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# 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

 ``

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`

## C

 `// C program to Compute the minimum``// or maximum of two integers without``// branching``#include` `/*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

 ``

## Javascript

 ``

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 ``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 "``<

## C

 `#include``#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

 ``

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 ``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

 ``

Time Complexity: O(1)

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

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