To find the greatest you need to look at exactly 3 ints, no more no less. You re looking at 6 with 3 compares. You should be able to do it in 3 and 2 compares.

int ret = max(i,j);
ret = max(ret, k);
return ret;

Pseudocode:

result = i
if j > result:
  result = j
if k > result:
  result = k
return result

How about

return i > j? (i > k? i: k): (j > k? j: k);

two comparisons, no use of transient temporary stack variables...

Your current method: http://ideone.com/JZEqZTlj (0.40s)

Chris s solution:

int ret = max(i,j);
ret = max(ret, k);
return ret;

http://ideone.com/hlnl7QZX (0.39s)

Solution by Ignacio Vazquez-Abrams:

result = i;
if (j > result)
  result = j;
if (k > result)
  result = k;
return result;

http://ideone.com/JKbtkgXi (0.40s)

And Charles Bretana s:

return i > j? (i > k? i: k): (j > k? j: k);

http://ideone.com/kyl0SpUZ (0.40s)

Of those tests, all the solutions take within 3% the amount of time to execute as the others. The code you are trying to optimize is extremely short as it is. Even if you re able to squeeze 1 instruction out of it, it s not likely to make a huge difference across the entirety of your program (modern compilers might catch that small optimization). Spend your time elsewhere.

EDIT: Updated the tests, turns out it was still optimizing parts of it out before. Hopefully it s not anymore.

For a question like this, there is no substitute for knowing just what your optimizing compiler is doing and just what s available on the hardware. If the fundamental tool you have is binary comparison or binary max, two comparisons or max s are both necessary and sufficient.

I prefer Ignacio s solution:

result = i;
if (j > result)
  result = j;
if (k > result)
  result = k;
return result;

because on the common modern Intel hardware, the compiler will find it extremely easy to emit just two comparisons and two cmov instructions, which place a smaller load on the I-cache and less stress on the branch predictor than conditional branches. (Also, the code is clear and easy to read.) If you are using x86-64, the compiler will even keep everything in registers.

Note you are going to be hard pressed to embed this code into a program where your choice makes a difference...

I like to eliminate conditional jumps as an intellectual exercise. Whether this has any measurable effect on performance I have no idea though :)

#include <iostream>
#include <limits>

inline int max(int a, int b)
{
    int difference = a - b;
    int b_greater = difference >> std::numeric_limits<int>::digits;
    return a - (difference & b_greater);
}

int max(int a, int b, int c)
{
    return max(max(a, b), c);
}

int main()
{
    std::cout << max(1, 2, 3) << std::endl;
    std::cout << max(1, 3, 2) << std::endl;
    std::cout << max(2, 1, 3) << std::endl;
    std::cout << max(2, 3, 1) << std::endl;
    std::cout << max(3, 1, 2) << std::endl;
    std::cout << max(3, 2, 1) << std::endl;
}

This bit twiddling is just for fun, the cmov solution is probably a lot faster.

Not sure if this is the most efficient or not, but it might be, and it s definitely shorter:

int maximum = max( max(i, j), k);

There is a proposal to include this into the C++ library under N2485. The proposal is simple, so I ve included the meaningful code below. Obviously, this assumes variadic templates

template < typename T >
const T & max ( const T & a )
{ return a ; }

template < typename T , typename ... Args >
const T & max( const T & a , const T & b , const Args &... args )
{ return  max ( b > a ? b : a , args ...); }

The easiest way to find a maximum or minimum of 2 or more numbers in c++ is:-

int a = 3, b = 4, c = 5;
int maximum = max({a, b, c});

int a = 3, b = 4, c = 5;
int minimum = min({a, b, c});

You can give as many variables as you want.

Interestingly enough it is also incredibly efficient, at least as efficient as Ignacio Vazquez-Abrams solution (https://godbolt.org/z/j1KM97):

        mov     eax, dword ptr [rsp + 8]
        mov     ecx, dword ptr [rsp + 4]
        cmp     eax, ecx
        cmovl   eax, ecx
        mov     ecx, dword ptr [rsp]
        cmp     eax, ecx
        cmovl   eax, ecx

Similar with GCC, while MSVC makes a mess with a loop.

public int maximum(int a,int b,int c){
    int max = a;
    if(b>max)
        max = b;
    if(c>max)
        max = c;
    return max;
}

I think by "most efficient" you are talking about performance, trying not to waste computing resources. But you could be referring to writing fewer lines of code or maybe about the readability of your source code. I am providing an example below, and you can evaluate if you find something useful or if you prefer another version from the answers you received.

