Compute the minimum or maximum of two integers without branching

On some rare machines where branching is expensive, the below obvious approach to find minimum can be slow as it uses branching.

C++

/* The obvious approach to find minimum (involves branching) */
int min(int x, int y)
{
  return (x < y) ? x : y
}

//This code is contributed by Shubham Singh

C

/* The obvious approach to find minimum (involves branching) */
int min(int x, int y)
{
  return (x < y) ? x : y
}

Java

/* The obvious approach to find minimum (involves branching) */
static int min(int x, int y)
{
  return (x < y) ? x : y;
}

// This code is contributed by rishavmahato348.

Python3

# The obvious approach to find minimum (involves branching)
def min(x, y):
    return x if x < y else y

  # This code is contributed by subham348.

C#

/* The obvious approach to find minimum (involves branching) */
static int min(int x, int y)
{
  return (x < y) ? x : y;
}

// This code is contributed by rishavmahato348.

Javascript

<script>

/* The obvious approach to find minimum (involves branching) */
function min(x, y)
{
  return (x < y) ? x : y;
}

// This code is contributed by subham348.
</script>

Below are the methods to get minimum(or maximum) without using branching. Typically, the obvious approach is best, though.

Method 1(Use XOR and comparison operator)
Minimum of x and y will be 

y ^ ((x ^ y) & -(x < y))

It works because if x < y, then -(x < y) will be -1 which is all ones(11111….), so r = y ^ ((x ^ y) & (111111…)) = y ^ x ^ y = x. 

And if x>y, then-(x<y) will be -(0) i.e -(zero) which is zero, so r = y^((x^y) & 0) = y^0 = y.

On some machines, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage.
To find the maximum, use 

x ^ ((x ^ y) & -(x < y));

C++

// C++ program to Compute the minimum
// or maximum of two integers without
// branching
#include<iostream>
using namespace std;

class gfg
{

    /*Function to find minimum of x and y*/
    public:
    int min(int x, int y)
    {
        return y ^ ((x ^ y) & -(x < y));
    }

    /*Function to find maximum of x and y*/
    int max(int x, int y)
    {
        return x ^ ((x ^ y) & -(x < y));
    }
    };

    /* Driver code */
    int main()
    {
        gfg g;
        int x = 15;
        int y = 6;
        cout << "Minimum of " << x <<
             " and " << y << " is ";
        cout << g. min(x, y);
        cout << "\nMaximum of " << x <<
                " and " << y << " is ";
        cout << g.max(x, y);
        getchar();
    }

// This code is contributed by SoM15242

C

// C program to Compute the minimum
// or maximum of two integers without
// branching
#include<stdio.h>

/*Function to find minimum of x and y*/
int min(int x, int y)
{
return y ^ ((x ^ y) & -(x < y));
}

/*Function to find maximum of x and y*/
int max(int x, int y)
{
return x ^ ((x ^ y) & -(x < y));
}

/* Driver program to test above functions */
int main()
{
int x = 15;
int y = 6;
printf("Minimum of %d and %d is ", x, y);
printf("%d", min(x, y));
printf("\nMaximum of %d and %d is ", x, y);
printf("%d", max(x, y));
getchar();
}

Java

// Java program to Compute the minimum
// or maximum of two integers without
// branching
public class AWS {

    /*Function to find minimum of x and y*/
    static int min(int x, int y)
    {
    return y ^ ((x ^ y) & -(x << y));
    }

    /*Function to find maximum of x and y*/
    static int max(int x, int y)
    {
    return x ^ ((x ^ y) & -(x << y));
    }

    /* Driver program to test above functions */
    public static void main(String[] args) {

        int x = 15;
        int y = 6;
        System.out.print("Minimum of "+x+" and "+y+" is ");
        System.out.println(min(x, y));
        System.out.print("Maximum of "+x+" and "+y+" is ");
        System.out.println( max(x, y));
    }

}

Python3

# Python3 program to Compute the minimum
# or maximum of two integers without
# branching

# Function to find minimum of x and y

def min(x, y):

    return y ^ ((x ^ y) & -(x < y))


# Function to find maximum of x and y
def max(x, y):

    return x ^ ((x ^ y) & -(x < y))


