On some rare machines where branching is expensive, the below obvious approach to find minimum can be slow as it uses branching.
C++
int min(int x, int y)
{
return (x < y) ? x : y
}
|
Java
static int min(int x, int y)
{
return (x < y) ? x : y;
}
|
Python3
def min(x, y):
return x if x < y else y
|
C#
static int min(int x, int y)
{
return (x < y) ? x : y;
}
|
Javascript
<script>
function min(x, y)
{
return (x < y) ? x : y;
}
</script>
|
C
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++
#include<iostream>
using namespace std;
class gfg
{
public:
int min(int x, int y)
{
return y ^ ((x ^ y) & -(x < y));
}
int max(int x, int y)
{
return x ^ ((x ^ y) & -(x < y));
}
};
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();
}
|
Java
public class AWS {
static int min(int x, int y)
{
return y ^ ((x ^ y) & -(x << y));
}
static int max(int x, int y)
{
return x ^ ((x ^ y) & -(x << y));
}
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
def min(x, y):
return y ^ ((x ^ y) & -(x < y))
def max(x, y):
return x ^ ((x ^ y) & -(x < y))
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))
|
C#
using System;
public class AWS
{
public static int min(int x, int y)
{
return y ^ ((x ^ y) & -(x << y));
}
public static int max(int x, int y)
{
return x ^ ((x ^ y) & -(x << y));
}
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));
}
}
|
Javascript
<script>
function min(x,y)
{
return y ^ ((x ^ y) & -(x << y));
}
function max(x,y)
{
return x ^ ((x ^ y) & -(x << y));
}
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");
</script>
|
C
#include<stdio.h>
int min(int x, int y)
{
return y ^ ((x ^ y) & -(x < y));
}
int max(int x, int y)
{
return x ^ ((x ^ y) & -(x < y));
}
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
<?php
function m_in($x, $y)
{
return $y ^ (($x ^ $y) &
- ($x < $y));
}
function m_ax($x, $y)
{
return $x ^ (($x ^ $y) &
- ($x < $y));
}
$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);
?>
|
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
int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHARBIT - 1)));
}
int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHARBIT - 1)));
}
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);
}
|
Java
class GFG
{
static int CHAR_BIT = 4;
static int INT_BIT = 8;
static int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
}
static int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
}
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));
}
}
|
Python3
import sys;
CHAR_BIT = 8;
INT_BIT = sys.getsizeof(int());
def Min(x, y):
return y + ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
def Max(x, y):
return x - ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
x = 15;
y = 6;
print("Minimum of", x, "and",
y, "is", Min(x, y));
print("Maximum of", x, "and",
y, "is", Max(x, y));
|
C#
using System;
class GFG
{
static int CHAR_BIT = 8;
static int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
static int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
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));
}
}
|
Javascript
<script>
var CHAR_BIT = 4;
var INT_BIT = 8;
function min(x , y) {
return y + ((x - y) & ((x - y) >> (INT_BIT * CHAR_BIT - 1)));
}
function max(x , y) {
return x - ((x - y) & ((x - y) >> (INT_BIT * CHAR_BIT - 1)));
}
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));
</script>
|
C
#include<stdio.h>
#define CHAR_BIT 8
int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
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)
