Compute the minimum or maximum of two integers without branching

Difficulty Level : Hard
Last Updated : 12 Jun, 2022

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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>