# Driver program to test above functions
x = 15
y = 6
print("Minimum of", x, "and", y, "is", end=" ")
print(min(x, y))
print("Maximum of", x, "and", y, "is", end=" ")
print(max(x, y))

# This code is contributed
# by Smitha Dinesh Semwal

C#

using System;

// C# program to Compute the minimum
// or maximum of two integers without
// branching
public class AWS
{

    /*Function to find minimum of x and y*/
    public  static int min(int x, int y)
    {
    return y ^ ((x ^ y) & -(x << y));
    }

    /*Function to find maximum of x and y*/
    public  static int max(int x, int y)
    {
    return x ^ ((x ^ y) & -(x << y));
    }

    /* Driver program to test above functions */
    public static void Main(string[] args)
    {

        int x = 15;
        int y = 6;
        Console.Write("Minimum of " + x + " and " + y + " is ");
        Console.WriteLine(min(x, y));
        Console.Write("Maximum of " + x + " and " + y + " is ");
        Console.WriteLine(max(x, y));
    }

}

  // This code is contributed by Shrikant13

PHP

<?php
// PHP program to Compute the minimum
// or maximum of two integers without
// branching

// Function to find minimum
// of x and y
function m_in($x, $y)
{
    return $y ^ (($x ^ $y) &
            - ($x < $y));
}

// Function to find maximum
// of x and y
function m_ax($x, $y)
{
    return $x ^ (($x ^ $y) &
            - ($x < $y));
}

// Driver Code
$x = 15;
$y = 6;
echo"Minimum of"," ", $x," ","and",
    " ",$y," "," is "," ";

echo m_in($x, $y);

echo "\nMaximum of"," ",$x," ",
    "and"," ",$y," ", " is ";

echo m_ax($x, $y);

// This code is contributed by anuj_67.
?>

Javascript

<script>
// Javascript program to Compute the minimum
// or maximum of two integers without
// branching

    /*Function to find minimum of x and y*/
    function min(x,y)
    {
        return y ^ ((x ^ y) & -(x << y));
    }

    /*Function to find maximum of x and y*/
    function max(x,y)
    {
        return x ^ ((x ^ y) & -(x << y));
    }

    /* Driver program to test above functions */
    let x = 15
    let y = 6
    document.write("Minimum of  "+ x + " and " + y + " is ");
    document.write(min(x, y) + "<br>");
    document.write("Maximum of " + x + " and " + y + " is ");
    document.write(max(x, y) + "\n");

    // This code is contributed by avanitrachhadiya2155
</script>

Output:

Minimum of 15 and 6 is 6
Maximum of 15 and 6 is 15

Time Complexity: O(1)

Auxiliary Space: O(1)

Method 2(Use subtraction and shift)
If we know that 

INT_MIN <= (x - y) <= INT_MAX

, then we can use the following, which are faster because (x – y) only needs to be evaluated once. 
Minimum of x and y will be 

y + ((x - y) & ((x - y) >>(sizeof(int) * CHAR_BIT - 1)))

This method shifts the subtraction of x and y by 31 (if size of integer is 32). If (x-y) is smaller than 0, then (x -y)>>31 will be 1. If (x-y) is greater than or equal to 0, then (x -y)>>31 will be 0. 
So if x >= y, we get minimum as y + (x-y)&0 which is y. 
If x < y, we get minimum as y + (x-y)&1 which is x.
Similarly, to find the maximum use 

x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1)))

C++

#include <bits/stdc++.h>
using namespace std;
#define CHARBIT 8

/*Function to find minimum of x and y*/
int min(int x, int y)
{
    return y + ((x - y) & ((x - y) >>
            (sizeof(int) * CHARBIT - 1)));
}

/*Function to find maximum of x and y*/
int max(int x, int y)
{
    return x - ((x - y) & ((x - y) >>
            (sizeof(int) * CHARBIT - 1)));
}

/* Driver code */
int main()
{
    int x = 15;
    int y = 6;
    cout<<"Minimum of "<<x<<" and "<<y<<" is ";
    cout<<min(x, y);
    cout<<"\nMaximum of"<<x<<" and "<<y<<" is ";
    cout<<max(x, y);
}