/* Java version, whose syntax is very similar to C++. Call this program "LargestOfThreeNumbers.java" */
class LargestOfThreeNumbers{
    public static void main(String args[]){
        int x, y, z, largest;
        x = 1;
        y = 2;
        z = 3;
        largest = x;
        if(y > x){
            largest = y;
            if(z > y){
                largest = z;
            }
        }else if(z > x){
            largest = z;
        }
        System.out.println("The largest number is: " + largest);
    }
}

#include<stdio.h>
int main()
{
    int a,b,c,d,e;
    scanf("%d %d %d",&a,&b,&c);
    d=(a+b+abs(a-b))/2;
    e=(d+c+abs(c-d))/2;
    printf("%d is Max
",e);
    return 0;
}

I Used This Way, It Took 0.01 Second

#include "iostream"
using std::cout;
using std::cin;
int main()
{
    int num1, num2, num3;
    cin>>num1>>num2>>num3;
    int cot {((num1>num2)?num1:num2)};
    int fnl {(num3>cot)?num3:cot};
    cout<<fnl;
}

Or This

#include "iostream"

using std::cout;
using std::cin;
int main()
{
    int num1, num2, num3;
    cin>>num1>>num2>>num3;
    int cot {(((num1>num2)?num1:num2)>((num3>cot)?num3:cot)?((num1>num2)?num1:num2):((num3>cot)?num3:cot))};
    cout<<cot;
}

The most efficient way to find the greatest among 3 numbers is by using max function. Here is a small example:

#include <iostream>
#include <algorithm>

using namespace std;
int main() {
   int x = 3, y = 4, z = 5;
   cout << max(x, max(y, z)) << endl;
   return 0;
}

If you have C++ 11, then you can do it as follow:

int main() {
   int x = 3, y = 4, z = 5;
   cout << std::max({x, y, z}); << endl;
   return 0;
}

If you are interested to use a function, so that you can call it easily multiple times, here is the code:

using namespace std;

int test(int x, int y, int z) { //created a test function

    //return std::max({x, y, z}); //if installed C++11
    return max(x, max(y, z));
}

int main() {
    cout << test(1, 2, 3) << endl;
    cout << test(1, 3, 2) << endl;
    cout << test(1, 1, 1) << endl;
    cout << test(1, 2, 2) << endl;
    return 0;
}

Here is a small function you can use:

int max3(int a, int b, int c=INT_MIN) {
    return max(a, max(b, c));
}

In C# finding the greatest and smallest number between 3 digit

static void recorrectFindSmallestNumber()
{
int x = 30, y = 22, z = 11;
if (x < y)
{
if (x < z)
{
Console.WriteLine("X is Smaller Numebr {0}.", x);
}
else
{
Console.WriteLine("z is Smaller Numebr {0}.", z);
}
}
else if (x > y)
{
if (y < z)
{
Console.WriteLine("y is Smaller number.{0}", y);
}
else
{
Console.WriteLine("z is Smaller number.{0}", z);
}
}
 else
{

}
}

=================================================================

static void recorrectFindLargeNumber()
{
int x, y, z;
Console.WriteLine("Enter the first number:");
x = int.Parse(Console.ReadLine());
Console.WriteLine("Enter the second number:");
y = int.Parse(Console.ReadLine());
Console.WriteLine("Enter the third nuumnber:");
z = int.Parse(Console.ReadLine());
if (x > y)
{
if (x > z)
{
Console.WriteLine("X is Greater numbaer: {0}.", x);
}
else
{
Console.WriteLine("Z is greatest number: {0}.", z);
}
}
else if (x < y)
{
if (y > z)
{
Console.WriteLine("y is Greater Number: {0}", y);
}
else 
{
Console.WriteLine("Z is Greater Number; {0}", z);                
}
}
else
{
                
}
}

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int num1,num2,num3,maximum;

cout<<"Enter 3 numbers one by one "<<endl;
cin>>num1;
cin>>num2;
cin>>num3;

maximum=max(max(num1,num2),num3);

cout<<"maximum of 3 numbers is "<<maximum<<endl;
}