smallest largest number

how do you find the smallest and largest number of 3 numbers.

without using logical operators and else statements, only if and relational operators ? is there any other way to find it out without checking every number against each other using if statements ?

There's more than one way. If you have your numbers in something like an array or a list that can be iterated through, you can do this:

Code:

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current_minimum = numbers[0]
for(i = 1; i < number count; ++i)
  if(numbers[i] < current_minimum) current_minimum = numbers[i];

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That'll work for any number of numbers. Finding a maximum is very similar. Additionally, if you always have 3 numbers, and you have a function min() that returns the minimum of 2 numbers, this will also work, and may be simpler:

Code:

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minimum_of_three = min( number1, min(number2, number3 ) )

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Code:

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vector<int> numbers;
...
vector<int>::iterator it = min_element( numbers.begin(), numbers.end() );
cout << "Min # = " << *it << endl;
it = max_element( numbers.begin(), numbers.end() );
cout << "Max # = " << *it << endl;

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or you could sort the numbers and just take the first and last numbers...

Since you're probably not allowed to do any of those methods (as it sounds like this is a contrived test/assignment problem), you can look at them and get a hint for a simpler way to do it.

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or you could sort the numbers and just take the first and last numbers...

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That is inefficient though, so I would go with the min_element and max_element solution instead, unless I wanted to get both of them in a single pass.

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you could sort the numbers and just take the first and last numbers...

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A sort is O(nlogn) at best. In this case, just running through the array (O(n)) will do.

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Originally Posted by cyberfish

A sort is O(nlogn) at best

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Bucket sort is O(n)

True. I was thinking about comparison sorts.

But still, why not just run through the array?

The difference between O(n) and O(nlogn) is relatively small when n is 3 ;)

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The difference between O(n) and O(nlogn) is relatively small when n is 3

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That's true... and in retrospect there is probably no array or other container, since Daved is probably right and the 3 numbers really are in three separate variables.

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Originally Posted by laserlight

That is inefficient though, so I would go with the min_element and max_element solution instead, unless I wanted to get both of them in a single pass.

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Yeah, that's why I only mentioned it in passing but gave examples for min_element() & max_element(). ;)