Compute the minimum or maximum of two integers without branching (original) (raw)

Last Updated : 26 Jun, 2023

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

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

C

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

`

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 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

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

JavaScript

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(); }

PHP

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

`

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

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

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(); }

`

Output

Minimum of 15 and 6 is 6 Maximum of15 and 6 is 15

**Time Complexity: O(1)
**Auxiliary Space: O(1)