// This code is contributed by rathbhupendra

C

#include<stdio.h>
#define CHAR_BIT 8

/*Function to find minimum of x and y*/
int min(int x, int y)
{
  return  y + ((x - y) & ((x - y) >>
            (sizeof(int) * CHAR_BIT - 1)));
}

/*Function to find maximum of x and y*/
int max(int x, int y)
{
  return x - ((x - y) & ((x - y) >>
            (sizeof(int) * CHAR_BIT - 1)));
}

/* Driver program to test above functions */
int main()
{
  int x = 15;
  int y = 6;
  printf("Minimum of %d and %d is ", x, y);
  printf("%d", min(x, y));
  printf("\nMaximum of %d and %d is ", x, y);
  printf("%d", max(x, y));
  getchar();
}

Java

// JAVA implementation of above approach
class GFG
{

static int CHAR_BIT = 4;
static int INT_BIT = 8;
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
    return y + ((x - y) & ((x - y) >>
                (INT_BIT * CHAR_BIT - 1)));
}

/*Function to find maximum of x and y*/
static int max(int x, int y)
{
    return x - ((x - y) & ((x - y) >>
            (INT_BIT * CHAR_BIT - 1)));
}

/* Driver code */
public static void main(String[] args)
{
    int x = 15;
    int y = 6;
    System.out.println("Minimum of "+x+" and "+y+" is "+min(x, y));
    System.out.println("Maximum of "+x+" and "+y+" is "+max(x, y));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
import sys;

CHAR_BIT = 8;
INT_BIT = sys.getsizeof(int());

#Function to find minimum of x and y
def Min(x, y):
    return y + ((x - y) & ((x - y) >>
                (INT_BIT * CHAR_BIT - 1)));

#Function to find maximum of x and y
def Max(x, y):
    return x - ((x - y) & ((x - y) >>
                (INT_BIT * CHAR_BIT - 1)));

# Driver code
x = 15;
y = 6;
print("Minimum of", x, "and",
                    y, "is", Min(x, y));
print("Maximum of", x, "and",
                    y, "is", Max(x, y));

# This code is contributed by PrinciRaj1992

C#

// C# implementation of above approach
using System;

class GFG
{

static int CHAR_BIT = 8;

/*Function to find minimum of x and y*/
static int min(int x, int y)
{
    return y + ((x - y) & ((x - y) >>
                (sizeof(int) * CHAR_BIT - 1)));
}

/*Function to find maximum of x and y*/
static int max(int x, int y)
{
    return x - ((x - y) & ((x - y) >>
            (sizeof(int) * CHAR_BIT - 1)));
}

/* Driver code */
static void Main()
{
    int x = 15;
    int y = 6;
    Console.WriteLine("Minimum of "+x+" and "+y+" is "+min(x, y));
    Console.WriteLine("Maximum of "+x+" and "+y+" is "+max(x, y));
}
}

// This code is contributed by mits

Javascript

<script>
// javascript implementation of above approach
var CHAR_BIT = 4;
    var INT_BIT = 8;

    /* Function to find minimum of x and y */
    function min(x , y) {
        return y + ((x - y) & ((x - y) >> (INT_BIT * CHAR_BIT - 1)));
    }

    /* Function to find maximum of x and y */
    function max(x , y) {
        return x - ((x - y) & ((x - y) >> (INT_BIT * CHAR_BIT - 1)));
    }

    /* Driver code */
        var x = 15;
        var y = 6;
        document.write("Minimum of " + x + " and " + y + " is " + min(x, y)+"<br/>");
        document.write("Maximum of " + x + " and " + y + " is " + max(x, y));

// This code is contributed by shikhasingrajput
</script>

Time Complexity: O(1)

Auxiliary Space: O(1)

Note that the 1989 ANSI C specification doesn’t specify the result of signed right-shift, so above method is not portable. If exceptions are thrown on overflows, then the values of x and y should be unsigned or cast to unsigned for the subtractions to avoid unnecessarily throwing an exception, however the right-shift needs a signed operand to produce all one bits when negative, so cast to signed there. 

