smallest largest number

This is a discussion on smallest largest number within the C++ Programming forums, part of the General Programming Boards category; how do you find the smallest and largest number of 3 numbers. without using logical operators and else statements, only ...

  1. #1
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    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 ?

  2. #2
    int x = *((int *) NULL); Cactus_Hugger's Avatar
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    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:
    current_minimum = numbers[0]
    for(i = 1; i < number count; ++i)
       if(numbers[i] < current_minimum) current_minimum = numbers[i];
    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:
    minimum_of_three = min( number1, min(number2, number3 ) )
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  3. #3
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    Code:
    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;
    or you could sort the numbers and just take the first and last numbers...

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    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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    C++ Witch laserlight's Avatar
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    or you could sort the numbers and just take the first and last numbers...
    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.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
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    you could sort the numbers and just take the first and last numbers...
    A sort is O(nlogn) at best. In this case, just running through the array (O(n)) will do.

  7. #7
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    Quote Originally Posted by cyberfish View Post
    A sort is O(nlogn) at best
    Bucket sort is O(n)
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    True. I was thinking about comparison sorts.

    But still, why not just run through the array?

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

  10. #10
    C++ Witch laserlight's Avatar
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    The difference between O(n) and O(nlogn) is relatively small when n is 3
    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.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  11. #11
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    Quote Originally Posted by laserlight View Post
    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.
    Yeah, that's why I only mentioned it in passing but gave examples for min_element() & max_element().

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