . . n] //number array, assuming unsorted

for (i=0, i < n, i++) {
  if (numList[i] > high)
    high = numList[i]

  else if (numList[i] < low)
    low = numList[i]

}

My problem is each iteration of the loop has one of three possibilities:

• low value is found - 1 comparison made
• high value is found - 2 comparisons made
• neither is found - 2 comparisons made

So for an entire array traversal, a maximum of 2n comparisons can be made, which is a far cry from the problem maximum requirement of 1.5n comparisons.

Answer 1

This is a question I had during an interview and I found the answer with a small hint from the interviewer which was "How do you compare two numbers?" (it really helped).

Here is the explanation:

Lets say I have 100 numbers (you can easily replace it by n but it work better for the example if n is an even number). What I do is that I split it into 50 lists of 2 numbers. For each couple I make one comparison and I'm done (which makes 50 comparisons by now) then I just have to find the minimum of the minimums (which is 49 comparisons) and the maximum of the maximums (which is 49 comparisons as well) such that we have to make 49+49+50=148 comparisons. We're done !

Remark: to find the minimum we proceed as follow (in pseudo code):

n = myList.size();
min = myList[0];
for (int i(1); i<n-1; i++)
{
    if (min>myList[i]) min = myList[i];
}
return min;

And we find it in (n-1) comparisons. The code is almost the same for maximum.

Answer 2

Start with a pairs of numbers and find local min and max (n/2 comparisons). Next, find global max from n/2 local maxes (n/2 comparisons), and similarly global min from local mins (n/2 comparisons). Total comparisons: 3*n/2 !

For i in 0 to n/2: #n/2 comparisons
    if x[2*i]>x[2*i+1]:
        swap(x,2*i,2*i+1)

global_min = min( x[0], x[2], ...) # n/2 comparisons
global_max = max( x[1], x[3], ...) # n/2 comparisons

Note that the above solution changes the array. Alternate solution:

Initialize min and max
For i = 0 to n/2:
    if x[2*i]<x[2*i+1]:
        if x[2*i]< min:
            min = x[2*i]
        if x[2*i+1]> max:
            max = x[2*i+1]
    else:
        if x[2*i+1]< min:
            min = x[2*i+1]
        if x[2*i]> max:
            max = x[2*i]