Method 3 (Use absolute value) 

A generalized formula to find the max/min number with absolute value is : 

(x + y + ABS(x-y) )/2

Find the min number is: 

(x + y - ABS(x-y) )/2

So, if we can use the bitwise operation to find the absolute value, we can find the max/min number without using if conditions. The way to find the absolute way with bitwise operation can be found here:

Step1) Set the mask as right shift of integer by 31 (assuming integers are stored as two’s-complement 32-bit values and that the right-shift operator does sign extension).

mask = n>>31

Step2) XOR the mask with number

mask ^ n

Step3) Subtract mask from result of step 2 and return the result.

(mask^n) - mask 

Therefore, we can conclude the solution as follows:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

int absbit32(int x, int y)
{
    int sub = x - y;
    int mask = (sub >> 31);
    return (sub ^ mask) - mask;
 }

int max(int x, int y)
{
    int abs = absbit32(x, y);
    return (x + y + abs) / 2;
 }

int min(int x, int y)
{
    int abs = absbit32(x, y);
    return (x + y - abs) / 2;
}

// Driver Code
int main()
{
    cout << max(2, 3) << endl; //3
    cout <<  max(2, -3) << endl; //2
    cout << max(-2, -3) << endl; //-2
    cout <<  min(2, 3) << endl; //2
    cout << min(2, -3) << endl; //-3
    cout << min(-2, -3) << endl; //-3

    return 0;
}

// This code is contributed by avijitmondal1998

Java

// Java program for the above approach

import java.io.*;

class GFG {
     public static void main(String []args){
        System.out.println( max(2,3) ); //3
        System.out.println( max(2,-3) ); //2
        System.out.println( max(-2,-3) ); //-2
        System.out.println( min(2,3) ); //2
        System.out.println( min(2,-3) ); //-3
        System.out.println( min(-2,-3) ); //-3
     }

     public static int max(int x, int y){
         int abs = absbit32(x,y);
         return (x + y + abs)/2;
     }

     public static int min(int x, int y){
         int abs = absbit32(x,y);
         return (x + y - abs)/2;
     }

     public static int absbit32(int x, int y){
         int sub = x - y;
         int mask = (sub >> 31);
         return (sub ^ mask) - mask;
     }
}

Python3

# Python3 program for the above approach
def max(x, y):
  abs = absbit32(x,y)
  return (x + y + abs)//2

def min(x, y):
  abs = absbit32(x,y)
  return (x + y - abs)//2

def absbit32( x, y):
  sub = x - y
  mask = (sub >> 31)
  return (sub ^ mask) - mask

# Driver code
print( max(2,3) ) #3
print( max(2,-3) ) #2
print( max(-2,-3) ) #-2
print( min(2,3) ) #2
print( min(2,-3) ) #-3
print( min(-2,-3) ) #-3

# This code is contributed by rohitsingh07052.

C#

// C# program for the above approach
using System;

class GFG{

public static void Main(String []args)
{
    Console.WriteLine(max(2, 3)); //3
    Console.WriteLine(max(2, -3)); //2
    Console.WriteLine(max(-2, -3)); //-2
    Console.WriteLine(min(2, 3)); //2
    Console.WriteLine(min(2, -3)); //-3
    Console.WriteLine(min(-2, -3)); //-3
}

public static int max(int x, int y)
{
    int abs = absbit32(x, y);
    return (x + y + abs) / 2;
}

public static int min(int x, int y)
{
    int abs = absbit32(x, y);
    return (x + y - abs) / 2;
}

public static int absbit32(int x, int y)
{
    int sub = x - y;
    int mask = (sub >> 31);
    return (sub ^ mask) - mask;
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript program for the above approach

 function max(x , y){
     var abs = absbit32(x,y);
     return (x + y + abs)/2;
 }

 function min(x , y){
     var abs = absbit32(x,y);
     return (x + y - abs)/2;
 }

 function absbit32(x , y){
     var sub = x - y;
     var mask = (sub >> 31);
     return (sub ^ mask) - mask;
 }
 // Drive code
 document.write( max(2,3)+"<br>" ); //3
 document.write( max(2,-3)+"<br>" ); //2
 document.write( max(-2,-3)+"<br>" ); //-2
 document.write( min(2,3)+"<br>" ); //2
 document.write( min(2,-3)+"<br>" ); //-3
 document.write( min(-2,-3) ); //-3

// This code is contributed by 29AjayKumar

</script>

Time Complexity: O(1)

Auxiliary Space: O(1)
Source:
http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax