c++ between three numbers sum two highers without loops and arrays

Question

How to write c++ program that will let user enters three numbers, and find between that three 2 higest and write smth like : 5,3,4 sum of 5 and 4 is 9!

I tried allready, but I get too many if elses and it looks really bad, and even not working well in all cases :/

#include <iostream>
#include<math.h>
using namespace std;
int main()
{
    float n1, n2, n3;
cout << "Enter three numbers: ";
    cin >> n1 >> n2 >> n3;

    if(n1 >= n2 && n1 >= n3)
    {
        if(n2>=n1 && n2>=3)
            cout << "Sum of 2 highest between this three numbers is: "  << cout<<n1+n2;
    }

    if(n1 >= n2 && n1 >= n3)
    {
        if(n3>=n1 && n3>=n2)
            cout << "Sum of 2 highest between this three numbers is: "  << cout<<n1+n3;
    }

    if(n2 >= n1 && n2 >= n3)
    {
        if(n3 >= n1 && n3 >= n2) {
            cout << "Sum of 2 highest between this three numbers is: " << cout<<n3+n2;
        }
    }

    if(n3 >= n1 && n3 >= n2)
    {
        if(n2 >= n1) {
            cout << "Sum of 2 highest between this three numbers is: "  << cout<<n3+n2;
        }
    }

    if(n3 >= n1 && n3 >= n2)
    {
        if(n1>= n2) {
            cout << "Sum of 2 highest between this three numbers is: "  << cout<<n3+n1;
        }
    }

        return 0;
}

Answer 1

A simple approach to answer the question as asked would be

  cout << "Sum of 2 highest between this three numbers is: ";
  cout << (long long) n1 + n2 + n3 - std::min(n1, std::min(n2, n3));

or (C++11)

  cout << "Sum of 2 highest between this three numbers is: ";
  cout << (long long) n1 + n2 + n3 - std::min({n1, n2, n3});

The (long long) conversion forces all values being added or subtracted to be promoted before the operation, to avoid overflow if int supports a smaller range of values than a long long.

Of course, this does embed a fixed assumption of three values being worked with.

Answer 2

Find the minimum first,

double min = std::min(n1, std:min(n2, n3)); // or std::min({n1, n2, n3})

then eliminate it

if(min == n1)
{ 
    std::cout << n2 << ' ' << n3 << ' ' << n2 + n3;
}
else if(min == n2) 
{ 
    std::cout << n1 << ' ' << n3 << ' ' << n1 + n3;
}
else 
{ 
    std::cout << n1 << ' ' << n2 << ' ' << n1 + n2;
}

Live example

Of course if you're really picky you can replace the call to std::min by

double min = n1;
if(n2 < min)
    min = n2;
if(n3 < min)
    min = n3;

Another solution is to just do the min finding and elimination in a single step, like

if(n1 < n2 && n1 < n3) // min is n1
{
    std::cout << n2 << ' ' << n3 << ' ' << n2 + n3;
}
else if(n2 < n1 && n2 < n3) // min is n2
{
    std::cout << n1 << ' ' << n3 << ' ' << n1 + n3;
}
else // min is n3
{
    std::cout << n1 << ' ' << n2 << ' ' << n1 + n2;
}

In terms of complexity, both solutions are identical, i.e. require 4 comparisons.

Answer 3

You only need to do a few checks to get the desired result.

int n1 = 0,n2 = 0,n3 = 0;
cin>>n1>>n2>>n3;

int highest = 0;
int second_highest = 0;

if(n1 > n2)
{
    if(n1 > n3)
    {
        highest = n1;
        if(n3 > n2)
        {
            second_highest = n3;
        }
        else
        {
            second_highest = n2;
        }
    }
    else
    {
        highest = n3;
        second_highest = n1;
    }
}
else if(n2 > n3) // if the first if is false it means that the n2 is highest or second highest
{
    highest = n2;
    if(n1 > n3)
    {
        second_highest = n1;
    }
    else
    {
        second_highest = n3;
    }
}
else
{
    highest = n3;
    second_highest = n2;
}

cout << "Highest: " << highest << "\nSecond highest: " << second_highest;

Hope that helps.

Answer 4

Try this:

max = (n1>n2) ? ( (n1>n3) ? n1 : n3) : ((n2>n3) ? n2 : n3);
second = (n1==max) ? ( (n2>n3) ? n2 : n3) : ((n2==max) ? ((n1>n3) ? n1 : n3) : ((n2>n1)? n2 : n1));