To find the minimum value into an array of items itsn’t difficult. There are not many options to do that. The most natural approach is to take the first item and to compare its value against the values of all other elements. Once we find a smaller element we continue the comparisons with its value. Finally we find the minimum.

First thing to note is that we pass through the array with n steps and we need exactly n-1 comparisons. It’s clear that this is the optimal solution, because we must check all the elements. For sure we can’t be sure that we’ve found the minimum (maximum) value without checking every single value.

Overview

The algorithm above is very simple and we’re sure that it is optimal. Obviously finding both the minimum and the maximum value is O(n) with n-1 comparisons, but what about combining these tasks into one single pass.

Since they both are O(n) and need n-1 comparisons it’s natural to think that combining the two tasks will be O(n) and 2n – 2 comparisons. However we can reduce the number of comparisons!

Instead of taking only one item from the array and comparing it against the minimum and maximum we can take a pair of items at each step. Thus we can first compare them and then compare the smaller value with the currently smallest value and the greater item with the currently greatest value. This will make only 3 comparisons instead of 4.

Implementation

It’s easy to implement the minimum (maximum) algorithms with a single loop.

The implementation of finding the maximum is practically the same.

Simply merging these two functions will lead us to a O(n) with 2n – 2 comparisons solution.

However we can take a pair of items on each step. First we’ll compare the items from that pair and after that we’ll compare them respectively with the minimum and the maximum value. Because on each iteration we jump by two items, in case the number of array items is even we must check for the array boundaries. This can be overcome by adding a sentinel. Thus the array items are always odd, but this will lead us to a “extra” memory solution.

Sentinel

Without sentinel

Complexity

The complexity of finding both minimum and maximum is O(n). Even after combining the both algorithms in one single pass the complexity remains O(n). However in the second case we can reduce the number of comparisons to 3 * ceil(n/2) instead of 2n – 2!

Application

This algorithm can be applied in various fields of the computer science, since its nature is so basic. However there are two reasons why this approach is so important.

First we can see how by combining two “algorithms” doesn’t mean that we combine their complexities or the number of operations. With a clever trick and with the observation that the two operations are related (minimum and maximum) we can reduce the number of comparisons.

In the other hand we see how using a sentinel can be very handy and can spare us some comparisons, just like the sequential search.

When a programmer sees a chunk of code he tends to evaluate it in a rather intuitive manner and to qualify it as “elegant” or not. This is quite easy, because it’s subjective and nobody knows what exactly elegant means. However behind this there is a powerful mathematical approach of measuring a program effectiveness.

It’s a pity that most of the developers still think of the big O notation as something from the university classes, but unusual in the practice and they barely use it their job. But before describing the big O notation, let me start from something really simple.

Let’s have the following example (note that all the examples are in PHP):

$n = 100;
$s = 0;

for ($i = 0; $i < $n; $i++) {
	for ($j = 0; $j < $n; $j++) {
		$s++;	
	}	
}

As you can see there are two assignments and two nested loops. This is really a widely used example from any algorithm book.

Constants, Languages, Compilers

First of all the time to assign a value to a variable, to compare two values and to increment a variable is constant. It depends on the computer resources, the compiler or the language, but it's constant on one machine if you compare two chunks of code. Now we can see that these operations take (add) constant time to the program, and we can assume this time is respectively a, b, c, d, e, f, g, h, i.

$n = 100; 	// a
$s = 0;		// b
$i = 0; 	// c
$i < $n; 	// d
$i++;		// e
$j = 0; 	// f
$j < $n;	// g
$j++;		// h
$s++;		// i

What Matters?

Actually the most important thing here is the value of n. By assigning greater values to n the more time will take the program to run. As we can see from the following table by multiplying the value of n by 10, the time became 100 times more.

n		time
10		0.00002
100		0.002
...		...

What happens in fact is that we can sum all these values.

a + b + c + n*d + n*e + n*(f + n*g + n*h + n*i)

and by substituting:

a + b + c = k
d + e + n = l
g + h + i = m

the result is:

m*n² + l*n + k

Conclusion

Here the most important thing is the degree of n, because it can change dramatically the program time consumption depending on the n value. Thus this chunk has a quadratic complexity or O(n²).

Of course there are constants, but in the practice they are not so important. Take a look at these two functions:

f = 2*n²
g = 200*n

OK, for n = 1 the first one will be faster, but as n increments the second function becomes to be faster and faster, thus after a given value of n the second function is really the fastest!